我试图告诉Python将整数转换为单词.
示例:(使用墙上的99瓶啤酒)
我用这段代码编写程序:
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
但我无法弄清楚如何编写程序,以便显示单词(即九十九,九十八等)而不是数字.
我一直令人头我的头在Python书我有,我明白,也许我只是不明白for/ if/ elif/ else循环,但是我只是纺纱我的车轮.
谁能提供任何见解？我不是在寻找一个直接的答案,虽然这可能有助于我看到我的问题,只要指出我正确方向的任何事情都会很棒.
Lee*_*Lee 61
inflect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
ran*_*dra 35
使用可在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'
Dev*_*per 20
我们改编了一个现有的解决方案(ref),用于将数字转换为单词,如下所示:
def numToWords(num,join=True):
'''words = {} convert an integer number into words'''
units = ['','one','two','three','four','five','six','seven','eight','nine']
teens = ['','eleven','twelve','thirteen','fourteen','fifteen','sixteen', \
'seventeen','eighteen','nineteen']
tens = ['','ten','twenty','thirty','forty','fifty','sixty','seventy', \
'eighty','ninety']
thousands = ['','thousand','million','billion','trillion','quadrillion', \
'quintillion','sextillion','septillion','octillion', \
'nonillion','decillion','undecillion','duodecillion', \
'tredecillion','quattuordecillion','sexdecillion', \
'septendecillion','octodecillion','novemdecillion', \
'vigintillion']
words = []
if num==0: words.append('zero')
else:
numStr = '%d'%num
numStrLen = len(numStr)
groups = (numStrLen+2)/3
numStr = numStr.zfill(groups*3)
for i in range(0,groups*3,3):
h,t,u = int(numStr[i]),int(numStr[i+1]),int(numStr[i+2])
g = groups-(i/3+1)
if h>=1:
words.append(units[h])
words.append('hundred')
if t>1:
words.append(tens[t])
if u>=1: words.append(units[u])
elif t==1:
if u>=1: words.append(teens[u])
else: words.append(tens[t])
else:
if u>=1: words.append(units[u])
if (g>=1) and ((h+t+u)>0): words.append(thousands[g]+',')
if join: return ' '.join(words)
return words
#example usages:
print numToWords(0)
print numToWords(11)
print numToWords(110)
print numToWords(1001000025)
print numToWords(123456789012)
结果:
zero
eleven
one hundred ten
one billion, one million, twenty five
one hundred twenty three billion, four hundred fifty six million, seven hundred
eighty nine thousand, twelve
请注意,它适用于整数.然而,将浮点数除以两个整数部分是微不足道的.
SiL*_*oNG 17
这是在Python 3中实现它的一种方法:
"""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
kin*_*all 10
嗯,简单的方法就是列出你感兴趣的所有数字:
numbers = ["zero", "one", "two", "three", "four", "five", ...
"ninety-eight", "ninety-nine"]
(...表示你在哪里键入其他数字的文本表示.不,Python不会为你神奇地填充它,你必须键入所有这些才能使用该技术.)
然后打印数字,只需打印numbers[i].十分简单.
当然,该列表是很多打字,所以你可能想知道一个简单的方法来生成它.不幸的是,英语有很多不规则因此你必须手动输入前20(0-19),但你可以使用规律来生成其余的99.(你也可以生成一些青少年,但只有其中一些,所以输入它们似乎最容易.)
numbers = "zero one two three four five six seven eight nine".split()
numbers.extend("ten eleven twelve thirteen fourteen fifteen sixteen".split())
numbers.extend("seventeen eighteen nineteen".split())
numbers.extend(tens if ones == "zero" else (tens + "-" + ones)
for tens in "twenty thirty forty fifty sixty seventy eighty ninety".split()
for ones in numbers[0:10])
print numbers[42] # "forty-two"
另一种方法是编写一个函数,每次将正确的字符串放在一起.同样,你必须对前20个数字进行硬编码,但之后你可以根据需要从头开始轻松生成它们.这将使用少的内存(一个很多少,一旦你开始有较大的数字工作).
您可以使用python-n2w库，只需这样做
pip install n2w
然后简单地
print(n2w.convert(your-number-here))
我的解决方案来了:)它是各种早期的解决方案，但是是我自己开发的 - 也许有人比其他提议更喜欢它。
TENS = {30: 'thirty', 40: 'forty', 50: 'fifty', 60: 'sixty', 70: 'seventy', 80: 'eighty', 90: 'ninety'}
ZERO_TO_TWENTY = (
'zero', 'one', 'two', 'three', 'four', 'five', 'six', 'seven', 'eight', 'nine', 'ten',
'eleven', 'twelve', 'thirteen', 'fourteen', 'fifteen', 'sixteen', 'seventeen', 'eighteen', 'nineteen', 'twenty'
)
def number_to_english(n):
if any(not x.isdigit() for x in str(n)):
return ''
if n <= 20:
return ZERO_TO_TWENTY[n]
elif n < 100 and n % 10 == 0:
return TENS[n]
elif n < 100:
return number_to_english(n - (n % 10)) + ' ' + number_to_english(n % 10)
elif n < 1000 and n % 100 == 0:
return number_to_english(n / 100) + ' hundred'
elif n < 1000:
return number_to_english(n / 100) + ' hundred ' + number_to_english(n % 100)
elif n < 1000000:
return number_to_english(n / 1000) + ' thousand ' + number_to_english(n % 1000)
return ''
它是递归解决方案，可以轻松扩展为更大的数字
您必须使用字典/数组。例如 ：
to_19= ['zero','one','two','three','four','five','six','seven','eight','nine'..'nineteen']
tens = ['twenty'...'ninety']
您可以通过执行以下操作来生成数字字符串：
if len(str(number)) == 2 and number > 20:
word_number = tens[str(number)[0]]+' '+units[str(number)[0]]
您必须检查最后一个数字是否不是零等等......经典的值检查。
它提醒了一个项目欧拉挑战（问题17）..你应该尝试找到一些关于它的解决方案
希望能帮助到你