Decimal to roman python
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6
$begingroup$
I'm a beginner as you can see and I would like to know how I can improve my code. Studying for 6 months now. Thank you.
roman_dict = {1: 'I', 4: 'IV', 5: 'V', 9: 'IX', 10: 'X', 40: 'XL', 50: 'L', 90: 'XC', 100: 'C', 400: 'CD',
500: 'D', 900: 'CM', 1000: 'M'}
divide_list = [1000, 100, 10, 1]
def not_in_dict(fixed_decimal, divide_num):
sub_count = 0
sub_roman_multi = roman_dict[divide_num]
temp_decimal = fixed_decimal
while temp_decimal not in roman_dict:
temp_decimal -= divide_num
sub_count += 1
return roman_dict[temp_decimal]+(sub_count*sub_roman_multi)
def decimal_to_roman(decimal):
original_decimal = decimal
roman = ""
for divide_num in divide_list:
if decimal >= divide_num:
reminder = decimal//divide_num
if(reminder >= 1) and ((reminder*divide_num) in roman_dict):
roman += roman_dict[(reminder*divide_num)]
decimal -= reminder*divide_num
else:
roman += not_in_dict(reminder*divide_num, divide_num)
decimal -= (reminder*divide_num)
return str(original_decimal)+' = '+roman
python roman-numerals
edited Apr 4 at 11:53
Graipher
26.8k54396
asked Apr 4 at 10:23
OfeksOfeks
333
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Ofeks is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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add a comment |
6
$begingroup$
I'm a beginner as you can see and I would like to know how I can improve my code. Studying for 6 months now. Thank you.
roman_dict = {1: 'I', 4: 'IV', 5: 'V', 9: 'IX', 10: 'X', 40: 'XL', 50: 'L', 90: 'XC', 100: 'C', 400: 'CD',
500: 'D', 900: 'CM', 1000: 'M'}
divide_list = [1000, 100, 10, 1]
def not_in_dict(fixed_decimal, divide_num):
sub_count = 0
sub_roman_multi = roman_dict[divide_num]
temp_decimal = fixed_decimal
while temp_decimal not in roman_dict:
temp_decimal -= divide_num
sub_count += 1
return roman_dict[temp_decimal]+(sub_count*sub_roman_multi)
def decimal_to_roman(decimal):
original_decimal = decimal
roman = ""
for divide_num in divide_list:
if decimal >= divide_num:
reminder = decimal//divide_num
if(reminder >= 1) and ((reminder*divide_num) in roman_dict):
roman += roman_dict[(reminder*divide_num)]
decimal -= reminder*divide_num
else:
roman += not_in_dict(reminder*divide_num, divide_num)
decimal -= (reminder*divide_num)
return str(original_decimal)+' = '+roman
python roman-numerals
edited Apr 4 at 11:53
Graipher
26.8k54396
asked Apr 4 at 10:23
OfeksOfeks
333
New contributor
Ofeks is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
add a comment |
6
6
6
$begingroup$
I'm a beginner as you can see and I would like to know how I can improve my code. Studying for 6 months now. Thank you.
roman_dict = {1: 'I', 4: 'IV', 5: 'V', 9: 'IX', 10: 'X', 40: 'XL', 50: 'L', 90: 'XC', 100: 'C', 400: 'CD',
500: 'D', 900: 'CM', 1000: 'M'}
divide_list = [1000, 100, 10, 1]
def not_in_dict(fixed_decimal, divide_num):
sub_count = 0
sub_roman_multi = roman_dict[divide_num]
temp_decimal = fixed_decimal
while temp_decimal not in roman_dict:
temp_decimal -= divide_num
sub_count += 1
return roman_dict[temp_decimal]+(sub_count*sub_roman_multi)
def decimal_to_roman(decimal):
original_decimal = decimal
roman = ""
for divide_num in divide_list:
if decimal >= divide_num:
reminder = decimal//divide_num
if(reminder >= 1) and ((reminder*divide_num) in roman_dict):
roman += roman_dict[(reminder*divide_num)]
decimal -= reminder*divide_num
else:
roman += not_in_dict(reminder*divide_num, divide_num)
decimal -= (reminder*divide_num)
return str(original_decimal)+' = '+roman
python roman-numerals
edited Apr 4 at 11:53
Graipher
26.8k54396
asked Apr 4 at 10:23
OfeksOfeks
333
New contributor
Ofeks is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
I'm a beginner as you can see and I would like to know how I can improve my code. Studying for 6 months now. Thank you.
