Question

我有一个文本，并且在文本中我有一个日期为22/09/1992的日期，我想将此日期转换为python中的22九月19夜二。我该怎么做。

Text- My name is edwin .I have registered the competition on 22/09/2018

预期输出为：

My name is edwin .I have registered the competition on twenty two september two thousand eighteen

Answer 1

ones = ["", "one ","two ","three ","four ", "five ", "six ","seven ","eight ","nine ","ten ","eleven ","twelve ", "thirteen ", "fourteen ", "fifteen ","sixteen ","seventeen ", "eighteen ","nineteen "]

twenties = ["","","twenty ","thirty ","forty ", "fifty ","sixty ","seventy ","eighty ","ninety "]

thousands = ["","thousand ","million ", "billion ", "trillion ", "quadrillion ", "quintillion ", "sextillion ", "septillion ","octillion ", "nonillion ", "decillion ", "undecillion ", "duodecillion ", "tredecillion ", "quattuordecillion ", "quindecillion", "sexdecillion ", "septendecillion ", "octodecillion ", "novemdecillion ", "vigintillion "]


def num999(n):
    c = n % 10 # singles digit
    b = ((n % 100) - c) / 10 # tens digit
    a = ((n % 1000) - (b * 10) - c) / 100 # hundreds digit
    t = ""
    h = ""
    if a != 0 and b == 0 and c == 0:
        t = ones[a] + "hundred "
    elif a != 0:
        t = ones[a] + "hundred and "
    if b <= 1:
        h = ones[n%100]
    elif b > 1:
        h = twenties[b] + ones[c]
    st = t + h
    return st

def num2word(num):
    if num == 0: return 'zero'
    i = 3
    n = str(num)
    word = ""
    k = 0
    while(i == 3):
        nw = n[-i:]
        n = n[:-i]
        if int(nw) == 0:
            word = num999(int(nw)) + thousands[int(nw)] + word
        else:
            word = num999(int(nw)) + thousands[k] + word
        if n == '':
            i = i+1
        k += 1
    return word[:-1]

在Python中将数字日期转换为文本日期格式

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