我如何告诉 Python 将整数转换为单词

Question

我试图告诉 Python 将整数转换为单词。

例子：（用墙上的歌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

但我不知道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。

我一直在 python 书中绞尽脑汁，我明白也许我只是不理解for / if / elif / else循环，但我只是在转动我的轮子。

任何人都可以提供任何见解吗？ 我不是在寻找直接的答案，尽管这可能会帮助我看到我的问题，但只要能指出我正确的方向就会很棒。

Answer 1

inflect 包可以做到这一点。

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

$ pip install inflect

接着：

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

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

这是在 Python 3 中执行此操作的一种方法：

"""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

Answer 4

我们采用了一个现有的不错的解决方案(ref)来将数字转换为单词，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。 尽管如此，将浮点数分成两个整数部分是微不足道的。

Answer 5

好吧，最简单的方法是列出您感兴趣的所有数字：

numbers = ["zero", "one", "two", "three", "four", "five", ... 
           "ninety-eight", "ninety-nine"]

（ ... 表示您在何处输入其他数字的文本表示。不，Python 不会神奇地为您填写，您必须输入所有这些内容才能使用该技术。）

然后打印数字，只需打印numbers[i] 。 十分简单。

当然，该列表需要大量输入，因此您可能想知道一种简单的生成方法。 不幸的是，英语有很多不规则之处，因此您必须手动输入前二十个 (0-19)，但您可以使用规则生成其余多达 99 个。（您也可以生成一些 teens，但仅限其中一些，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，您必须对前二十个数字进行硬编码，但之后您可以根据需要轻松地从头开始生成它们。 这会使用更少的内存（一旦您开始使用更大的数字，就会少很多）。

Answer 6

您可以使用python-n2w库，只需执行

pip install n2w

然后简单地

print(n2w.convert(your-number-here))

Answer 7

这是上面发布的几个代码示例的重构版本（主要是“开发人员”粘贴的代码。

def int2words(num):
    """Given an int32 number, print it in English.

    Parameters
    ----------
    num : int

    Returns
    -------
    words : str
    """
    assert (0 <= 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'
    }
    h = [100, 'hundred', 'hundred and']
    k = [h[0] * 10, 'thousand', 'thousand,']
    m = [k[0] * 1000, 'million', 'million,']
    b = [m[0] * 1000, 'billion', 'billion,']
    t = [b[0] * 1000, 'trillion', 'trillion,']
    if num < 20:
        return d[num]
    if num < 100:
        div_, mod_ = divmod(num, 10)
        return d[num] if mod_ == 0 else d[div_ * 10] + '-' + d[mod_]
    else:
        if num < k[0]:
            divisor, word1, word2 = h
        elif num < m[0]:
            divisor, word1, word2 = k
        elif num < b[0]:
            divisor, word1, word2 = m
        elif num < t[0]:
            divisor, word1, word2 = b
        else:
            divisor, word1, word2 = t
        div_, mod_ = divmod(num, divisor)
        if mod_ == 0:
            return '{} {}'.format(int2words(div_), word1)
        else:
            return '{} {} {}'.format(int2words(div_), word2, int2words(mod_))

Answer 8

这是我的解决方案:) 这是各种早期的解决方案，但我自己开发的 - 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

Answer 9

听起来您需要使用一个数组，其中num[1] = "one" 、 num[2] = "two"等等。 然后你可以像你一样循环遍历每个

num = array(["one","two","three","four","five","six","seven","eight","nine","ten"])
for i in range(10,0,-1):
    print num[i], "Bottles of beer on the wall,"
    print num[i], "bottles of beer."
    print "Take one down and pass it around,"
    print num[i-1], "bottles of beer on the wall."
    print ""

Answer 10

这也适用于前 1000 个数字（Python 3）。 该问题的初学者解决方案，但仍然是一个解决方案......