roman_dict = {1: 'I', 4: 'IV', 5: 'V', 9: 'IX', 10: 'X', 40: 'XL', 50: 'L', 90: 'XC', 100: 'C', 400: 'CD',
500: 'D', 900: 'CM', 1000: 'M'}
divide_list = [1000, 100, 10, 1]
def not_in_dict(fixed_decimal, divide_num):
sub_count = 0
sub_roman_multi = roman_dict[divide_num]
temp_decimal = fixed_decimal
while temp_decimal not in roman_dict:
temp_decimal -= divide_num
sub_count += 1
return roman_dict[temp_decimal]+(sub_count*sub_roman_multi)
def decimal_to_roman(decimal):
original_decimal = decimal
roman = ""
for divide_num in divide_list:
if decimal >= divide_num:
reminder = decimal//divide_num
if(reminder >= 1) and ((reminder*divide_num) in roman_dict):
roman += roman_dict[(reminder*divide_num)]
decimal -= reminder*divide_num
else:
roman += not_in_dict(reminder*divide_num, divide_num)
decimal -= (reminder*divide_num)
return str(original_decimal)+' = '+roman
python roman-numerals
python roman-numerals
edited Apr 4 at 11:53
Graipher
26.8k54396
asked Apr 4 at 10:23
OfeksOfeks
333
New contributor
Ofeks is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited Apr 4 at 11:53
Graipher
26.8k54396
asked Apr 4 at 10:23
OfeksOfeks
333
New contributor
Ofeks is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited Apr 4 at 11:53
Graipher
26.8k54396
edited Apr 4 at 11:53
Graipher
26.8k54396
edited Apr 4 at 11:53
Graipher
26.8k54396
26.8k54396
asked Apr 4 at 10:23
OfeksOfeks
333
New contributor
Ofeks is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked Apr 4 at 10:23
OfeksOfeks
333
asked Apr 4 at 10:23
OfeksOfeks
333
333
New contributor
Ofeks is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
Ofeks is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Ofeks is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
10
$begingroup$
If you use a list of tuples instead of a dictionary and reverse the order, you can simply iterate over it. Your while loop also becomes a lot easier to understand and there is no longer any need to outsource it to another function that returns the literal and its count.
Instead of manually adding strings (something you should basically never do in in Python), use str.join.
ROMAN_LITERALS = [(1000, 'M'), (900, 'CM'), (500, 'D'), (400, 'CD'), (100, 'C'),
(90, 'XC'), (50, 'L'), (40, 'XL'), (10, 'X'), (9, 'IX'),
(5, 'V'), (4, 'IV'), (1, 'I')]
def decimal_to_roman(x):
out =
for value, literal in ROMAN_LITERALS:
while x >= value:
x -= value
out.append(literal)
return "".join(out)
Instead of the while loop you can also use integer division like you did:
def decimal_to_roman(x):
out =
for value, literal in ROMAN_LITERALS:
n = x // value # will be 0 if value is too large
out.extend([literal] * n) # will not do anything if n == 0
x -= n * value # will also not do anything if n == 0
return "".join(out)
answered Apr 4 at 12:01
GraipherGraipher
26.8k54396
$endgroup$
$begingroup$
wow. looks so easy now, thank you. that's great.
$endgroup$
– Ofeks
Apr 4 at 13:36
$begingroup$
@Ofeks: If this helped you, consider accepting it as the correct answer (by clicking the checkmark to the left of the answer). It is customary to wait about 24 hours, though, to give everyon on the globe a chance to answer and not discourage other people from commenting.
