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The multiplication of list of matrices



The Next CEO of Stack Overflowexporting list of matrices in mathematicaMatrix multiplicationEasy way to perform multiplication of two 2x2 matrices, that contain list elements?Multiplying block matricesMatrix multiplication for higher dimensional matricesMultiply a matrix by a vectorouter product of matricesAdding two matrices together results in a weird resultMultiply two listsUsing ReplaceAll on matrix to produce new list of complete matrices










3












$begingroup$


I need to multiply 3 lists of matrices (b.a.b) as the following code 



a = -17.8227277373099`, -1.6565234964560602`, 
-1.6565234954649242`, 
 5.298073701591974`, -17.812203521929003`, 
-1.5013126607114478`, -1.5013126574896714`, 
 4.384050851253119`, -17.801677045750512`, 
 -1.4055541329078751`, -1.405554138172727`, 
 3.869511752542245`;

b = 0.8409518416651456`, 0, 0, 
 0.1274293000222242`, 0.8409815693580924`, 0, 0, 
 0.14187218616724442`, 0.841011296000238`, 0, 0, 
 0.15290209433231844`;


I used the following:



 mat = b.a.b


But, I think this way is not correct because I didn't get the result. Also, I make a test for the first matrices as the following code, that what I wanted to get for the whole matrices multiplication. 



 a1 = -17.8227277373099`, -1.6565234964560602`, 
-1.6565234954649242`, 5.298073701591974`;

b1 = 0.8409518416651456`, 0, 0, 0.1274293000222242`;

 mat1 = b1.a1.b1


 -12.6042, -0.177516, -0.177516, 0.0860313

Thanks.





list-manipulation matrix







share|improve this question









$endgroup$







  • 1




    $begingroup$
    Table[b[[i]].a[[i]].b[[i]], i, 1, 3] or #[[2]].#[[1]].#[[2]] & /@ Transpose[a, b].
    $endgroup$
    – corey979
    Mar 18 at 21:06











  • $begingroup$
    Thanks a lot corey979!
    $endgroup$
    – Ghady
    Mar 19 at 0:30















3












$begingroup$


I need to multiply 3 lists of matrices (b.a.b) as the following code 



a = -17.8227277373099`, -1.6565234964560602`, 
-1.6565234954649242`, 
 5.298073701591974`, -17.812203521929003`, 
-1.5013126607114478`, -1.5013126574896714`, 
 4.384050851253119`, -17.801677045750512`, 
 -1.4055541329078751`, -1.405554138172727`, 
 3.869511752542245`;

b = 0.8409518416651456`, 0, 0, 
 0.1274293000222242`, 0.8409815693580924`, 0, 0, 
 0.14187218616724442`, 0.841011296000238`, 0, 0, 
 0.15290209433231844`;


I used the following:



 mat = b.a.b


But, I think this way is not correct because I didn't get the result. Also, I make a test for the first matrices as the following code, that what I wanted to get for the whole matrices multiplication. 



 a1 = -17.8227277373099`, -1.6565234964560602`, 
-1.6565234954649242`, 5.298073701591974`;

b1 = 0.8409518416651456`, 0, 0, 0.1274293000222242`;

 mat1 = b1.a1.b1


 -12.6042, -0.177516, -0.177516, 0.0860313

Thanks.





list-manipulation matrix







share|improve this question









$endgroup$







  • 1




    $begingroup$
    Table[b[[i]].a[[i]].b[[i]], i, 1, 3] or #[[2]].#[[1]].#[[2]] & /@ Transpose[a, b].
    $endgroup$
    – corey979
    Mar 18 at 21:06











  • $begingroup$
    Thanks a lot corey979!
    $endgroup$
    – Ghady
    Mar 19 at 0:30













3












3








3





$begingroup$


I need to multiply 3 lists of matrices (b.a.b) as the following code 



a = -17.8227277373099`, -1.6565234964560602`, 
-1.6565234954649242`, 
 5.298073701591974`, -17.812203521929003`, 
-1.5013126607114478`, -1.5013126574896714`, 
 4.384050851253119`, -17.801677045750512`, 
 -1.4055541329078751`, -1.405554138172727`, 
 3.869511752542245`;

b = 0.8409518416651456`, 0, 0, 
 0.1274293000222242`, 0.8409815693580924`, 0, 0, 
 0.14187218616724442`, 0.841011296000238`, 0, 0, 
 0.15290209433231844`;


I used the following:



 mat = b.a.b


But, I think this way is not correct because I didn't get the result. Also, I make a test for the first matrices as the following code, that what I wanted to get for the whole matrices multiplication. 



