Question

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

但是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。

我一直在I py py，，，，，，，，，，，，，maybe maybe maybe maybe maybe maybe maybe maybe maybe maybe maybe maybe maybe maybe maybe maybe maybe maybe maybe maybe maybe maybe maybe maybe maybe但我只是在转动轮子。

有人能提供任何见解吗？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，只要指出正确方向的任何事情都会很棒。

Answer 1

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

Answer 2

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'

Answer 3

我们调整了现有的优秀解决方案(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。

Answer 4

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

Answer 5

嗯，简单易行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，你必须输入所有这些才能使用该技术。）

然后打印该号码，只需打印numbers[i]即可。容易腻。

当然，这个列表是很多打字，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，但只有它们中的一些，因此最简单的方法就是输入它们。）

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，每次都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。这使用少一点内存（一旦你开始使用更大的数字， lot 就更少了。）

Answer 6

听起来您需要使用数组，其中num[1] = "one"，num[2] = "two"等等。然后你可以像现在一样循环遍历每个

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 7

以上是上面发布的几个代码示例的重构版本（主要是代码粘贴在＆＃34;开发人员＆＃34;。

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

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 9

这是我的解决方案:)它是早期的各种解决方案，但都是我自己开发的-也许有人会比其他主张更喜欢它。

 @Test
fun runColorMatchDetails():Unit {

    val timeout = 2000
    val addresses = InetAddress.getAllByName("www.google.com")
    for (address in addresses) {
        if (address.isReachable(timeout))
            System.out.printf("%s is reachable%n", address)
        else
            System.out.printf("%s could not be contacted%n", address)
    }


}

java.net.ConnectException: No route to host (connect failed)
at java.net.Inet6AddressImpl.isReachable0(Native Method)
at java.net.Inet6AddressImpl.isReachable(Inet6AddressImpl.java:77)
at java.net.InetAddress.isReachable(InetAddress.java:502)
at java.net.InetAddress.isReachable(InetAddress.java:461)

这是一种折价解决方案，可以轻松扩展为更大的数字

Answer 10

此代码：

>>>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 11

这也适用于前1000个数字（Python 3）。该问题的初步解决方案，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 12

你必须使用字典/数组。例如：

 to_19= ['zero','one','two','three','four','five','six','seven','eight','nine'..'nineteen']
 tens = ['twenty'...'ninety']

你可以通过这样做来生成一个数字的字符串，例如：

 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为零，依此类推。经典值检查。

它提醒项目欧拉挑战（问题17）..你应该试着找到一些关于它的解决方案

希望有所帮助

Answer 13

func stringToDate(_ str: String)->Date{
    let formatter = DateFormatter()
    formatter.dateFormat="yyyy-MM-dd hh:mm:ss Z"
    return formatter.date(from: str)!
}
func dateToString(_ str: Date)->String{
    var dateFormatter = DateFormatter()
    dateFormatter.timeStyle=DateFormatter.Style.short
    return dateFormatter.string(from: str)
}

Answer 14

这可以完成没有任何库的工作。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 15

将数字转换为单词：
这是一个使用字典将数字转换成单词的示例。

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 16

我会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

如何告诉Python将整数转换为单词

16 个答案: