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How to make a list of partial sums using forEach



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24















I have an array of arrays which looks like this:



changes = [ [1, 1, 1, -1], [1, -1, -1], [1, 1] ];


I want to get the next value in the array by adding the last value



values = [ [1, 2, 3, 2], [1, 0, -1], [1, 2] ];


so far I have tried to use a forEach:



changes.forEach(change => 
let i = changes.indexOf(change);
let newValue = change[i] + change[i + 1]
);


I think I am on the right lines but I cannot get this approach to work, or maybe there is a better way to do it.










share|improve this question



















  • 1





    Please elaborate this "I want to be able to increment the values based on the next value in the array to get this result:"

    – Syed Mehtab Hassan
    Mar 20 at 10:51






  • 1





    This is just a cumulative sum wrapped in .map. The answer by Thomas is better than anything you'll find there though.

    – JollyJoker
    Mar 20 at 12:21






  • 2





    @JollyJoker Thomas simulate .reduce method, why not use .reduce directly ?

    – R3tep
    Mar 20 at 12:26






  • 1





    @Thomas Look my answer, you can use an array as accumulator.

    – R3tep
    Mar 20 at 12:52






  • 1





    Never use forEach if you want to produce a result.

    – Bergi
    Mar 20 at 13:55

















24















I have an array of arrays which looks like this:



changes = [ [1, 1, 1, -1], [1, -1, -1], [1, 1] ];


I want to get the next value in the array by adding the last value



values = [ [1, 2, 3, 2], [1, 0, -1], [1, 2] ];


so far I have tried to use a forEach:



changes.forEach(change => 
let i = changes.indexOf(change);
let newValue = change[i] + change[i + 1]
);


I think I am on the right lines but I cannot get this approach to work, or maybe there is a better way to do it.










share|improve this question



















  • 1





    Please elaborate this "I want to be able to increment the values based on the next value in the array to get this result:"

    – Syed Mehtab Hassan
    Mar 20 at 10:51






  • 1





    This is just a cumulative sum wrapped in .map. The answer by Thomas is better than anything you'll find there though.

    – JollyJoker
    Mar 20 at 12:21






  • 2





    @JollyJoker Thomas simulate .reduce method, why not use .reduce directly ?

    – R3tep
    Mar 20 at 12:26






  • 1





    @Thomas Look my answer, you can use an array as accumulator.

    – R3tep
    Mar 20 at 12:52






  • 1





    Never use forEach if you want to produce a result.

    – Bergi
    Mar 20 at 13:55













24












24








24


2






I have an array of arrays which looks like this:



changes = [ [1, 1, 1, -1], [1, -1, -1], [1, 1] ];


I want to get the next value in the array by adding the last value



values = [ [1, 2, 3, 2], [1, 0, -1], [1, 2] ];


so far I have tried to use a forEach:



changes.forEach(change => 
let i = changes.indexOf(change);
let newValue = change[i] + change[i + 1]
);


I think I am on the right lines but I cannot get this approach to work, or maybe there is a better way to do it.










share|improve this question
















I have an array of arrays which looks like this:



changes = [ [1, 1, 1, -1], [1, -1, -1], [1, 1] ];


I want to get the next value in the array by adding the last value



values = [ [1, 2, 3, 2], [1, 0, -1], [1, 2] ];


so far I have tried to use a forEach:



changes.forEach(change => 
let i = changes.indexOf(change);
let newValue = change[i] + change[i + 1]
);


I think I am on the right lines but I cannot get this approach to work, or maybe there is a better way to do it.







javascript arrays ecmascript-6 foreach






share|improve this question















share|improve this question













share|improve this question




share|improve this question








edited Mar 20 at 14:51









Solomon Ucko

88121122




88121122










asked Mar 20 at 10:47









Team CafeTeam Cafe

1365




1365







  • 1





    Please elaborate this "I want to be able to increment the values based on the next value in the array to get this result:"

    – Syed Mehtab Hassan
    Mar 20 at 10:51






  • 1





    This is just a cumulative sum wrapped in .map. The answer by Thomas is better than anything you'll find there though.

    – JollyJoker
    Mar 20 at 12:21






  • 2





    @JollyJoker Thomas simulate .reduce method, why not use .reduce directly ?

    – R3tep
    Mar 20 at 12:26






  • 1





    @Thomas Look my answer, you can use an array as accumulator.

    – R3tep
    Mar 20 at 12:52






  • 1





    Never use forEach if you want to produce a result.

    – Bergi
    Mar 20 at 13:55












  • 1





    Please elaborate this "I want to be able to increment the values based on the next value in the array to get this result:"

    – Syed Mehtab Hassan
    Mar 20 at 10:51






  • 1





    This is just a cumulative sum wrapped in .map. The answer by Thomas is better than anything you'll find there though.

    – JollyJoker
    Mar 20 at 12:21






  • 2





    @JollyJoker Thomas simulate .reduce method, why not use .reduce directly ?

    – R3tep
    Mar 20 at 12:26






  • 1





    @Thomas Look my answer, you can use an array as accumulator.

    – R3tep
    Mar 20 at 12:52






  • 1





    Never use forEach if you want to produce a result.

    – Bergi
    Mar 20 at 13:55







1




1





Please elaborate this "I want to be able to increment the values based on the next value in the array to get this result:"

– Syed Mehtab Hassan
Mar 20 at 10:51





Please elaborate this "I want to be able to increment the values based on the next value in the array to get this result:"

– Syed Mehtab Hassan
Mar 20 at 10:51




1




1





This is just a cumulative sum wrapped in .map. The answer by Thomas is better than anything you'll find there though.

– JollyJoker
Mar 20 at 12:21





This is just a cumulative sum wrapped in .map. The answer by Thomas is better than anything you'll find there though.

– JollyJoker
Mar 20 at 12:21




2




2





@JollyJoker Thomas simulate .reduce method, why not use .reduce directly ?

– R3tep
Mar 20 at 12:26





@JollyJoker Thomas simulate .reduce method, why not use .reduce directly ?

– R3tep
Mar 20 at 12:26




1




1





@Thomas Look my answer, you can use an array as accumulator.