 a1 = -17.8227277373099`, -1.6565234964560602`, 
-1.6565234954649242`, 5.298073701591974`;

b1 = 0.8409518416651456`, 0, 0, 0.1274293000222242`;

 mat1 = b1.a1.b1


 -12.6042, -0.177516, -0.177516, 0.0860313

Thanks.





list-manipulation matrix







share|improve this question









$endgroup$




I need to multiply 3 lists of matrices (b.a.b) as the following code 



a = -17.8227277373099`, -1.6565234964560602`, 
-1.6565234954649242`, 
 5.298073701591974`, -17.812203521929003`, 
-1.5013126607114478`, -1.5013126574896714`, 
 4.384050851253119`, -17.801677045750512`, 
 -1.4055541329078751`, -1.405554138172727`, 
 3.869511752542245`;

b = 0.8409518416651456`, 0, 0, 
 0.1274293000222242`, 0.8409815693580924`, 0, 0, 
 0.14187218616724442`, 0.841011296000238`, 0, 0, 
 0.15290209433231844`;


I used the following:



 mat = b.a.b


But, I think this way is not correct because I didn't get the result. Also, I make a test for the first matrices as the following code, that what I wanted to get for the whole matrices multiplication. 



 a1 = -17.8227277373099`, -1.6565234964560602`, 
-1.6565234954649242`, 5.298073701591974`;

b1 = 0.8409518416651456`, 0, 0, 0.1274293000222242`;

 mat1 = b1.a1.b1


 -12.6042, -0.177516, -0.177516, 0.0860313

Thanks.





list-manipulation matrix




list-manipulation matrix






share|improve this question













share|improve this question











share|improve this question




share|improve this question










asked Mar 18 at 20:56









GhadyGhady

696




696







  • 1




    $begingroup$
    Table[b[[i]].a[[i]].b[[i]], i, 1, 3] or #[[2]].#[[1]].#[[2]] & /@ Transpose[a, b].
    $endgroup$
    – corey979
    Mar 18 at 21:06











  • $begingroup$
    Thanks a lot corey979!
    $endgroup$
    – Ghady
    Mar 19 at 0:30












  • 1




    $begingroup$
    Table[b[[i]].a[[i]].b[[i]], i, 1, 3] or #[[2]].#[[1]].#[[2]] & /@ Transpose[a, b].
    $endgroup$
    – corey979
    Mar 18 at 21:06











  • $begingroup$
    Thanks a lot corey979!
    $endgroup$
    – Ghady
    Mar 19 at 0:30







1




1




$begingroup$
Table[b[[i]].a[[i]].b[[i]], i, 1, 3] or #[[2]].#[[1]].#[[2]] & /@ Transpose[a, b].
$endgroup$
– corey979
Mar 18 at 21:06





$begingroup$
Table[b[[i]].a[[i]].b[[i]], i, 1, 3] or #[[2]].#[[1]].#[[2]] & /@ Transpose[a, b].
$endgroup$
– corey979
Mar 18 at 21:06













$begingroup$
Thanks a lot corey979!
$endgroup$
– Ghady
Mar 19 at 0:30




$begingroup$
Thanks a lot corey979!
$endgroup$
– Ghady
Mar 19 at 0:30










2 Answers
2






active

oldest

votes


















8












$begingroup$

Using MapThread and Dot:



MapThread[Dot, b, a, b]





share|improve this answer









$endgroup$












  • $begingroup$
    Thank you very much, It is the easiest!
    $endgroup$
    – Ghady
    Mar 19 at 0:33


















6












$begingroup$

If you need it really fast, then use Compile:



n = 1000000;
a = RandomReal[-1, 1, n, 2, 2];
b = RandomReal[-1, 1, n, 2, 2];

cf = Compile[a, _Real, 2, b, _Real, 2,
 b.a.b,
 RuntimeAttributes -> Listable,
 Parallelization -> True
 ];

MapThread[Dot, b, a, b]; // AbsoluteTiming // First
cf[a, b]; // AbsoluteTiming // First



1.46343



0.075722







share|improve this answer











$endgroup$












  • $begingroup$
    Thank you very much!
    $endgroup$
    – Ghady
    Mar 19 at 0:31










  • $begingroup$
    You're welcome.
    $endgroup$
    – Henrik Schumacher
    Mar 19 at 8:00











Your Answer





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2 Answers
2






active

oldest

votes








2 Answers
2






active

oldest

votes









active

oldest

votes






active

oldest

votes









8












$begingroup$

Using MapThread and Dot:



MapThread[Dot, b, a, b]





share|improve this answer









$endgroup$












  • $begingroup$
    Thank you very much, It is the easiest!
    $endgroup$
    – Ghady
    Mar 19 at 0:33















8












$begingroup$

Using MapThread and Dot:



