convertir un nombre en Lettre en python

Bonjour à tous,

existe-t-il en python une fonction pour convertir un nombre en Lettre?

je travail sur libreoffice et je veux ajouter la phrase suivante en bas de la facture : " arrêté la présente facture à la somme : Mille quatre cents quarante dirhams."

aidez moi svp

slts

Bonjour,
j'ai fait un script pour les nombres entiers. Voici les fichiers avec un petit bonus pour passer des lettres aux chiffres.

config_fr.py

Code:

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# -*- coding: utf-8 -*- #!/usr/bin/env python   SPECIAL_NUMBERS_NAMES = { '0': "zéro", '1': "un", '2': "deux", '3': "trois", '4': "quatre", '5': "cinq", '6': "six", '7': "sept", '8': "huit", '9': "neuf",   '10': "dix", '11': "onze", '12': "douze", '13': "treize", '14': "quatorze", '15': "quinze", '16': "seize",   '20': "vingt", '30': "trente", '40': "quarante", '50': "cinquante", '60': "soixante", '70': "soixante-dix", '80': "quatre-vingts", '90': "quatre-vingt-dix",   '100': "cent" }   TEN_POWERS_NAMES = {}   # 10^3 and 10 9 are given at the end.   TEN_POWERS_NAMES['everyday'] = { 9: "milliard" # 10^9 }   TEN_POWERS_NAMES['chuquet'] = { 12: "billion", # 10^12 18: "trillion", # 10^18 ... 24: "quadrillion", 30: "quintillion", 36: "sextillion", 42: "septillion", 48: "octillion", 54: "nonillion", 60: "decillion" }   TEN_POWERS_NAMES['rowlett'] = { 9: "milliard", # 10^9 12: "tetrillion", # 10^12 15: "pentillion", # 10^15 ... 18: "hexillion", 21: "eptillion", 24: "oktillion", 27: "ennillion", 30: "dekillion", 33: "hendekillion", 36: "dodekillion", 39: "trisdekillion", 42: "tetradekillion", 45: "pentadekillion", 48: "hexadekillion", 51: "heptadekillion", 54: "oktadekillion", 57: "enneadekillion",   60: "icosillion", 63: "icosihenillion", 66: "icosidillion", 69: "icositrillion", 72: "icositetrillion", 75: "icosipentillion", 78: "icosihexillion", 81: "icosiheptillion", 84: "icosioktillion", 87: "icosiennillion",   90: "triacontillion" }   for oneConvention in ['everyday', 'chuquet', 'rowlett']: TEN_POWERS_NAMES[oneConvention][3] = "mille" TEN_POWERS_NAMES[oneConvention][6] = "million"

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integerToWords_fr.py

Code:

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# -*- coding: utf-8 -*- #!/usr/bin/env python   # **NOTES :** good improvings have been proposed in the following page. # http://www.developpez.net/forums/d864956/autres-langages/python-zope/general-python/ecrire-nombre-toutes-lettres   # There are three conventions used for very big numbers: # # 1) The first one named "everyday" is the every day convention : # 132*10^9 = "cent-trent-deux milliards" # 10^12 = "mille milliards" # 124*10^6 * 10*9 = "cent-vingt-quatre millions de milliards" ... etc # # 2) The secund convention named "chuquet", which is very easy to learn, # is defined in the following link : # http://fr.wikipedia.org/wiki/Nom_des_grands_nombres#Famille_des_-llions # For number bigger or equal to 10^60 , we use a convention similar to the first one. # # 3) The third convention named "rowlett" is defined here : # http://fr.wikipedia.org/wiki/Nom_des_grands_nombres#Syst.C3.A8me_Gillion # For number bigger or equal to 10^90 , we use a convention similar to the first one. # # **NOTE :** we use to write "huit-cent-quatre" for "804". # The dashes are used ONLY for numbers between 101 and 999. # This is not the official convention but by doing this # we make the groups of three digits more visible.   import config_fr   SPECIAL_NUMBERS_NAMES = config_fr.SPECIAL_NUMBERS_NAMES TEN_POWERS_NAMES = config_fr.TEN_POWERS_NAMES   # We add 01, 02, 03,... for i in range(0,10): SPECIAL_NUMBERS_NAMES['0' + str(i)] = SPECIAL_NUMBERS_NAMES[str(i)]     #################################### # SPECIFIC NAMES FOR FRENCH == START # # We add 'dix-sept', 'dix-huit' and 'dix-neuf'. for i in range(7, 10): SPECIAL_NUMBERS_NAMES['1' + str(i)] = "dix-" + SPECIAL_NUMBERS_NAMES[str(i)]   # We add 'vingt-et-un', ... , 'soixante-et-un'. for i in range(2, 7): SPECIAL_NUMBERS_NAMES[str(i) + '1'] = SPECIAL_NUMBERS_NAMES[str(i) + '0'] + "-et-un"   # We add 'soixante-et-onze'. SPECIAL_NUMBERS_NAMES['71'] = "soixante-et-onze"   # We add 'deux cents', 'trois cents', ... such as # to not treat the very boring gramatical rules. for i in range(2, 10): SPECIAL_NUMBERS_NAMES[str(i) + '00'] = SPECIAL_NUMBERS_NAMES[str(i)] + "-cents"   # For "trente-..." and co. TEN_PREFIXES = {} for i in range(2, 10): if i == 7: TEN_PREFIXES[str(i)] = SPECIAL_NUMBERS_NAMES[str(i-1) + '0'] + "-" elif i == 8: TEN_PREFIXES[str(i)] = SPECIAL_NUMBERS_NAMES[str(i) + '0'][:-1] + "-" elif i == 9: TEN_PREFIXES[str(i)] = SPECIAL_NUMBERS_NAMES[str(i-1) + '0'][:-1] + "-" else: TEN_PREFIXES[str(i)] = SPECIAL_NUMBERS_NAMES[str(i) + '0'] + "-"   # # SPECIFIC NAMES FOR FRENCH == END ##################################     # We build the range to build so as to split big numbers. THE_POWERS = {} MAX_POWER = {} THE_BIGGER_NAME = {}   for oneConvention in TEN_POWERS_NAMES: # We add zero so as to facilitate the procedures. THE_POWERS[oneConvention] = sorted([0] + [ x for x in TEN_POWERS_NAMES[oneConvention] if x != 3 ]) MAX_POWER[oneConvention] = THE_POWERS[oneConvention][-1] THE_BIGGER_NAME[oneConvention] = TEN_POWERS_NAMES[oneConvention][ THE_POWERS[oneConvention][-1] ]     ############### # THE FUNCTIONS   #TODO orderMagnitude à améliorer.... def orderMagnitude(number): """     For example, 123456 becomes 123000, and 12345 becomes 12000     """ l = len(number) // 3 i = len(number) - 3*l   if i == 0: i += 3 l -= 1   return number[:i] + '0'*l*3     def floor(number, tenPowerPrecision = 0): """     This function changes the tenPowerPrecision right digits with zeros.     """ if type(tenPowerPrecision) != int or \ tenPowerPrecision < 0: raise ValueError("""tenPowerPrecision = "' + str(tenPowerPrecision) + '" is not allowed.   tenPowerPrecision can only be equal to a natural n with 10^n is the precision needed. """) number = str(number)   if tenPowerPrecision > 0 and len(number) > tenPowerPrecision: number = number[:len(number) - tenPowerPrecision] + '0'*(tenPowerPrecision)   return number     def cleanInteger(number): """     None is return when the number is not an integer.     """   number = str(number).replace(' ', '')   test = number for i in range(10): test = test.replace(str(i), '')   if test: return None   return number     def boringFrenchGrammaticalRuleForCENT(litteralNumber): if litteralNumber[-5:] == 'cents': litteralNumber = litteralNumber[:-1] return litteralNumber     def printer(number, checkValidity = True, convention = "everyday"): if checkValidity: if convention not in TEN_POWERS_NAMES: raise ValueError('convention = "' + str(convention) + '" is unknown.')   # We work with a string. number = cleanInteger(number) if not number: raise ValueError('number = "' + str(number) + '" is not an integer.')   # We have directly the name of the number. if number in SPECIAL_NUMBERS_NAMES: # We have to take care of 10*6 answer = SPECIAL_NUMBERS_NAMES[number]   if answer == TEN_POWERS_NAMES[convention][6]: answer = 'un ' + answer   return answer   # We have a number lower than 100. # # 0, 1, ... , 9 have been already treated. # 10, 11, 12, 13, 14, 15, 16, 17, 18, 19 have been already treated. # 20, 21, 30, 31, ... , 70, 71, 80 and 90 have been already treated. if len(number) == 2: if number[0] in ['7', '9']: return TEN_PREFIXES[number[0]] + SPECIAL_NUMBERS_NAMES['1' + number[1]] else: return TEN_PREFIXES[number[0]] + SPECIAL_NUMBERS_NAMES[number[1]]   # We have a number between 101 and 999. # # 100, 200, 300, ... , 800 and 900 has been already treated. # So we do not have to take care of french boring rules. if len(number) == 3: if number[0] == '0': hundredName = '' else: hundredName = SPECIAL_NUMBERS_NAMES['100'] if number[0] != '1': hundredName = SPECIAL_NUMBERS_NAMES[number[0]] + '-' + hundredName hundredName += '-'   return hundredName + printer( number = number[1:], checkValidity = False )   # We begin to treat number bigger than 1000. ten_powers_names = TEN_POWERS_NAMES[convention]   # We have a number between 1000 and 999 999. # # We do that because later, "6 digits" is the bigger size of # the "intermediate" part like "77" and "124 345" in 77 124 345 000 000. if len(number) <= 6: hundredPart = printer( number = number[-3:], checkValidity = False ) # We can have 123000. if hundredPart == SPECIAL_NUMBERS_NAMES['0']: hundredPart = '' else: hundredPart = ' ' + hundredPart   thousandPart = printer( number = number[:-3], checkValidity = False )   # We can have 000123 because of the recursive calls. if thousandPart == SPECIAL_NUMBERS_NAMES['0']: thousandPart = '' elif thousandPart == SPECIAL_NUMBERS_NAMES['1']: thousandPart = ten_powers_names[3] else: # Gramatical french boring rules for 'cent' like in 'quatre-cent mille'. thousandPart = boringFrenchGrammaticalRuleForCENT(thousandPart) + ' ' + ten_powers_names[3]   return thousandPart + hundredPart   # We have a number between 10^6 and 10^Pmax-1. # # For example with the convention "everyday", # we have Pmax = 9 and a number like # 123 456 789 # must be treated as # 123 ---> Numbers of 10^6 # 345678 ---> Lower than 10^6 # # With the convention "chuquet", # we have Pmax = 60 and a number like # 123 456 789 012 345 678 # must be treated as # 123456 ---> Numbers of 10^12 # 789012 ---> Numbers of 10^6 # 345678 ---> Lower than 10^6 # # With the convention "rowlett", # we have Pmax = 90 and a number like # 123 456 789 012 345 678 # must be treated as # 123 ---> Numbers of 10^15 # 456 ---> Numbers of 10^12 # 789 ---> Numbers of 10^9 # 012 ---> Numbers of 10^6 # 345678 ---> Lower than 10^6 the_powers = THE_POWERS[convention] max_power = MAX_POWER[convention] len_number = len(number)   if len_number <= max_power: answer = printer( number = number[-the_powers[1]:], checkValidity = False ) # We can have ...000. if answer == SPECIAL_NUMBERS_NAMES['0']: answer = ''   for i in range(1, len(the_powers) - 1): if the_powers[i] > len_number: break   numberOfIntermediatePart = printer( number = number[-the_powers[i+1]:-the_powers[i]], checkValidity = False ) # We can have ...000... if numberOfIntermediatePart not in [SPECIAL_NUMBERS_NAMES['0'], '']: # Gramatical french boring rules for 'cent' like in 'quatre-cent millions'. numberOfIntermediatePart = boringFrenchGrammaticalRuleForCENT(numberOfIntermediatePart)   if numberOfIntermediatePart == SPECIAL_NUMBERS_NAMES['1']: numberOfIntermediatePart += ' ' + ten_powers_names[the_powers[i]] else: numberOfIntermediatePart += ' ' + ten_powers_names[the_powers[i]] + 's'   if answer: answer = numberOfIntermediatePart + ' ' + answer else: answer = numberOfIntermediatePart   return answer     # We have a number bigger or equal to 10^Pmax. # # For example with the convention "everyday", # we have Pmax = 9 and a number like # 123 456789012 345678901 234567890 # must be treated as # 123 ---> Numbers of 10^9 of 10^9 of 10^9 # 456789012 ---> Numbers of 10^9 of 10^9 # 345678901 ---> Numbers of 10^9 # 234567890 ---> Lower than 10^9 theBiggerName = THE_BIGGER_NAME[convention] currentBigPartName = ''   answer = printer( number = number[-max_power:], checkValidity = False ) number = number[:-max_power] # We can have ...000. if answer == SPECIAL_NUMBERS_NAMES['0']: answer = ''   while(number): numberOfIntermediatePart = printer( number = number[-max_power:], checkValidity = False ) number = number[:-max_power]   # We can have ...000... if numberOfIntermediatePart not in [SPECIAL_NUMBERS_NAMES['0'], '']: # Gramatical french boring rules for 'cent' like in 'quatre-cent millions'. numberOfIntermediatePart = boringFrenchGrammaticalRuleForCENT(numberOfIntermediatePart)   if numberOfIntermediatePart == SPECIAL_NUMBERS_NAMES['1']: numberOfIntermediatePart += ' ' + theBiggerName + currentBigPartName # We have to take care of case like "un million de milliards" = 10^12 else: for onePower in ten_powers_names: if onePower != 3: nameToTest = ten_powers_names[onePower] l = len(nameToTest)   if numberOfIntermediatePart[-l:] == nameToTest or \ numberOfIntermediatePart[-l-1:] == nameToTest + 's': numberOfIntermediatePart += ' de' break   numberOfIntermediatePart += ' ' + theBiggerName + 's' + currentBigPartName   if answer: answer = numberOfIntermediatePart + ' ' + answer else: answer = numberOfIntermediatePart   currentBigPartName += ' de ' + theBiggerName + 's'   return answer     ############### ############### ## ## ## FOR TESTS ## ## ## ############### ###############   if __name__ == '__main__': myConvention ="everyday" myConvention ="rowlett" myConvention ="chuquet"   mytenPowerPrecision = 0 # mytenPowerPrecision = 3   onlyTheOrderOfMagnitude = False # onlyTheOrderOfMagnitude = True   randomTest = True randomTest = False nMin = 0 nMax = 10**18-1 nbOfTest = 5   test = [ 4, 400, 5000, 600000, 107, 80, 1080, 91, 71, 184, 1840, 18400, 181000, 400567, "1 200 000 567", "123 456 789", "123 456 789 012 345", 10**6, 134*10**6, 10**9,   #### mille millards "1000" + '0'*9, #### un million de millards "1" + '0'*6 + '0'*9, #### deux millions de millards "2" + '0'*6 + '0'*9, #### sept milliards de milliards de milliards "7" + '0'*(9*4), ]   if randomTest: import random nMax += 1 test = [random.randint(nMin, nMax) for x in range(nbOfTest)]   for oneNumber in test: print(str(oneNumber))   oneNumber = str(oneNumber).strip().replace(' ', '')   if onlyTheOrderOfMagnitude: oneNumber = orderMagnitude(number = oneNumber) print('\torder of magnitude : ' + str(oneNumber))   elif mytenPowerPrecision: oneNumber = floor(number = oneNumber, tenPowerPrecision = mytenPowerPrecision) print('\tfloor : ' + str(oneNumber))   print('\t' + printer( number = oneNumber, convention = myConvention ))

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wordsToInteger_fr.py

Code:

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#s!/usr/bin/env python   # This code only works with Python 3.   # CODE TO CLEAN AND SIMPLIFY !!!!!!   import re # For cleaning the spaces between digits. import difflib # For finding closest number names to a mistaped word.   import config_fr   # We don't keep 70, 80 and 90 because we are going to transform "soixante dix" to "70"... # We need to do that so have to be more flexible. SPECIAL_NAMES_NUMBERS = dict([ (val, key) for (key, val) in config_fr.SPECIAL_NUMBERS_NAMES.items() \ if key not in ['80', '90', '70'] ])   NAMES_TEN_POWERS = {} THE_BIGGEST_NUMBER = {} THE_TEN_POWERS = {}   for oneConvention in config_fr.TEN_POWERS_NAMES: NAMES_TEN_POWERS[oneConvention] = {} THE_TEN_POWERS[oneConvention] = ['100']   for (key, val) in config_fr.TEN_POWERS_NAMES[oneConvention].items(): NAMES_TEN_POWERS[oneConvention][val] = '1' + '0'*key THE_TEN_POWERS[oneConvention].append('1' + '0'*key)   # We have to sort the powers from the smaller to the bigger. THE_TEN_POWERS[oneConvention] = sorted( THE_TEN_POWERS[oneConvention], key=(lambda x: len(x)) )   THE_BIGGEST_NUMBER[oneConvention] = THE_TEN_POWERS[oneConvention][-1]   # We must import integerToWords_fr after config_fr because # SPECIAL_NUMBERS_NAMES is changed in integerToWords_fr. import integerToWords_fr # We need this to indicate errors...     ############### # THE FUNCTIONS   def wellFormattedTextWithDigits(text): # Source : http://www.developpez.net/forums/d867083/autres-langages/python-zope/general-python/nettoyer-espaces-regex return list( map(lambda x: x.replace(' ',''),re.findall('(\d[\d ]*|[^ \d-]+)',text)) )     def printer(text, allowMistype = False, allowNumber = False, convention = "everyday"): """     If sympaMode = True, then we try to find the closest word possible     when mistyping are done. For example 'katr" will be seen as "quatre".       **NOTE :** this functionnality is the work of the module 'difflib'.     """ if convention not in NAMES_TEN_POWERS: raise ValueError('convention = "' + str(convention) + '" is unknown.')   text = text.lower()   if allowNumber: # "3456 7890" must become "34567890". theWords = wellFormattedTextWithDigits(text) else: theWords = text.replace('-', ' ').split()   # We verify that the names used are allowed. names_ten_powers = NAMES_TEN_POWERS[convention]   for index, oneWord in enumerate(theWords): # "cents", "millions",... must be seen as "cent", "million",... # # WARNING ! "trois" must stay "trois" if oneWord[-1] == 's' and oneWord != 'trois': oneWord = oneWord[:-1] weCleanAnS = True # Only for error detected. else: weCleanAnS = False   # WARNING ! "trente et un milliards de milliards" # "deux millions de milliards" if oneWord in ['et', 'de']: pass # We convert each word to digit numbers. elif oneWord in SPECIAL_NAMES_NUMBERS: theWords[index] = SPECIAL_NAMES_NUMBERS[oneWord] elif oneWord in names_ten_powers: theWords[index] = names_ten_powers[oneWord] else: integerCleaned = integerToWords_fr.cleanInteger(oneWord)   if integerCleaned: if allowNumber: theWords[index] = integerCleaned else: raise ValueError( 'The integer "{0}" must be written using words !.'.format(str(oneWord)) )   elif allowMistype: # 'difflib' does all the job ! nearestWords = difflib.get_close_matches(oneWord, SPECIAL_NAMES_NUMBERS)   if len(nearestWords) == 1: theWords[index] = SPECIAL_NAMES_NUMBERS[nearestWords[0]] else: if weCleanAnS: oneWord += 's' raise ValueError( '"{0}" does not look closer to a name numbers with the "{1}" convention.'.format(str(oneWord), convention) )   else: if weCleanAnS: oneWord += 's' raise ValueError( '"{0}" is not used to name numbers with the "{1}" convention.'.format(str(oneWord), convention) )   # For the moment we have ['4', '20'] instead of '80'. # We have a similar problem with '90'. # # We have to treat that... i = 0 lookFor = '4' in theWords   while lookFor: i = theWords.index('4', i)   if i < len(theWords) - 1: if theWords[i + 1] == '20': theWords.pop(i + 1) if i == len(theWords) - 1: theWords[i] = '80' elif theWords[i + 1] == '10': theWords.pop(i + 1) theWords[i] = '90' else: theWords[i] = '80'   try: i = theWords.index('4', i + 1) lookFor = True except: lookFor = False   # We look for a good use of the word "et". while('et' in theWords): i = theWords.index('et')   if i == 0: raise ValueError('"et" can not be at the beggining of the name of the number.')   if i == len(theWords) - 1: raise ValueError('"et" can not be at the end of the name of the number.')   beforeET = theWords[i-1] afterET = theWords[i+1]   oneError = False   if afterET == '1': if not beforeET in ['20', '30', '40', '50', '60', '80']: oneError = True else: theWords[i-1] = theWords[i-1][0] + '1'   elif afterET == '11': if not beforeET == '60': oneError = True else: theWords[i-1] = str( int(theWords[i-1][0]) + 1 ) + '1'   else: oneError = True   if oneError: beforeET = integerToWords_fr.printer( number = beforeET, convention = convention ) afterET = integerToWords_fr.printer( number = afterET, convention = convention ) raise ValueError( '"{0}" can not be used to name numbers with the "{1}" convention.'.format(str(beforeET +'-et-' + afterET), convention) )   theWords.pop(i) theWords.pop(i)   # We look for a good use of the word "de". the_biggest_number = THE_BIGGEST_NUMBER[convention] the_ten_powers = THE_TEN_POWERS[convention]   while('de' in theWords): i = theWords.index('de')   if i == 0: raise ValueError('"de" can not be at the beggining of the name of the number.')   if i == len(theWords) - 1: raise ValueError('"de" can not be at the end of the name of the number.')   beforeDE = theWords[i - 1] afterDE = theWords[i + 1]   if (beforeDE not in the_ten_powers and beforeDE != '*') or afterDE != the_biggest_number: beforeDE = integerToWords_fr.printer( number = beforeDE, convention = convention ) afterDE = integerToWords_fr.printer( number = afterDE, convention = convention ) raise ValueError( '"{0}" can not be used to name numbers with the "{1}" convention.'.format(str(beforeDE +' de ' + afterDE.replace('un ', '')), convention) )   # '*' means '.... * 10^max * ...' if beforeDE == the_biggest_number: theWords[i] = '*' theWords.pop(i + 1) else: theWords.pop(i)   # We build the two digits numbers like "30-5 == > 35", "80-15 == > 95"... i = 0 while i < len(theWords): addOne = True   # ERRORS TO FIND : "4-30" or "5-6" !!!! if len(theWords[i]) <= 2 and i > 0 and theWords[i] != '*': if len(theWords[i - 1]) == 1 and theWords[i-1] != '*': theWords[i - 1] = integerToWords_fr.printer( number = theWords[i - 1], convention = convention ) theWords[i] = integerToWords_fr.printer( number = theWords[i], convention = convention ) raise ValueError( '"{0}" can not be used to name numbers with the "{1}" convention.'.format(str(theWords[i - 1] + ' ' + theWords[i]), convention) )   if len(theWords[i]) == 2: if i < len(theWords) - 1: if theWords[i + 1] != '*': # "30-5 == > 35" if len(theWords[i + 1]) == 1: if theWords[i][1] != '0': theWords[i] = integerToWords_fr.printer( number = theWords[i], convention = convention ) theWords[i + 1] = integerToWords_fr.printer( number = theWords[i+1], convention = convention ) raise ValueError( '"{0}" can not be used to name numbers with the "{1}" convention.'.format(str(theWords[i] + ' ' + theWords[i + 1]), convention) )   theWords[i] = theWords[i][0] + theWords[i + 1] theWords.pop(i + 1) addOne = False # "80-15 == > 95" elif len(theWords[i + 1]) == 2: if theWords[i] not in ['60', '80'] or theWords[i + 1][0] != '1': theWords[i] = integerToWords_fr.printer( number = theWords[i], convention = convention ) theWords[i + 1] = integerToWords_fr.printer( number = theWords[i+1], convention = convention ) raise ValueError( '"{0}" can not be used to name numbers with the "{1}" convention.'.format(str(theWords[i] + ' ' + theWords[i + 1]), convention) )   theWords[i] = str(int(theWords[i][0]) + 1) + theWords[i + 1][1] theWords.pop(i + 1) addOne = False   # ERRORS TO FIND : "30-100" !!!! elif theWords[i + 1] == '100': theWords[i] = integerToWords_fr.printer( number = theWords[i], convention = convention ) theWords[i + 1] = integerToWords_fr.printer( number = theWords[i+1], convention = convention ) raise ValueError( '"{0}" can not be used to name numbers with the "{1}" convention.'.format(str(theWords[i] + ' ' + theWords[i + 1]), convention) )   if addOne: i += 1   # We build the mutiples of 100, 1 000, 1 000 000 and 10^p for p > 6 . # # Suppose that we have theWords equal to # ['100', '34', '1000', '7', '100', '95', '1000000000', '*'] # The following procedures will produce first # ['134', '1000', '795', '1000000000', '*'] # and then # ['134 795', '1000000000', '*'] # We do the analyse from the smaller power of ten to the bigger one. # # The treatment of [..., '*'] will be done after. for currentTenPower in the_ten_powers: # The forbidden writing "trente cents" have been treated before. nbOfDigitsAllowed = len(currentTenPower) - 1   # SPECIAL RULES ===> ?00 "un cent" NO ! # ??? 000 "un mille" NO ! # ??? 000 000 "un million" YES ! # The same is true for the other "big" powers of ten. if currentTenPower in ['100', '1000']: oneAllowed = False else: oneAllowed = True   i = 0 while i < len(theWords): addOne = True   if theWords[i] == currentTenPower: if i < len(theWords) - 1: if theWords[i + 1][0] != '*': # "?00-45 == > ?45" if len(theWords[i + 1]) <= nbOfDigitsAllowed: theWords[i + 1] = '0'*(nbOfDigitsAllowed - len(theWords[i + 1])) + theWords[i + 1] theWords[i] = theWords[i][0] + theWords[i + 1][-nbOfDigitsAllowed:] theWords.pop(i + 1) addOne = False   # "5-1?? == > 5??" if i > 0: if theWords[i - 1][0] != '*': if len(theWords[i - 1]) <= nbOfDigitsAllowed: theWords[i - 1] = theWords[i - 1] + theWords[i][1:] theWords.pop(i) addOne = False   if addOne: i += 1   # A single word ? if len(theWords) == 1: return theWords[0]   # ERRORS TO FIND : "3456-1000" , "3456-6789" ,... for i in range(len(theWords)-1): if theWords[i + 1] != '*' \ and theWords[i + 1] != '*' \ and len(theWords[i]) >= len(theWords[i + 1]): theWords[i] = integerToWords_fr.printer( number = theWords[i], convention = convention ) theWords[i + 1] = integerToWords_fr.printer( number = theWords[i+1], convention = convention ) raise ValueError( '"{0}" can not be used to name numbers with the "{1}" convention.'