convertir un nombre en Lettre en python

**hirondellle** · 09/03/2012, 12h20

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

**rambc** · 09/03/2012, 12h45

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 :

Sélectionner tout - Visualiser dans une fenêtre à part

1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
46
47
48
49
50
51
52
53
54
55
56
57
58
59
60
61
62
63
64
65
66
67
68
69
70
71
72
73
74
75
76
77
78
79
80
81
82
83
84
85
86
87
88
89
90
91
# -*- 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"

integerToWords_fr.py

Code :

Sélectionner tout - Visualiser dans une fenêtre à part

1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
46
47
48
49
50
51
52
53
54
55
56
57
58
59
60
61
62
63
64
65
66
67
68
69
70
71
72
73
74
75
76
77
78
79
80
81
82
83
84
85
86
87
88
89
90
91
92
93
94
95
96
97
98
99
100
101
102
103
104
105
106
107
108
109
110
111
112
113
114
115
116
117
118
119
120
121
122
123
124
125
126
127
128
129
130
131
132
133
134
135
136
137
138
139
140
141
142
143
144
145
146
147
148
149
150
151
152
153
154
155
156
157
158
159
160
161
162
163
164
165
166
167
168
169
170
171
172
173
174
175
176
177
178
179
180
181
182
183
184
185
186
187
188
189
190
191
192
193
194
195
196
197
198
199
200
201
202
203
204
205
206
207
208
209
210
211
212
213
214
215
216
217
218
219
220
221
222
223
224
225
226
227
228
229
230
231
232
233
234
235
236
237
238
239
240
241
242
243
244
245
246
247
248
249
250
251
252
253
254
255
256
257
258
259
260
261
262
263
264
265
266
267
268
269
270
271
272
273
274
275
276
277
278
279
280
281
282
283
284
285
286
287
288
289
290
291
292
293
294
295
296
297
298
299
300
301
302
303
304
305
306
307
308
309
310
311
312
313
314
315
316
317
318
319
320
321
322
323
324
325
326
327
328
329
330
331
332
333
334
335
336
337
338
339
340
341
342
343
344
345
346
347
348
349
350
351
352
353
354
355
356
357
358
359
360
361
362
363
364
365
366
367
368
369
370
371
372
373
374
375
376
377
378
379
380
381
382
383
384
385
386
387
388
389
390
391
392
393
394
395
396
397
398
399
400
401
402
403
404
405
406
407
# -*- 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 ))

wordsToInteger_fr.py

Code :

Sélectionner tout - Visualiser dans une fenêtre à part

1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
46
47
48
49
50
51
52
53
54
55
56
57
58
59
60
61
62
63
64
65
66
67
68
69
70
71
72
73
74
75
76
77
78
79
80
81
82
83
84
85
86
87
88
89
90
91
92
93
94
95
96
97
98
99
100
101
102
103
104
105
106
107
108
109
110
111
112
113
114
115
116
117
118
119
120
121
122
123
124
125
126
127
128
129
130
131
132
133
134
135
136
137
138
139
140
141
142
143
144
145
146
147
148
149
150
151
152
153
154
155
156
157
158
159
160
161
162
163
164
165
166
167
168
169
170
171
172
173
174
175
176
177
178
179
180
181
182
183
184
185
186
187
188
189
190
191
192
193
194
195
196
197
198
199
200
201
202
203
204
205
206
207
208
209
210
211
212
213
214
215
216
217
218
219
220
221
222
223
224
225
226
227
228
229
230
231
232
233
234
235
236
237
238
239
240
241
242
243
244
245
246
247
248
249
250
251
252
253
254
255
256
257
258
259
260
261
262
263
264
265
266
267
268
269
270
271
272
273
274
275
276
277
278
279
280
281
282
283
284
285
286
287
288
289
290
291
292
293
294
295
296
297
298
299
300
301
302
303
304
305
306
307
308
309
310
311
312
313
314
315
316
317
318
319
320
321
322
323
324
325
326
327
328
329
330
331
332
333
334
335
336
337
338
339
340
341
342
343
344
345
346
347
348
349
350
351
352
353
354
355
356
357
358
359
360
361
362
363
364
365
366
367
368
369
370
371
372
373
374
375
376
377
378
379
380
381
382
383
384
385
386
387
388
389
390
391
392
393
394
395
396
397
398
399
400
401
402
403
404
405
406
407
408
409
410
411
412
413
414
415
416
417
418
419
420
421
422
423
424
425
426
427
428
429
430
431
432
433
434
435
436
437
438
439
440
441
442
443
444
445
446
447
#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 !')

convertir un nombre en Lettre en python

Python

Vue hybride

Discussions similaires

Partager

Partager