MapThread[Dot, b, a, b]





share|improve this answer









$endgroup$












  • $begingroup$
    Thank you very much, It is the easiest!
    $endgroup$
    – Ghady
    Mar 19 at 0:33













8












8








8





$begingroup$

Using MapThread and Dot:



MapThread[Dot, b, a, b]





share|improve this answer









$endgroup$



Using MapThread and Dot:



MapThread[Dot, b, a, b]






share|improve this answer












share|improve this answer



share|improve this answer










answered Mar 18 at 21:21









swishswish

4,1811536




4,1811536











  • $begingroup$
    Thank you very much, It is the easiest!
    $endgroup$
    – Ghady
    Mar 19 at 0:33
















  • $begingroup$
    Thank you very much, It is the easiest!
    $endgroup$
    – Ghady
    Mar 19 at 0:33















$begingroup$
Thank you very much, It is the easiest!
$endgroup$
– Ghady
Mar 19 at 0:33




$begingroup$
Thank you very much, It is the easiest!
$endgroup$
– Ghady
Mar 19 at 0:33











6












$begingroup$

If you need it really fast, then use Compile:



n = 1000000;
a = RandomReal[-1, 1, n, 2, 2];
b = RandomReal[-1, 1, n, 2, 2];

cf = Compile[a, _Real, 2, b, _Real, 2,
 b.a.b,
 RuntimeAttributes -> Listable,
 Parallelization -> True
 ];

MapThread[Dot, b, a, b]; // AbsoluteTiming // First
cf[a, b]; // AbsoluteTiming // First



1.46343



0.075722







share|improve this answer











$endgroup$












  • $begingroup$
    Thank you very much!
    $endgroup$
    – Ghady
    Mar 19 at 0:31










  • $begingroup$
    You're welcome.
    $endgroup$
    – Henrik Schumacher
    Mar 19 at 8:00















6












$begingroup$

If you need it really fast, then use Compile:



n = 1000000;
a = RandomReal[-1, 1, n, 2, 2];
b = RandomReal[-1, 1, n, 2, 2];

cf = Compile[a, _Real, 2, b, _Real, 2,
 b.a.b,
 RuntimeAttributes -> Listable,
 Parallelization -> True
 ];

MapThread[Dot, b, a, b]; // AbsoluteTiming // First
cf[a, b]; // AbsoluteTiming // First



1.46343



0.075722







share|improve this answer











$endgroup$












  • $begingroup$
    Thank you very much!
    $endgroup$
    – Ghady
    Mar 19 at 0:31










  • $begingroup$
    You're welcome.
    $endgroup$
    – Henrik Schumacher
    Mar 19 at 8:00













6












6








6





$begingroup$

If you need it really fast, then use Compile:



n = 1000000;
a = RandomReal[-1, 1, n, 2, 2];
b = RandomReal[-1, 1, n, 2, 2];

cf = Compile[a, _Real, 2, b, _Real, 2,
 b.a.b,
 RuntimeAttributes -> Listable,
 Parallelization -> True
 ];

MapThread[Dot, b, a, b]; // AbsoluteTiming // First
cf[a, b]; // AbsoluteTiming // First



1.46343



0.075722







share|improve this answer











$endgroup$



If you need it really fast, then use Compile:



n = 1000000;
a = RandomReal[-1, 1, n, 2, 2];
b = RandomReal[-1, 1, n, 2, 2];

cf = Compile[a, _Real, 2, b, _Real, 2,
 b.a.b,
 RuntimeAttributes -> Listable,
 Parallelization -> True
 ];

MapThread[Dot, b, a, b]; // AbsoluteTiming // First
cf[a, b]; // AbsoluteTiming // First



1.46343



0.075722








share|improve this answer














share|improve this answer



share|improve this answer








edited Mar 19 at 8:00

























answered Mar 18 at 23:36









Henrik SchumacherHenrik Schumacher

58.6k581162




58.6k581162











  • $begingroup$
    Thank you very much!
    $endgroup$
    – Ghady
    Mar 19 at 0:31










  • $begingroup$
    You're welcome.
    $endgroup$
    – Henrik Schumacher
    Mar 19 at 8:00
















  • $begingroup$
    Thank you very much!
    $endgroup$
    – Ghady
    Mar 19 at 0:31










  • $begingroup$
    You're welcome.
    $endgroup$
    – Henrik Schumacher
    Mar 19 at 8:00















$begingroup$
Thank you very much!
$endgroup$
– Ghady
Mar 19 at 0:31




$begingroup$
Thank you very much!
$endgroup$
– Ghady
Mar 19 at 0:31












$begingroup$
You're welcome.
$endgroup$
– Henrik Schumacher
Mar 19 at 8:00




$begingroup$
You're welcome.
$endgroup$
– Henrik Schumacher
Mar 19 at 8:00

















draft saved

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