$endgroup$
– Graipher
Apr 4 at 14:01
add a comment |
Your Answer
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
10
$begingroup$
If you use a list of tuples instead of a dictionary and reverse the order, you can simply iterate over it. Your while loop also becomes a lot easier to understand and there is no longer any need to outsource it to another function that returns the literal and its count.
Instead of manually adding strings (something you should basically never do in in Python), use str.join.
ROMAN_LITERALS = [(1000, 'M'), (900, 'CM'), (500, 'D'), (400, 'CD'), (100, 'C'),
(90, 'XC'), (50, 'L'), (40, 'XL'), (10, 'X'), (9, 'IX'),
(5, 'V'), (4, 'IV'), (1, 'I')]
def decimal_to_roman(x):
out =
for value, literal in ROMAN_LITERALS:
while x >= value:
x -= value
out.append(literal)
return "".join(out)
Instead of the while loop you can also use integer division like you did:
def decimal_to_roman(x):
out =
for value, literal in ROMAN_LITERALS:
n = x // value # will be 0 if value is too large
out.extend([literal] * n) # will not do anything if n == 0
x -= n * value # will also not do anything if n == 0
return "".join(out)
answered Apr 4 at 12:01
GraipherGraipher
26.8k54396
$endgroup$
$begingroup$
wow. looks so easy now, thank you. that's great.
$endgroup$
– Ofeks
Apr 4 at 13:36
$begingroup$
@Ofeks: If this helped you, consider accepting it as the correct answer (by clicking the checkmark to the left of the answer). It is customary to wait about 24 hours, though, to give everyon on the globe a chance to answer and not discourage other people from commenting.
$endgroup$
– Graipher
Apr 4 at 14:01
add a comment |
10
$begingroup$
If you use a list of tuples instead of a dictionary and reverse the order, you can simply iterate over it. Your while loop also becomes a lot easier to understand and there is no longer any need to outsource it to another function that returns the literal and its count.
Instead of manually adding strings (something you should basically never do in in Python), use str.join.
ROMAN_LITERALS = [(1000, 'M'), (900, 'CM'), (500, 'D'), (400, 'CD'), (100, 'C'),
(90, 'XC'), (50, 'L'), (40, 'XL'), (10, 'X'), (9, 'IX'),
(5, 'V'), (4, 'IV'), (1, 'I')]
def decimal_to_roman(x):
out =
for value, literal in ROMAN_LITERALS:
while x >= value:
x -= value
out.append(literal)
return "".join(out)
Instead of the while loop you can also use integer division like you did:
def decimal_to_roman(x):
out =
for value, literal in ROMAN_LITERALS:
n = x // value # will be 0 if value is too large
out.extend([literal] * n) # will not do anything if n == 0
x -= n * value # will also not do anything if n == 0
return "".join(out)
answered Apr 4 at 12:01
GraipherGraipher
26.8k54396
$endgroup$
$begingroup$
wow. looks so easy now, thank you. that's great.
$endgroup$
– Ofeks
Apr 4 at 13:36
$begingroup$
@Ofeks: If this helped you, consider accepting it as the correct answer (by clicking the checkmark to the left of the answer). It is customary to wait about 24 hours, though, to give everyon on the globe a chance to answer and not discourage other people from commenting.
$endgroup$
– Graipher
Apr 4 at 14:01
add a comment |
10
10
10
$begingroup$
If you use a list of tuples instead of a dictionary and reverse the order, you can simply iterate over it. Your while loop also becomes a lot easier to understand and there is no longer any need to outsource it to another function that returns the literal and its count.
Instead of manually adding strings (something you should basically never do in in Python), use str.join.
ROMAN_LITERALS = [(1000, 'M'), (900, 'CM'), (500, 'D'), (400, 'CD'), (100, 'C'),
(90, 'XC'), (50, 'L'), (40, 'XL'), (10, 'X'), (9, 'IX'),
(5, 'V'), (4, 'IV'), (1, 'I')]
def decimal_to_roman(x):
out =
for value, literal in ROMAN_LITERALS:
while x >= value:
x -= value
out.append(literal)
return "".join(out)
Instead of the while loop you can also use integer division like you did:
def decimal_to_roman(x):
out =
for value, literal in ROMAN_LITERALS:
n = x // value # will be 0 if value is too large
out.extend([literal] * n) # will not do anything if n == 0
x -= n * value # will also not do anything if n == 0
return "".join(out)
answered Apr 4 at 12:01
GraipherGraipher
26.8k54396
$endgroup$
If you use a list of tuples instead of a dictionary and reverse the order, you can simply iterate over it. Your while loop also becomes a lot easier to understand and there is no longer any need to outsource it to another function that returns the literal and its count.
Instead of manually adding strings (something you should basically never do in in Python), use str.join.
ROMAN_LITERALS = [(1000, 'M'), (900, 'CM'), (500, 'D'), (400, 'CD'), (100, 'C'),
(90, 'XC'), (50, 'L'), (40, 'XL'), (10, 'X'), (9, 'IX'),
(5, 'V'), (4, 'IV'), (1, 'I')]
def decimal_to_roman(x):
out =
for value, literal in ROMAN_LITERALS:
while x >= value:
x -= value
out.append(literal)
return "".join(out)
Instead of the while loop you can also use integer division like you did:
def decimal_to_roman(x):
out =
for value, literal in ROMAN_LITERALS:
n = x // value # will be 0 if value is too large
out.extend([literal] * n) # will not do anything if n == 0
x -= n * value # will also not do anything if n == 0
return "".join(out)
answered Apr 4 at 12:01
GraipherGraipher
26.8k54396
edited Apr 4 at 14:44
answered Apr 4 at 12:01
GraipherGraipher
26.8k54396
answered Apr 4 at 12:01
GraipherGraipher
26.8k54396
answered Apr 4 at 12:01
GraipherGraipher
26.8k54396
26.8k54396
$begingroup$
wow. looks so easy now, thank you. that's great.
$endgroup$
– Ofeks
Apr 4 at 13:36
$begingroup$
@Ofeks: If this helped you, consider accepting it as the correct answer (by clicking the checkmark to the left of the answer). It is customary to wait about 24 hours, though, to give everyon on the globe a chance to answer and not discourage other people from commenting.
$endgroup$
– Graipher
Apr 4 at 14:01
add a comment |
$begingroup$
wow. looks so easy now, thank you. that's great.
$endgroup$
– Ofeks
Apr 4 at 13:36
$begingroup$
@Ofeks: If this helped you, consider accepting it as the correct answer (by clicking the checkmark to the left of the answer). It is customary to wait about 24 hours, though, to give everyon on the globe a chance to answer and not discourage other people from commenting.
$endgroup$
– Graipher
Apr 4 at 14:01
$begingroup$
wow. looks so easy now, thank you. that's great.
$endgroup$
– Ofeks
Apr 4 at 13:36
$begingroup$
wow. looks so easy now, thank you. that's great.
$endgroup$
– Ofeks
Apr 4 at 13:36
$begingroup$
@Ofeks: If this helped you, consider accepting it as the correct answer (by clicking the checkmark to the left of the answer). It is customary to wait about 24 hours, though, to give everyon on the globe a chance to answer and not discourage other people from commenting.
$endgroup$
– Graipher
Apr 4 at 14:01
$begingroup$
@Ofeks: If this helped you, consider accepting it as the correct answer (by clicking the checkmark to the left of the answer). It is customary to wait about 24 hours, though, to give everyon on the globe a chance to answer and not discourage other people from commenting.
$endgroup$
– Graipher
Apr 4 at 14:01
add a comment |
Ofeks is a new contributor. Be nice, and check out our Code of Conduct.
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Ofeks is a new contributor. Be nice, and check out our Code of Conduct.
Ofeks is a new contributor. Be nice, and check out our Code of Conduct.
Ofeks is a new contributor. Be nice, and check out our Code of Conduct.
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