– R3tep
Mar 20 at 12:52





@Thomas Look my answer, you can use an array as accumulator.

– R3tep
Mar 20 at 12:52




1




1





Never use forEach if you want to produce a result.

– Bergi
Mar 20 at 13:55





Never use forEach if you want to produce a result.

– Bergi
Mar 20 at 13:55












8 Answers
8






active

oldest

votes


















22














You could save a sum and add the values.






var array = [[1, 1, 1, -1], [1, -1, -1], [1, 1]],
result = array.map(a => a.map((s => v => s += v)(0)));

console.log(result);





By using forEach, you need to take the object reference and the previous value or zero.






var array = [[1, 1, 1, -1], [1, -1, -1], [1, 1]];

array.forEach(a => a.forEach((v, i, a) => a[i] = (a[i - 1] || 0) + v));

console.log(array);








share|improve this answer


















  • 1





    That first approach looks nice, but is really inefficient.

    – T.J. Crowder
    Mar 20 at 10:58






  • 13





    but man does it look nice

    – Jeremy Thille
    Mar 20 at 10:58






  • 1





    @T.J.Crowder could you explain or point me in the right direction, why the first approach is inefficient?

    – Thomas
    Mar 20 at 11:01






  • 1





    @Thomas - Look at all the functions it's creating and executing. Your solution is great -- as concise (way more so than mine) and still efficient.

    – T.J. Crowder
    Mar 20 at 11:01







  • 2





    really, for three items in the outer array? if you really want to speed up the execution, you never use some array methods, instead use nested for loops and create new arrays.

    – Nina Scholz
    Mar 20 at 11:05



















13














A version with map.






const changes = [
[1, 1, 1, -1],
[1, -1, -1],
[1, 1]
];

const values = changes.map(array =>
let acc = 0;
return array.map(v => acc += v);
);

console.log(values);

.as-console-wrappertop:0;max-height:100%!important





And this doesn't change the source Array.






share|improve this answer

























  • Note that the array of arrays structure is just a needless complication. The question is actually about the cumulative sum of an array. I'd split out the outer mapping function into something called cumulativeSum

    – JollyJoker
    Mar 20 at 12:25


















4














New ESNext features of generators are nice for this.



Here I've created a simple sumpUp generator that you can re-use.






function* sumUp(a) 
let sum = 0;
for (const v of a) yield sum += v;


const changes = [ [1, 1, 1, -1], [1, -1, -1], [1, 1] ];
const values = changes.map(a => [...sumUp(a)]);

console.log(values);








share|improve this answer
































    3

















    const changes = [ [1, 1, 1, -1], [1, -1, -1], [1, 1] ]
    let values = []
    changes.forEach(arr =>
    let accu = 0
    let nestedArr = []
    arr.forEach(n =>
    accu += n
    nestedArr.push(accu)
    )
    values.push(nestedArr)
    )
    console.log(values)








    share|improve this answer






























      3














      You may use map function of Array




      const changes = [ [1, 1, 1, -1], [1, -1, -1], [1, 1] ]; 
      const result = changes.map((v) => v.slice(0).map((t, i, arr) => i === 0 ? t : (arr[i] += arr[i - 1])))
      console.log(changes);
      console.log(result);





      Update



      Use slice to clone array. This will prevent changes to the original array.






      share|improve this answer




















      • 3





        This modifies the source array, which is probably not a good idea.

        – T.J. Crowder
        Mar 20 at 10:56











      • Yes. And slice will help to prevent that. Thanks for the tip

        – Alexander
        Mar 20 at 11:02


















      2














      Here is an easier to read way that iterates over the outer list of arrays. A copy of the inner array is made to keep the initial values (like [1, 1, 1, -1]). It then iterates over each value in the copied array and adds it to each index after it in the original array.






      var changes = [[1, 1, 1, -1], [1, -1, -1], [1, 1]];
      changes.forEach(subArray =>
      var subArrayCopy = subArray.slice(); // Create a copy of the current sub array (i.e. subArrayCopy = [1, 1, 1, -1];)
      subArrayCopy.forEach((val, index) => // Iterate through each value in the copy
      for (var i = subArray.length - 1; i > index; i--) // For each element from the end to the current index
      subArray[i] += val; // Add the copy's current index value to the original array

      );
      )
      console.log(changes);








      share|improve this answer






























        2














        Another way,



        You can use .map to return your new array with the desired results. By using .reduce with an array as an accumulator, you can generate the subarray.






        var array = [[1, 1, 1, -1], [1, -1, -1], [1, 1]],
        result = array.map(a => a.reduce((ac, v, i) => , []));

        console.log(result);

        // shorter
        result = array.map(a => a.reduce((ac, v, i) => [...ac, (ac[i-1] || 0) + v], []));
        console.log(result);








        share|improve this answer
































          0














          You have an array of arrays, but each constituent array is independent from the next so let's tackle them individually. Let's talk about [1, 1, 1, -1].



          Your use of the phrase "partial sum" is very informative. The full sum could be implemented using reduce:



          [1, 1, 1, -1].reduce((x, y) => x + y);
          // 2


          But you want an array of partial sums. That's very similar to this usage of reduce, but instead of retaining only the last computed result, we retain every intermediate value too. In other languages, this is called scan (cf. F#, Haskell).



