如何让Python将数字转换为单词

Question

我想让Python把整数转换成文字。

举个例子：（用歌曲《墙上的99瓶啤酒》来说明）

我用这段代码来写程序：

for i in range(99,0,-1):
    print i, "Bottles of beer on the wall,"
    print i, "bottles of beer."
    print "Take one down and pass it around,"
    print i-1, "bottles of beer on the wall."
    print

但是我不知道怎么写程序才能让它显示文字（比如“九十九”，“九十八”等），而不是数字。

我一直在翻阅我手上的Python书，虽然我明白可能是我还不太懂for/if/elif/else这些循环和条件语句，但我感觉自己在原地踏步。

有没有人能给我一些建议？我并不想要直接的答案，虽然那可能会让我更清楚我的问题，只要能给我一些指引就很好了。

Answer 1

这里有一种在Python 3中实现的方法：

"""Given an int32 number, print it in English."""
def int_to_en(num):
    d = { 0 : 'zero', 1 : 'one', 2 : 'two', 3 : 'three', 4 : 'four', 5 : 'five',
          6 : 'six', 7 : 'seven', 8 : 'eight', 9 : 'nine', 10 : 'ten',
          11 : 'eleven', 12 : 'twelve', 13 : 'thirteen', 14 : 'fourteen',
          15 : 'fifteen', 16 : 'sixteen', 17 : 'seventeen', 18 : 'eighteen',
          19 : 'nineteen', 20 : 'twenty',
          30 : 'thirty', 40 : 'forty', 50 : 'fifty', 60 : 'sixty',
          70 : 'seventy', 80 : 'eighty', 90 : 'ninety' }
    k = 1000
    m = k * 1000
    b = m * 1000
    t = b * 1000

    assert(0 <= num)

    if (num < 20):
        return d[num]

    if (num < 100):
        if num % 10 == 0: return d[num]
        else: return d[num // 10 * 10] + '-' + d[num % 10]

    if (num < k):
        if num % 100 == 0: return d[num // 100] + ' hundred'
        else: return d[num // 100] + ' hundred and ' + int_to_en(num % 100)

    if (num < m):
        if num % k == 0: return int_to_en(num // k) + ' thousand'
        else: return int_to_en(num // k) + ' thousand, ' + int_to_en(num % k)

    if (num < b):
        if (num % m) == 0: return int_to_en(num // m) + ' million'
        else: return int_to_en(num // m) + ' million, ' + int_to_en(num % m)

    if (num < t):
        if (num % b) == 0: return int_to_en(num // b) + ' billion'
        else: return int_to_en(num // b) + ' billion, ' + int_to_en(num % b)

    if (num % t == 0): return int_to_en(num // t) + ' trillion'
    else: return int_to_en(num // t) + ' trillion, ' + int_to_en(num % t)

    raise AssertionError('num is too large: %s' % str(num))

结果是：

0 zero
3 three
10 ten
11 eleven
19 nineteen
20 twenty
23 twenty-three
34 thirty-four
56 fifty-six
80 eighty
97 ninety-seven
99 ninety-nine
100 one hundred
101 one hundred and one
110 one hundred and ten
117 one hundred and seventeen
120 one hundred and twenty
123 one hundred and twenty-three
172 one hundred and seventy-two
199 one hundred and ninety-nine
200 two hundred
201 two hundred and one
211 two hundred and eleven
223 two hundred and twenty-three
376 three hundred and seventy-six
767 seven hundred and sixty-seven
982 nine hundred and eighty-two
999 nine hundred and ninety-nine
1000 one thousand
1001 one thousand, one
1017 one thousand, seventeen
1023 one thousand, twenty-three
1088 one thousand, eighty-eight
1100 one thousand, one hundred
1109 one thousand, one hundred and nine
1139 one thousand, one hundred and thirty-nine
1239 one thousand, two hundred and thirty-nine
1433 one thousand, four hundred and thirty-three
2000 two thousand
2010 two thousand, ten
7891 seven thousand, eight hundred and ninety-one
89321 eighty-nine thousand, three hundred and twenty-one
999999 nine hundred and ninety-nine thousand, nine hundred and ninety-nine
1000000 one million
2000000 two million
2000000000 two billion

Answer 2

使用可以在sourceforge找到的 pynum2word 模块。

>>> import num2word
>>> num2word.to_card(15)
'fifteen'
>>> num2word.to_card(55)
'fifty-five'
>>> num2word.to_card(1555)
'one thousand, five hundred and fifty-five'

Answer 3

inflect这个包可以做到这一点。

https://pypi.python.org/pypi/inflect

$ pip install inflect

然后：

>>>import inflect
>>>p = inflect.engine()
>>>p.number_to_words(99)
ninety-nine