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Decimal to roman python


Roman Numeral to Decimal ConversionConverting Roman numerals to decimalRoman numeral converter in RubyRoman numerals to decimalRoman numeral to decimal converterCurrency converter in Python 2.7“Merchants Guide to Galaxy” challengeGreed Dice Scoring Game expanded - Python KoansArea and volume calculatorNumber of possible numbers in roman number string






.everyoneloves__top-leaderboard:empty,.everyoneloves__mid-leaderboard:empty,.everyoneloves__bot-mid-leaderboard:empty margin-bottom:0;








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









share|improve this question









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.







$endgroup$


















    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









    share|improve this question









    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.







    $endgroup$














      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









      share|improve this question









      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






      share|improve this question









      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.











      share|improve this question









      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.









      share|improve this question




      share|improve this question








      edited Apr 4 at 11:53









      Graipher

      26.8k54396




      26.8k54396






      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




      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.




















          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)





          share|improve this answer











          $endgroup$












          • $begingroup$
            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











          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)





          share|improve this answer











          $endgroup$












          • $begingroup$
            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















          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)





          share|improve this answer











          $endgroup$












          • $begingroup$
            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













          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)





          share|improve this answer











          $endgroup$



          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)






          share|improve this answer














          share|improve this answer



          share|improve this answer








          edited Apr 4 at 14:44

























          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
















          • $begingroup$
            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










          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.














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