          A javascript implementation of a generic scan would probably look a lot like reduce. In fact, you can implement it with reduce, with just a little extra work:



          function scan(array, callback) 
          const results = [array[0]];

          // reduce does all the heavy lifting for us, but we define a wrapper to pass to it.
          array.reduce((...args) =>
          // The wrapper forwards all those arguments to the callback, but captures the result...
          const result = callback(...args);
          // ...storing that intermediate result in our results array...
          results.push(result);
          // ...then passes it back to reduce to continue what it was doing.
          return result;
          );

          return results;


          // scan([1, 1, 1, -1], (x, y) => x + y) -> [1, 2, 3, 2]


          A more robust implementation would hew closer to the standard library's reduce, especially around initial values:



          function scan(array, callback, initialValue) 
          const results = [];

          const reducer = (...args) =>
          const result = callback(...args);
          results.push(result);
          return result;
          ;

          if (arguments.length === 2)
          results.push(array[0]);
          array.reduce(reducer);
          else
          results.push(initialValue);
          array.reduce(reducer, initialValue);


          return results;



          Bringing it back together, if you'd like to do this for your arrays of arrays, it would just be a map over scan:



          [[1, 1, 1, -1], [1, -1, -1], [1, 1]].map(a => scan(a, (x, y) => x + y));
          // [[1, 2, 3, 2], [1, 0, -1], [1, 2]]


          No copying necessary or side effects to watch out for, and you get a handy higher-order function out of it to boot!






          share|improve this answer























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            8 Answers
            8






            active

            oldest

            votes








            8 Answers
            8






            active

            oldest

            votes









            active

            oldest

            votes






            active

            oldest

            votes









            22














            You could save a sum and add the values.






            var array = [[1, 1, 1, -1], [1, -1, -1], [1, 1]],
            result = array.map(a => a.map((s => v => s += v)(0)));

            console.log(result);





            By using forEach, you need to take the object reference and the previous value or zero.






            var array = [[1, 1, 1, -1], [1, -1, -1], [1, 1]];

            array.forEach(a => a.forEach((v, i, a) => a[i] = (a[i - 1] || 0) + v));

            console.log(array);








            share|improve this answer


















            • 1





              That first approach looks nice, but is really inefficient.

              – T.J. Crowder
              Mar 20 at 10:58






            • 13





              but man does it look nice

              – Jeremy Thille
              Mar 20 at 10:58






            • 1





              @T.J.Crowder could you explain or point me in the right direction, why the first approach is inefficient?

              – Thomas
              Mar 20 at 11:01






            • 1





              @Thomas - Look at all the functions it's creating and executing. Your solution is great -- as concise (way more so than mine) and still efficient.

              – T.J. Crowder
              Mar 20 at 11:01







            • 2





              really, for three items in the outer array? if you really want to speed up the execution, you never use some array methods, instead use nested for loops and create new arrays.

              – Nina Scholz
              Mar 20 at 11:05
















            22














            You could save a sum and add the values.






            var array = [[1, 1, 1, -1], [1, -1, -1], [1, 1]],
            result = array.map(a => a.map((s => v => s += v)(0)));

            console.log(result);





            By using forEach, you need to take the object reference and the previous value or zero.






            var array = [[1, 1, 1, -1], [1, -1, -1], [1, 1]];

            array.forEach(a => a.forEach((v, i, a) => a[i] = (a[i - 1] || 0) + v));

            console.log(array);








            share|improve this answer


















            • 1





              That first approach looks nice, but is really inefficient.

              – T.J. Crowder
              Mar 20 at 10:58






            • 13





              but man does it look nice

              – Jeremy Thille
              Mar 20 at 10:58






            • 1





              @T.J.Crowder could you explain or point me in the right direction, why the first approach is inefficient?

              – Thomas
              Mar 20 at 11:01






            • 1





              @Thomas - Look at all the functions it's creating and executing. Your solution is great -- as concise (way more so than mine) and still efficient.

              – T.J. Crowder
              Mar 20 at 11:01







            • 2





              really, for three items in the outer array? if you really want to speed up the execution, you never use some array methods, instead use nested for loops and create new arrays.

              – Nina Scholz
              Mar 20 at 11:05














            22












            22








            22







            You could save a sum and add the values.






            var array = [[1, 1, 1, -1], [1, -1, -1], [1, 1]],
            result = array.map(a => a.map((s => v => s += v)(0)));

            console.log(result);





            By using forEach, you need to take the object reference and the previous value or zero.






            var array = [[1, 1, 1, -1], [1, -1, -1], [1, 1]];

            array.forEach(a => a.forEach((v, i, a) => a[i] = (a[i - 1] || 0) + v));

            console.log(array);








            share|improve this answer













            You could save a sum and add the values.






            var array = [[1, 1, 1, -1], [1, -1, -1], [1, 1]],
            result = array.map(a => a.map((s => v => s += v)(0)));

            console.log(result);





            By using forEach, you need to take the object reference and the previous value or zero.






            var array = [[1, 1, 1, -1], [1, -1, -1], [1, 1]];

            array.forEach(a => a.forEach((v, i, a) => a[i] = (a[i - 1] || 0) + v));

            console.log(array);








            var array = [[1, 1, 1, -1], [1, -1, -1], [1, 1]],
            result = array.map(a => a.map((s => v => s += v)(0)));

            console.log(result);





            var array = [[1, 1, 1, -1], [1, -1, -1], [1, 1]],
            result = array.map(a => a.map((s => v => s += v)(0)));

            console.log(result);





            var array = [[1, 1, 1, -1], [1, -1, -1], [1, 1]];

            array.forEach(a => a.forEach((v, i, a) => a[i] = (a[i - 1] || 0) + v));

            console.log(array);





            var array = [[1, 1, 1, -1], [1, -1, -1], [1, 1]];

            array.forEach(a => a.forEach((v, i, a) => a[i] = (a[i - 1] || 0) + v));

            console.log(array);






            share|improve this answer












            share|improve this answer



            share|improve this answer










            answered Mar 20 at 10:54









            Nina ScholzNina Scholz

            200k15112182




            200k15112182







            • 1





              That first approach looks nice, but is really inefficient.

              – T.J. Crowder
              Mar 20 at 10:58






            • 13





              but man does it look nice

              – Jeremy Thille
              Mar 20 at 10:58






            • 1





              @T.J.Crowder could you explain or point me in the right direction, why the first approach is inefficient?

              – Thomas
              Mar 20 at 11:01






            • 1





              @Thomas - Look at all the functions it's creating and executing. Your solution is great -- as concise (way more so than mine) and still efficient.

              – T.J. Crowder
              Mar 20 at 11:01







            • 2





              really, for three items in the outer array? if you really want to speed up the execution, you never use some array methods, instead use nested for loops and create new arrays.

              – Nina Scholz
              Mar 20 at 11:05













            • 1





              That first approach looks nice, but is really inefficient.