first_nums = ["", "one", "two", "three", "four", "five", "six", "seven", "eight", "nine", "ten", "eleven", "twelve",
              "thirteen", "fourteen", "fifteen", "sixteen", "seventeen", "eighteen", "nineteen"]
tens = ["ten", "twenty", "thirty","forty","fifty", "sixty","seventy","eighty","ninety"]

for num in range(1,1001):
    if num < 20:
        print(first_nums[num])
    elif num < 100 and num % 10 == 0:
        print(tens[int(num/10 - 1)])
    elif num < 1000 and num % 100 == 0:
        print(first_nums[int(num/100)] + " hundred")
    elif num == 1000:
        print("one thousand")
    elif num < 100 and num % 10 != 0:
        print(tens[int(num//10 - 1)] + " " + first_nums[int(num%10)])
    elif num < 1000:
        if num%100 < 20:
            print(first_nums[int(num//100)] + " hundred and " + first_nums[num%100])
        else:
            print(first_nums[int(num//100)] + " hundred and " + tens[int(num%100//10 - 1)] + " " + first_nums[int(num%10)])

Answer 11

这在没有任何库的情况下完成了这项工作。 使用递归，它是印度风格。 ——拉维。

def spellNumber(no):
    # str(no) will result in  56.9 for 56.90 so we used the method which is given below.
    strNo = "%.2f" %no
    n = strNo.split(".")
    rs = numberToText(int(n[0])).strip()
    ps =""
    if(len(n)>=2):
        ps = numberToText(int(n[1])).strip()
        rs = "" + ps+ " paise"  if(rs.strip()=="") else  (rs + " and " + ps+ " paise").strip()
    return rs
print(spellNumber(0.67))
print(spellNumber(5858.099))
print(spellNumber(5083754857380.50))    

def numberToText(no):
    ones = " ,one,two,three,four,five,six,seven,eight,nine,ten,eleven,tweleve,thirteen,fourteen,fifteen,sixteen,seventeen,eighteen,nineteen,twenty".split(',')
    tens = "ten,twenty,thirty,fourty,fifty,sixty,seventy,eighty,ninety".split(',')
    text = ""
    if len(str(no))<=2:
        if(no<20):
            text = ones[no]
        else:
            text = tens[no//10-1] +" " + ones[(no %10)]
    elif len(str(no))==3:
        text = ones[no//100] +" hundred " + numberToText(no- ((no//100)* 100))
    elif len(str(no))<=5:
        text = numberToText(no//1000) +" thousand " + numberToText(no- ((no//1000)* 1000))
    elif len(str(no))<=7:
        text = numberToText(no//100000) +" lakh " + numberToText(no- ((no//100000)* 100000))
    else:
        text = numberToText(no//10000000) +" crores " + numberToText(no- ((no//10000000)* 10000000))
    return text

Answer 12

代码：

>>>def handel_upto_99(number):
predef={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:"fourty",50:"fifty",60:"sixty",70:"seventy",80:"eighty",90:"ninety",100:"hundred",100000:"lakh",10000000:"crore",1000000:"million",1000000000:"billion"}
if number in predef.keys():
    return predef[number]
else:
    return predef[(number/10)*10]+' '+predef[number%10]

---

>>>def return_bigdigit(number,devideby):
predef={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:"fourty",50:"fifty",60:"sixty",70:"seventy",80:"eighty",90:"ninety",100:"hundred",1000:"thousand",100000:"lakh",10000000:"crore",1000000:"million",1000000000:"billion"}
if devideby in predef.keys():
    return predef[number/devideby]+" "+predef[devideby]
else:
    devideby/=10
    return handel_upto_99(number/devideby)+" "+predef[devideby]