.format(str(theWords[i] + ' ' + theWords[i + 1]), convention) )   # To finish we have to treat things like "... milliars de milliards..." # # '*' means '.... * 10^max * ...' # For example, we have : # "un milliard de milliards" # ---> ['1000000000', '*'] # "deux milliards de milliards deux cents" # ---> ['2000000000', '*', '200'] answer = '' maxiNbOfZeros = len(the_biggest_number) - 1 i = len(theWords) - 1   while i >= 0: if theWords[i] == '*': nbOfDE = 2 i -= 1   while theWords[i] == '*': nbOfDE += 1 i -= 1 # "deux cent trois milliards de milliards quatre-vingt" # ---> ['203000000000', '*', '80'] theWords[i] = theWords[i][:-maxiNbOfZeros] # We have then : ['203', '000000080']   if len(answer) > maxiNbOfZeros*nbOfDE: answer = integerToWords_fr.printer( number = answer, convention = convention ) the_biggest_number = integerToWords_fr.printer( number = the_biggest_number, convention = convention ).replace('un', '') raise ValueError( 'You misuse "... {0}" before "{1}".'.format('de' + the_biggest_number, answer) ) answer = '0'*(maxiNbOfZeros*nbOfDE - len(answer)) + answer   else: answer = theWords[i] + answer i -= 1   return answer     ############### ############### ## ## ## FOR TESTS ## ## ## ############### ###############   if __name__ == '__main__': myConvention ="everyday" # myConvention ="rowlett" # myConvention ="chuquet"   weAllowMistype = False weAllowMistype = True   weAllowNumber = False weAllowNumber = True   randomTest = True randomTest = False nMin = 0 nMax = 10**50 nbOfTest = 100   test = [ 'quatre six', 'quatre cinquante', '3456 mille', 'mille 7890', '3456 sept mille 890', '3456 7890', 'sans douse', 'troiiii cen douse', 'cinquante six', '30et un mille quatre cents quatre-vingt-onze millions soixante-et-onze', 'deux 100 mille trois', 'trois millions de milliards', 'trente-quatre mille sept cents quatre-vingt-quinze milliards de milliards deux cents milliards', 'sept milliards de milliards de milliards deux cents milliards', 'sept milliards de milliards de milliards ' ]   if randomTest: import random nMax += 1   for x in range(nbOfTest): print()   oneNumber = str( random.randint(nMin, nMax) ) print( 'Test No {0} : {1}'.format(str(x+1), oneNumber) )   words = integerToWords_fr.printer( number = oneNumber, convention = myConvention ) print(words) print()   if oneNumber != printer( text = words, allowMistype = weAllowMistype, allowNumber = weAllowNumber, convention = myConvention ): print( printer( text = words, allowMistype = weAllowMistype, allowNumber = weAllowNumber, convention = myConvention ) ) raise Exception('PB WITH : ' + '\n\t' + oneNumber + '\n\t' + printer(words) + '\n\t' + words) print() print('END OF THE RANDOM TESTS.')   else: for oneNumber in test: print() print(str(oneNumber)) try: print('\t' + printer( text = oneNumber, allowMistype = weAllowMistype, allowNumber = weAllowNumber, convention = myConvention )) except: print('\tUNCORRECT NUMBER !')

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