              – T.J. Crowder
              Mar 20 at 10:58






            • 13





              but man does it look nice

              – Jeremy Thille
              Mar 20 at 10:58






            • 1





              @T.J.Crowder could you explain or point me in the right direction, why the first approach is inefficient?

              – Thomas
              Mar 20 at 11:01






            • 1





              @Thomas - Look at all the functions it's creating and executing. Your solution is great -- as concise (way more so than mine) and still efficient.

              – T.J. Crowder
              Mar 20 at 11:01







            • 2





              really, for three items in the outer array? if you really want to speed up the execution, you never use some array methods, instead use nested for loops and create new arrays.

              – Nina Scholz
              Mar 20 at 11:05








            1




            1





            That first approach looks nice, but is really inefficient.

            – T.J. Crowder
            Mar 20 at 10:58





            That first approach looks nice, but is really inefficient.

            – T.J. Crowder
            Mar 20 at 10:58




            13




            13





            but man does it look nice

            – Jeremy Thille
            Mar 20 at 10:58





            but man does it look nice

            – Jeremy Thille
            Mar 20 at 10:58




            1




            1





            @T.J.Crowder could you explain or point me in the right direction, why the first approach is inefficient?

            – Thomas
            Mar 20 at 11:01





            @T.J.Crowder could you explain or point me in the right direction, why the first approach is inefficient?

            – Thomas
            Mar 20 at 11:01




            1




            1





            @Thomas - Look at all the functions it's creating and executing. Your solution is great -- as concise (way more so than mine) and still efficient.

            – T.J. Crowder
            Mar 20 at 11:01






            @Thomas - Look at all the functions it's creating and executing. Your solution is great -- as concise (way more so than mine) and still efficient.

            – T.J. Crowder
            Mar 20 at 11:01





            2




            2





            really, for three items in the outer array? if you really want to speed up the execution, you never use some array methods, instead use nested for loops and create new arrays.

            – Nina Scholz
            Mar 20 at 11:05






            really, for three items in the outer array? if you really want to speed up the execution, you never use some array methods, instead use nested for loops and create new arrays.

            – Nina Scholz
            Mar 20 at 11:05














            13














            A version with map.






            const changes = [
            [1, 1, 1, -1],
            [1, -1, -1],
            [1, 1]
            ];

            const values = changes.map(array =>
            let acc = 0;
            return array.map(v => acc += v);
            );

            console.log(values);

            .as-console-wrappertop:0;max-height:100%!important





            And this doesn't change the source Array.






            share|improve this answer

























            • Note that the array of arrays structure is just a needless complication. The question is actually about the cumulative sum of an array. I'd split out the outer mapping function into something called cumulativeSum

              – JollyJoker
              Mar 20 at 12:25















            13














            A version with map.






            const changes = [
            [1, 1, 1, -1],
            [1, -1, -1],
            [1, 1]
            ];

            const values = changes.map(array =>
            let acc = 0;
            return array.map(v => acc += v);
            );

            console.log(values);

            .as-console-wrappertop:0;max-height:100%!important





            And this doesn't change the source Array.






            share|improve this answer

























            • Note that the array of arrays structure is just a needless complication. The question is actually about the cumulative sum of an array. I'd split out the outer mapping function into something called cumulativeSum

              – JollyJoker
              Mar 20 at 12:25













            13












            13








            13







            A version with map.






            const changes = [
            [1, 1, 1, -1],
            [1, -1, -1],
            [1, 1]
            ];

            const values = changes.map(array =>
            let acc = 0;
            return array.map(v => acc += v);
            );

            console.log(values);

            .as-console-wrappertop:0;max-height:100%!important





            And this doesn't change the source Array.






            share|improve this answer















            A version with map.






            const changes = [
            [1, 1, 1, -1],
            [1, -1, -1],
            [1, 1]
            ];

            const values = changes.map(array =>
            let acc = 0;
            return array.map(v => acc += v);
            );

            console.log(values);

            .as-console-wrappertop:0;max-height:100%!important





            And this doesn't change the source Array.






            const changes = [
            [1, 1, 1, -1],
            [1, -1, -1],
            [1, 1]
            ];

            const values = changes.map(array =>
            let acc = 0;
            return array.map(v => acc += v);
            );

            console.log(values);

            .as-console-wrappertop:0;max-height:100%!important





            const changes = [
            [1, 1, 1, -1],
            [1, -1, -1],
            [1, 1]
            ];

            const values = changes.map(array =>
            let acc = 0;
            return array.map(v => acc += v);
            );

            console.log(values);

            .as-console-wrappertop:0;max-height:100%!important






            share|improve this answer














            share|improve this answer



            share|improve this answer








            edited Mar 20 at 11:01









            T.J. Crowder

            703k12412501343




            703k12412501343










            answered Mar 20 at 10:59









            ThomasThomas

            5,2261510




            5,2261510












            • Note that the array of arrays structure is just a needless complication. The question is actually about the cumulative sum of an array. I'd split out the outer mapping function into something called cumulativeSum

              – JollyJoker
              Mar 20 at 12:25

















            • Note that the array of arrays structure is just a needless complication. The question is actually about the cumulative sum of an array. I'd split out the outer mapping function into something called cumulativeSum

              – JollyJoker
              Mar 20 at 12:25
















            Note that the array of arrays structure is just a needless complication. The question is actually about the cumulative sum of an array. I'd split out the outer mapping function into something called cumulativeSum

            – JollyJoker
            Mar 20 at 12:25





            Note that the array of arrays structure is just a needless complication. The question is actually about the cumulative sum of an array. I'd split out the outer mapping function into something called cumulativeSum

            – JollyJoker
            Mar 20 at 12:25











            4














            New ESNext features of generators are nice for this.



            Here I've created a simple sumpUp generator that you can re-use.






            function* sumUp(a) 
            let sum = 0;
            for (const v of a) yield sum += v;


            const changes = [ [1, 1, 1, -1], [1, -1, -1], [1, 1] ];
            const values = changes.map(a => [...sumUp(a)]);

            console.log(values);








            share|improve this answer





























              4














              New ESNext features of generators are nice for this.



              Here I've created a simple sumpUp generator that you can re-use.






              function* sumUp(a) 
              let sum = 0;
              for (const v of a) yield sum += v;


              const changes = [ [1, 1, 1, -1], [1, -1, -1], [1, 1] ];
              const values = changes.map(a => [...sumUp(a)]);

              console.log(values);








              share|improve this answer



























                4












                4








                4







                New ESNext features of generators are nice for this.



                Here I've created a simple sumpUp generator that you can re-use.






                function* sumUp(a) 
                let sum = 0;
                for (const v of a) yield sum += v;


                const changes = [ [1, 1, 1, -1], [1, -1, -1], [1, 1] ];
                const values = changes.map(a => [...sumUp(a)]);

                console.log(values);








                share|improve this answer















                New ESNext features of generators are nice for this.



                Here I've created a simple sumpUp generator that you can re-use.






                function* sumUp(a) 
                let sum = 0;
                for (const v of a) yield sum += v;


                const changes = [ [1, 1, 1, -1], [1, -1, -1], [1, 1] ];
                const values = changes.map(a => [...sumUp(a)]);

                console.log(values);








                function* sumUp(a) 
                let sum = 0;
                for (const v of a) yield sum += v;


                const changes = [ [1, 1, 1, -1], [1, -1, -1], [1, 1] ];
                const values = changes.map(a => [...sumUp(a)]);

                console.log(values);





                function* sumUp(a) 
                let sum = 0;
                for (const v of a) yield sum += v;


                const changes = [ [1, 1, 1, -1], [1, -1, -1], [1, 1] ];
                const values = changes.map(a => [...sumUp(a)]);

                console.log(values);






                share|improve this answer














                share|improve this answer



                share|improve this answer








                edited Mar 20 at 11:18

























                answered Mar 20 at 11:13









                KeithKeith

                9,3631821




                9,3631821





















                    3

















                    const changes = [ [1, 1, 1, -1], [1, -1, -1], [1, 1] ]
                    let values = []
                    changes.forEach(arr =>
                    let accu = 0
                    let nestedArr = []
                    arr.forEach(n =>
                    accu += n
                    nestedArr.push(accu)
                    )
                    values.push(nestedArr)
                    )
                    console.log(values)








                    share|improve this answer



























                      3

















                      const changes = [ [1, 1, 1, -1], [1, -1, -1], [1, 1] ]
                      let values = []
                      changes.forEach(arr =>
                      let accu = 0
                      let nestedArr = []
                      arr.forEach(n =>
                      accu += n
                      nestedArr.push(accu)
                      )
                      values.push(nestedArr)
                      )
                      console.log(values)








                      share|improve this answer

























                        3












                        3








                        3










                        const changes = [ [1, 1, 1, -1], [1, -1, -1], [1, 1] ]
                        let values = []
                        changes.forEach(arr =>
                        let accu = 0
                        let nestedArr = []
                        arr.forEach(n =>
                        accu += n
                        nestedArr.push(accu)
                        )
                        values.push(nestedArr)
                        )
                        console.log(values)








                        share|improve this answer
















                        const changes = [ [1, 1, 1, -1], [1, -1, -1], [1, 1] ]
                        let values = []
                        changes.forEach(arr =>
                        let accu = 0
                        let nestedArr = []
                        arr.forEach(n =>
                        accu += n
                        nestedArr.push(accu)
                        )
                        values.push(nestedArr)
                        )
                        console.log(values)








                        const changes = [ [1, 1, 1, -1], [1, -1, -1], [1, 1] ]
                        let values = []
                        changes.forEach(arr =>
                        let accu = 0
                        let nestedArr = []
                        arr.forEach(n =>
                        accu += n
                        nestedArr.push(accu)
                        )
                        values.push(nestedArr)
                        )
                        console.log(values)





                        const changes = [ [1, 1, 1, -1], [1, -1, -1], [1, 1] ]
                        let values = []
                        changes.forEach(arr =>
                        let accu = 0
                        let nestedArr = []
                        arr.forEach(n =>
                        accu += n
                        nestedArr.push(accu)
                        )
                        values.push(nestedArr)
                        )
                        console.log(values)






                        share|improve this answer












                        share|improve this answer



                        share|improve this answer










                        answered Mar 20 at 10:52









                        holydragonholydragon

                        2,80321330




                        2,80321330





















                            3














                            You may use map function of Array




                            const changes = [ [1, 1, 1, -1], [1, -1, -1], [1, 1] ]; 
                            const result = changes.map((v) => v.slice(0).map((t, i, arr) => i === 0 ? t : (arr[i] += arr[i - 1])))
                            console.log(changes);
                            console.log(result);





                            Update



                            Use slice to clone array. This will prevent changes to the original array.






                            share|improve this answer




















                            • 3





                              This modifies the source array, which is probably not a good idea.

                              – T.J. Crowder
                              Mar 20 at 10:56











                            • Yes. And slice will help to prevent that. Thanks for the tip

                              – Alexander
                              Mar 20 at 11:02















                            3














                            You may use map function of Array




                            const changes = [ [1, 1, 1, -1], [1, -1, -1], [1, 1] ]; 
                            const result = changes.map((v) => v.slice(0).map((t, i, arr) => i === 0 ? t : (arr[i] += arr[i - 1])))
                            console.log(changes);
                            console.log(result);





                            Update



                            Use slice to clone array. This will prevent changes to the original array.






                            share|improve this answer




















                            • 3





                              This modifies the source array, which is probably not a good idea.

                              – T.J. Crowder
                              Mar 20 at 10:56











                            • Yes. And slice will help to prevent that. Thanks for the tip

                              – Alexander
                              Mar 20 at 11:02













                            3












                            3








                            3







                            You may use map function of Array




                            const changes = [ [1, 1, 1, -1], [1, -1, -1], [1, 1] ]; 
                            const result = changes.map((v) => v.slice(0).map((t, i, arr) => i === 0 ? t : (arr[i] += arr[i - 1])))
                            console.log(changes);
                            console.log(result);





                            Update



                            Use slice to clone array. This will prevent changes to the original array.






                            share|improve this answer















                            You may use map function of Array




                            const changes = [ [1, 1, 1, -1], [1, -1, -1], [1, 1] ]; 
                            const result = changes.map((v) => v.slice(0).map((t, i, arr) => i === 0 ? t : (arr[i] += arr[i - 1])))
                            console.log(changes);
                            console.log(result);





                            Update



                            Use slice to clone array. This will prevent changes to the original array.






                            const changes = [ [1, 1, 1, -1], [1, -1, -1], [1, 1] ]; 
                            const result = changes.map((v) => v.slice(0).map((t, i, arr) => i === 0 ? t : (arr[i] += arr[i - 1])))
                            console.log(changes);
                            console.log(result);





                            const changes = [ [1, 1, 1, -1], [1, -1, -1], [1, 1] ]; 
                            const result = changes.map((v) => v.slice(0).map((t, i, arr) => i === 0 ? t : (arr[i] += arr[i - 1])))
                            console.log(changes);
                            console.log(result);






                            share|improve this answer














                            share|improve this answer



                            share|improve this answer








                            edited Mar 20 at 11:01

























                            answered Mar 20 at 10:51









                            AlexanderAlexander

                            1,155314




                            1,155314







                            • 3





                              This modifies the source array, which is probably not a good idea.

                              – T.J. Crowder
                              Mar 20 at 10:56











                            • Yes. And slice will help to prevent that. Thanks for the tip

                              – Alexander
                              Mar 20 at 11:02












                            • 3





                              This modifies the source array, which is probably not a good idea.

                              – T.J. Crowder
                              Mar 20 at 10:56











                            • Yes. And slice will help to prevent that. Thanks for the tip

                              – Alexander
                              Mar 20 at 11:02







                            3




                            3





                            This modifies the source array, which is probably not a good idea.

                            – T.J. Crowder
                            Mar 20 at 10:56





                            This modifies the source array, which is probably not a good idea.

                            – T.J. Crowder
                            Mar 20 at 10:56













                            Yes. And slice will help to prevent that. Thanks for the tip

                            – Alexander
                            Mar 20 at 11:02





                            Yes. And slice will help to prevent that. Thanks for the tip

                            – Alexander
                            Mar 20 at 11:02











                            2














                            Here is an easier to read way that iterates over the outer list of arrays. A copy of the inner array is made to keep the initial values (like [1, 1, 1, -1]). It then iterates over each value in the copied array and adds it to each index after it in the original array.






                            var changes = [[1, 1, 1, -1], [1, -1, -1], [1, 1]];
                            changes.forEach(subArray =>
                            var subArrayCopy = subArray.slice(); // Create a copy of the current sub array (i.e. subArrayCopy = [1, 1, 1, -1];)
                            subArrayCopy.forEach((val, index) => // Iterate through each value in the copy
                            for (var i = subArray.length - 1; i > index; i--) // For each element from the end to the current index
                            subArray[i] += val; // Add the copy's current index value to the original array

                            );
                            )
                            console.log(changes);








                            share|improve this answer



























                              2














                              Here is an easier to read way that iterates over the outer list of arrays. A copy of the inner array is made to keep the initial values (like [1, 1, 1, -1]). It then iterates over each value in the copied array and adds it to each index after it in the original array.






                              var changes = [[1, 1, 1, -1], [1, -1, -1], [1, 1]];
                              changes.forEach(subArray =>
                              var subArrayCopy = subArray.slice(); // Create a copy of the current sub array (i.e. subArrayCopy = [1, 1, 1, -1];)
                              subArrayCopy.forEach((val, index) => // Iterate through each value in the copy
                              for (var i = subArray.length - 1; i > index; i--) // For each element from the end to the current index
                              subArray[i] += val; // Add the copy's current index value to the original array

                              );
                              )
                              console.log(changes);








                              share|improve this answer

























                                2












                                2








                                2







                                Here is an easier to read way that iterates over the outer list of arrays. A copy of the inner array is made to keep the initial values (like [1, 1, 1, -1]). It then iterates over each value in the copied array and adds it to each index after it in the original array.






                                var changes = [[1, 1, 1, -1], [1, -1, -1], [1, 1]];
                                changes.forEach(subArray =>
                                var subArrayCopy = subArray.slice(); // Create a copy of the current sub array (i.e. subArrayCopy = [1, 1, 1, -1];)
                                subArrayCopy.forEach((val, index) => // Iterate through each value in the copy
                                for (var i = subArray.length - 1; i > index; i--) // For each element from the end to the current index
                                subArray[i] += val; // Add the copy's current index value to the original array

                                );
                                )
                                console.log(changes);








                                share|improve this answer













                                Here is an easier to read way that iterates over the outer list of arrays. A copy of the inner array is made to keep the initial values (like [1, 1, 1, -1]). It then iterates over each value in the copied array and adds it to each index after it in the original array.






                                var changes = [[1, 1, 1, -1], [1, -1, -1], [1, 1]];
                                changes.forEach(subArray =>
                                var subArrayCopy = subArray.slice(); // Create a copy of the current sub array (i.e. subArrayCopy = [1, 1, 1, -1];)
                                subArrayCopy.forEach((val, index) => // Iterate through each value in the copy
                                for (var i = subArray.length - 1; i > index; i--) // For each element from the end to the current index
                                subArray[i] += val; // Add the copy's current index value to the original array

                                );
                                )
                                console.log(changes);








                                var changes = [[1, 1, 1, -1], [1, -1, -1], [1, 1]];
                                changes.forEach(subArray =>
                                var subArrayCopy = subArray.slice(); // Create a copy of the current sub array (i.e. subArrayCopy = [1, 1, 1, -1];)
                                subArrayCopy.forEach((val, index) => // Iterate through each value in the copy
                                for (var i = subArray.length - 1; i > index; i--) // For each element from the end to the current index
                                subArray[i] += val; // Add the copy's current index value to the original array

                                );
                                )
                                console.log(changes);





                                var changes = [[1, 1, 1, -1], [1, -1, -1], [1, 1]];
                                changes.forEach(subArray =>
                                var subArrayCopy = subArray.slice(); // Create a copy of the current sub array (i.e. subArrayCopy = [1, 1, 1, -1];)
                                subArrayCopy.forEach((val, index) => // Iterate through each value in the copy
                                for (var i = subArray.length - 1; i > index; i--) // For each element from the end to the current index
                                subArray[i] += val; // Add the copy's current index value to the original array

                                );
                                )
                                console.log(changes);






                                share|improve this answer












                                share|improve this answer



                                share|improve this answer










                                answered Mar 20 at 13:06









                                Nick GNick G

                                1,029614




                                1,029614





















                                    2














                                    Another way,



                                    You can use .map to return your new array with the desired results. By using .reduce with an array as an accumulator, you can generate the subarray.






