Question

有没有办法将数字，字符串中的ether或int转换为单词而不会丢失前导零？我正在尝试将日期和时间以及电话号码转换为单词，但第二次我将字符串值转换为int我失去了我的前导零。

这里是我的代码，用于对数字进行数字处理，只要没有前导零，它就会很好用。这是我的问题的一个例子...... 假设我正在约会08-02-2004我不想输出零八零二......等等但是对于我来说，在目前状态下，我必须对方法做一些...除非我遗漏了一些东西。

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 "]

def numToWords(self, num):
    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 = int(numStr[i])
            t = int(numStr[i+1])
            u = 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])
    return ' '.join([w for w in words])

对此的任何帮助或建议将不胜感激。

Answer 1

确保首先提供字符串，并在必要时仅将其评估为int。 E.g：

def numToWords(self, numStr):
    words = []
    if int(numStr) == 0:
        words.append("zero")
    else:
        # no longer needed, it's already a string
        # numStr = "%d" % num
        numStrLen = len(numStr)
        groups = (numStrLen + 2) /
        ...

Answer 2

使用％d将int格式化为字符串时，会删除任何前导零。要保留它们，您需要指定最小位数，如下所示：

numStr = "%03d" % num

这会将前导零附加到任何小于3位的数字（在这种情况下，最小位数为3）。但是，在你肆无忌惮地使用前导零之前，首先需要确定你想要看多少位数。

Answer 3

使用递归方法的解决方案怎么样？

def numToWords(i):
    if i < 20:
        result = 'zero,one,two,three,four,five,six,\
                  seven,eight,nine,ten,eleven,twelve,\
                  thirteen,fourteen,fifteen,sixteen,\
                  seventeen,eighteen,nineteen'.split(',')[i]
    elif i < 100:
        result = ',,twenty,thirty,forty,fifty,sixty,seventy,\
                    eighty,ninety'.split(',')[i//10]
        if i % 10:
            result += ' ' + numToWords(i % 10)
    elif i < 1000:
        result = checkio(i // 100) + ' hundred'
        if i % 100:
            result += ' ' + numToWords(i % 100)
    return result

如何在保持前导零的同时将数字转换为单词？

3 个答案: