Decimal to roman python
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5
$begingroup$
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 11 hours ago
Graipher
26.7k54092
asked 12 hours ago
OfeksOfeks
283
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.
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add a comment |
5
$begingroup$
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 11 hours ago
Graipher
26.7k54092
asked 12 hours ago
OfeksOfeks
283
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 |
5
5
5
$begingroup$
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 11 hours ago
Graipher
26.7k54092
asked 12 hours ago
OfeksOfeks
283
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 11 hours ago
Graipher
26.7k54092
asked 12 hours ago
OfeksOfeks
283
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 11 hours ago
Graipher
26.7k54092
asked 12 hours ago
OfeksOfeks
283
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 11 hours ago
Graipher
26.7k54092
edited 11 hours ago
Graipher
26.7k54092
edited 11 hours ago
Graipher
26.7k54092
26.7k54092
asked 12 hours ago
OfeksOfeks
283
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 12 hours ago
OfeksOfeks
283
asked 12 hours ago
OfeksOfeks
283
283
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
8
$begingroup$
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 11 hours ago
GraipherGraipher
26.7k54092
$endgroup$
$begingroup$
wow. looks so easy now, thank you. that's great.
$endgroup$
– Ofeks
9 hours ago
$begingroup$
@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
9 hours ago
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
8
$begingroup$
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 11 hours ago
GraipherGraipher
26.7k54092
$endgroup$
$begingroup$
wow. looks so easy now, thank you. that's great.
$endgroup$
– Ofeks
9 hours ago
$begingroup$
@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
9 hours ago
add a comment |
8
$begingroup$
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 11 hours ago
GraipherGraipher
26.7k54092
$endgroup$
$begingroup$
wow. looks so easy now, thank you. that's great.
$endgroup$
– Ofeks
9 hours ago
$begingroup$
@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
9 hours ago
add a comment |
8
8
8
$begingroup$
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 11 hours ago
GraipherGraipher
26.7k54092
$endgroup$
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 11 hours ago
GraipherGraipher
26.7k54092
edited 8 hours ago
answered 11 hours ago
GraipherGraipher
26.7k54092
answered 11 hours ago
GraipherGraipher
26.7k54092
answered 11 hours ago
GraipherGraipher
26.7k54092
26.7k54092
$begingroup$
wow. looks so easy now, thank you. that's great.
$endgroup$
– Ofeks
9 hours ago
$begingroup$
@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
9 hours ago
add a comment |
$begingroup$
wow. looks so easy now, thank you. that's great.
$endgroup$
– Ofeks
9 hours ago
$begingroup$
@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
9 hours ago
$begingroup$
wow. looks so easy now, thank you. that's great.
$endgroup$
– Ofeks
9 hours ago
$begingroup$
wow. looks so easy now, thank you. that's great.
$endgroup$
– Ofeks
9 hours ago
$begingroup$
@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
9 hours ago
$begingroup$
@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
9 hours ago
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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