                                    var array = [[1, 1, 1, -1], [1, -1, -1], [1, 1]],
                                    result = array.map(a => a.reduce((ac, v, i) => , []));

                                    console.log(result);

                                    // shorter
                                    result = array.map(a => a.reduce((ac, v, i) => [...ac, (ac[i-1] || 0) + v], []));
                                    console.log(result);








                                    share|improve this answer





























                                      2














                                      Another way,



                                      You can use .map to return your new array with the desired results. By using .reduce with an array as an accumulator, you can generate the subarray.






                                      var array = [[1, 1, 1, -1], [1, -1, -1], [1, 1]],
                                      result = array.map(a => a.reduce((ac, v, i) => , []));

                                      console.log(result);

                                      // shorter
                                      result = array.map(a => a.reduce((ac, v, i) => [...ac, (ac[i-1] || 0) + v], []));
                                      console.log(result);








                                      share|improve this answer



























                                        2












                                        2








                                        2







                                        Another way,



                                        You can use .map to return your new array with the desired results. By using .reduce with an array as an accumulator, you can generate the subarray.






                                        var array = [[1, 1, 1, -1], [1, -1, -1], [1, 1]],
                                        result = array.map(a => a.reduce((ac, v, i) => , []));

                                        console.log(result);

                                        // shorter
                                        result = array.map(a => a.reduce((ac, v, i) => [...ac, (ac[i-1] || 0) + v], []));
                                        console.log(result);








                                        share|improve this answer















                                        Another way,



                                        You can use .map to return your new array with the desired results. By using .reduce with an array as an accumulator, you can generate the subarray.






                                        var array = [[1, 1, 1, -1], [1, -1, -1], [1, 1]],
                                        result = array.map(a => a.reduce((ac, v, i) => , []));

                                        console.log(result);

                                        // shorter
                                        result = array.map(a => a.reduce((ac, v, i) => [...ac, (ac[i-1] || 0) + v], []));
                                        console.log(result);








                                        var array = [[1, 1, 1, -1], [1, -1, -1], [1, 1]],
                                        result = array.map(a => a.reduce((ac, v, i) => , []));

                                        console.log(result);

                                        // shorter
                                        result = array.map(a => a.reduce((ac, v, i) => [...ac, (ac[i-1] || 0) + v], []));
                                        console.log(result);





                                        var array = [[1, 1, 1, -1], [1, -1, -1], [1, 1]],
                                        result = array.map(a => a.reduce((ac, v, i) => , []));

                                        console.log(result);

                                        // shorter
                                        result = array.map(a => a.reduce((ac, v, i) => [...ac, (ac[i-1] || 0) + v], []));
                                        console.log(result);






                                        share|improve this answer














                                        share|improve this answer



                                        share|improve this answer








                                        edited Mar 20 at 14:37

























                                        answered Mar 20 at 12:12









                                        R3tepR3tep

                                        8,33882962




                                        8,33882962





















                                            0














                                            You have an array of arrays, but each constituent array is independent from the next so let's tackle them individually. Let's talk about [1, 1, 1, -1].



                                            Your use of the phrase "partial sum" is very informative. The full sum could be implemented using reduce:



                                            [1, 1, 1, -1].reduce((x, y) => x + y);
                                            // 2


                                            But you want an array of partial sums. That's very similar to this usage of reduce, but instead of retaining only the last computed result, we retain every intermediate value too. In other languages, this is called scan (cf. F#, Haskell).



                                            A javascript implementation of a generic scan would probably look a lot like reduce. In fact, you can implement it with reduce, with just a little extra work:



                                            function scan(array, callback) 
                                            const results = [array[0]];

                                            // reduce does all the heavy lifting for us, but we define a wrapper to pass to it.
                                            array.reduce((...args) =>
                                            // The wrapper forwards all those arguments to the callback, but captures the result...
                                            const result = callback(...args);
                                            // ...storing that intermediate result in our results array...
                                            results.push(result);
                                            // ...then passes it back to reduce to continue what it was doing.
                                            return result;
                                            );

                                            return results;


                                            // scan([1, 1, 1, -1], (x, y) => x + y) -> [1, 2, 3, 2]


                                            A more robust implementation would hew closer to the standard library's reduce, especially around initial values:



                                            function scan(array, callback, initialValue) 
                                            const results = [];

                                            const reducer = (...args) =>
                                            const result = callback(...args);
                                            results.push(result);
                                            return result;
                                            ;

                                            if (arguments.length === 2)
                                            results.push(array[0]);
                                            array.reduce(reducer);
                                            else
                                            results.push(initialValue);
                                            array.reduce(reducer, initialValue);


                                            return results;



                                            Bringing it back together, if you'd like to do this for your arrays of arrays, it would just be a map over scan:



                                            [[1, 1, 1, -1], [1, -1, -1], [1, 1]].map(a => scan(a, (x, y) => x + y));
                                            // [[1, 2, 3, 2], [1, 0, -1], [1, 2]]


                                            No copying necessary or side effects to watch out for, and you get a handy higher-order function out of it to boot!






                                            share|improve this answer



























                                              0














                                              You have an array of arrays, but each constituent array is independent from the next so let's tackle them individually. Let's talk about [1, 1, 1, -1].



                                              Your use of the phrase "partial sum" is very informative. The full sum could be implemented using reduce:



                                              [1, 1, 1, -1].reduce((x, y) => x + y);
                                              // 2


                                              But you want an array of partial sums. That's very similar to this usage of reduce, but instead of retaining only the last computed result, we retain every intermediate value too. In other languages, this is called scan (cf. F#, Haskell).



                                              A javascript implementation of a generic scan would probably look a lot like reduce. In fact, you can implement it with reduce, with just a little extra work:



                                              function scan(array, callback) 
                                              const results = [array[0]];

                                              // reduce does all the heavy lifting for us, but we define a wrapper to pass to it.
                                              array.reduce((...args) =>
                                              // The wrapper forwards all those arguments to the callback, but captures the result...
                                              const result = callback(...args);
                                              // ...storing that intermediate result in our results array...
                                              results.push(result);
                                              // ...then passes it back to reduce to continue what it was doing.
                                              return result;
                                              );

                                              return results;


                                              // scan([1, 1, 1, -1], (x, y) => x + y) -> [1, 2, 3, 2]


                                              A more robust implementation would hew closer to the standard library's reduce, especially around initial values:



                                              function scan(array, callback, initialValue) 
                                              const results = [];

                                              const reducer = (...args) =>
                                              const result = callback(...args);
                                              results.push(result);
                                              return result;
                                              ;

                                              if (arguments.length === 2)
                                              results.push(array[0]);
                                              array.reduce(reducer);
                                              else
                                              results.push(initialValue);
                                              array.reduce(reducer, initialValue);


                                              return results;



                                              Bringing it back together, if you'd like to do this for your arrays of arrays, it would just be a map over scan:



                                              [[1, 1, 1, -1], [1, -1, -1], [1, 1]].map(a => scan(a, (x, y) => x + y));
                                              // [[1, 2, 3, 2], [1, 0, -1], [1, 2]]


                                              No copying necessary or side effects to watch out for, and you get a handy higher-order function out of it to boot!






                                              share|improve this answer

























                                                0












                                                0








                                                0







                                                You have an array of arrays, but each constituent array is independent from the next so let's tackle them individually. Let's talk about [1, 1, 1, -1].



                                                Your use of the phrase "partial sum" is very informative. The full sum could be implemented using reduce:



                                                [1, 1, 1, -1].reduce((x, y) => x + y);
                                                // 2


                                                But you want an array of partial sums. That's very similar to this usage of reduce, but instead of retaining only the last computed result, we retain every intermediate value too. In other languages, this is called scan (cf. F#, Haskell).



                                                A javascript implementation of a generic scan would probably look a lot like reduce. In fact, you can implement it with reduce, with just a little extra work:



                                                function scan(array, callback) 
                                                const results = [array[0]];

                                                // reduce does all the heavy lifting for us, but we define a wrapper to pass to it.
                                                array.reduce((...args) =>
                                                // The wrapper forwards all those arguments to the callback, but captures the result...
                                                const result = callback(...args);
                                                // ...storing that intermediate result in our results array...
                                                results.push(result);
                                                // ...then passes it back to reduce to continue what it was doing.
                                                return result;
                                                );

                                                return results;


                                                // scan([1, 1, 1, -1], (x, y) => x + y) -> [1, 2, 3, 2]


                                                A more robust implementation would hew closer to the standard library's reduce, especially around initial values:



                                                function scan(array, callback, initialValue) 
                                                const results = [];

                                                const reducer = (...args) =>
                                                const result = callback(...args);
                                                results.push(result);
                                                return result;
                                                ;

                                                if (arguments.length === 2)
                                                results.push(array[0]);
                                                array.reduce(reducer);
                                                else
                                                results.push(initialValue);
                                                array.reduce(reducer, initialValue);


                                                return results;



                                                Bringing it back together, if you'd like to do this for your arrays of arrays, it would just be a map over scan:



                                                [[1, 1, 1, -1], [1, -1, -1], [1, 1]].map(a => scan(a, (x, y) => x + y));
                                                // [[1, 2, 3, 2], [1, 0, -1], [1, 2]]


                                                No copying necessary or side effects to watch out for, and you get a handy higher-order function out of it to boot!






                                                share|improve this answer













                                                You have an array of arrays, but each constituent array is independent from the next so let's tackle them individually. Let's talk about [1, 1, 1, -1].



                                                Your use of the phrase "partial sum" is very informative. The full sum could be implemented using reduce:



                                                [1, 1, 1, -1].reduce((x, y) => x + y);
                                                // 2


                                                But you want an array of partial sums. That's very similar to this usage of reduce, but instead of retaining only the last computed result, we retain every intermediate value too. In other languages, this is called scan (cf. F#, Haskell).



                                                A javascript implementation of a generic scan would probably look a lot like reduce. In fact, you can implement it with reduce, with just a little extra work:



                                                function scan(array, callback) 
                                                const results = [array[0]];

                                                // reduce does all the heavy lifting for us, but we define a wrapper to pass to it.
                                                array.reduce((...args) =>
                                                // The wrapper forwards all those arguments to the callback, but captures the result...
                                                const result = callback(...args);
                                                // ...storing that intermediate result in our results array...
                                                results.push(result);
                                                // ...then passes it back to reduce to continue what it was doing.
                                                return result;
                                                );

                                                return results;


                                                // scan([1, 1, 1, -1], (x, y) => x + y) -> [1, 2, 3, 2]


                                                A more robust implementation would hew closer to the standard library's reduce, especially around initial values:



                                                function scan(array, callback, initialValue) 
                                                const results = [];

                                                const reducer = (...args) =>
                                                const result = callback(...args);
                                                results.push(result);
                                                return result;
                                                ;

                                                if (arguments.length === 2)
                                                results.push(array[0]);
                                                array.reduce(reducer);
                                                else
                                                results.push(initialValue);
                                                array.reduce(reducer, initialValue);


                                                return results;



                                                Bringing it back together, if you'd like to do this for your arrays of arrays, it would just be a map over scan:



                                                [[1, 1, 1, -1], [1, -1, -1], [1, 1]].map(a => scan(a, (x, y) => x + y));
                                                // [[1, 2, 3, 2], [1, 0, -1], [1, 2]]


                                                No copying necessary or side effects to watch out for, and you get a handy higher-order function out of it to boot!







                                                share|improve this answer












                                                share|improve this answer



                                                share|improve this answer










                                                answered Apr 11 at 7:00









                                                stuffystuffy

                                                10918




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