---

>>>def mainfunction(number):
dev={100:"hundred",1000:"thousand",100000:"lakh",10000000:"crore",1000000000:"billion"}
if number is 0:
    return "Zero"
if number<100:
    result=handel_upto_99(number)

else:
    result=""
    while number>=100:
        devideby=1
        length=len(str(number))
        for i in range(length-1):
            devideby*=10
        if number%devideby==0:
            if devideby in dev:
                return handel_upto_99(number/devideby)+" "+ dev[devideby]
            else:
                return handel_upto_99(number/(devideby/10))+" "+ dev[devideby/10]
        res=return_bigdigit(number,devideby)
        result=result+' '+res
        if devideby not in dev:
            number=number-((devideby/10)*(number/(devideby/10)))
        number=number-devideby*(number/devideby)

    if number <100:
        result = result + ' '+ handel_upto_99(number)
return result

---

结果：

>>>mainfunction(12345)
' twelve thousand three hundred fourty five'
>>>mainfunction(2000)
'two thousand'

Answer 13

#This valid till 4 digit number
numbers={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', 100:'hundred', 1000:'thousand'}

def my_fun(num):
    list = []
    num_len = len(str(num)) - 1
    while num_len > 0 and num > 0:
       while num_len > 0 and num > 0:
        if num in numbers and num < 1000:
            list.append(numbers[num])
            num_len = 0
        elif num < 100:
            list.extend([numbers[num - num%10], numbers[num%10]])
            num_len = 0
        else:
            quotent = num//10**num_len # 4567//1000= 4
            num = num % 10**num_len #4567%1000 =567
            if quotent != 0 :
                list.append(numbers[quotent])
                list.append(numbers[10**num_len])
            else:
                list.append(numbers[num])
            num_len -= 1
    return ' '.join(list)

Answer 14

将数字转换为单词：
这是一个使用字典将数字转换为单词的示例。

string = input("Enter a string: ")
my_dict = {'0': 'zero', '1': 'one', '2': 'two', '3': 'three', '4': 'four', '5': 'five', '6': 'six', '7': 'seven', '8': 'eight', '9': 'nine'}
for item in string:
  if item in my_dict.keys():
    string = string.replace(item, my_dict[item])
print(string)

Answer 15

我会简单地通过这样做来解决这个问题：

numberText = {
    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',
    100: 'hundred', 1000: 'thousand', 1000000: 'million'
}


def numberToEnglishText(n):
    if n == 0:
        return 'zero'
    if n < 0:
        return 'negative ' + numberToEnglishText(-n)

    result = ''

    for num in sorted(numberText.keys(), reverse=True):
        count = int(n/num)

        if (count < 1):
            continue

        if (num >= 100):
            result += numberToEnglishText(count) + ' '

        result += numberText[num]
        n -= count * num
        if (n > 0):
            result += ' '

    return result

Answer 16

我在 Python 3 中创建了一个快速版本的数字到单词转换器，它可以在 [0-99] 范围内工作，而无需使用任何库。 如果您不需要更高的值，此版本可能会很方便。

# converts number-to-word in range [0-99]
def number_to_word(num):
    sub_twenty = ('zero', 'one', 'two', 'three', 'four', 'five', 'six', 'seven', 'eight',
                  'nine', 'ten', 'eleven', 'twelve', 'thirteen', 'fourteen', 'fifteen',
                  'sixteen', 'seventeen', 'eighteen', 'nineteen')
    tens = {20: 'twenty', 30: 'thirty', 40: 'forty', 50: 'fifty', 
            60: 'sixty', 70: 'seventy', 80: 'eighty', 90: 'ninety'}

    # if it is less than 20 directly converts its value from the list
    if num < 20:
        return sub_twenty[num]

    # if its order of 10 gets from the dict
    elif num % 10 == 0:
        return tens[num]

    # if it has any value in units digit gets both
    else:
        quo = num // 10             # quotient
        quo = tens[quo * 10]        # scale and get value from dict

        rem = num % 10              # remainder
        rem = sub_twenty[rem]       # get its word format
        return quo + " " + rem

Answer 17

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，例如：

 if len(str(number)) == 2 and  number > 20:
       word_number = tens[str(number)[0]]+' '+units[str(number)[0]]

你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解决方案

希望能帮助到你