Преобразовать числовую дату в текстовый формат даты в Python

Question

У меня есть текст, и в тексте у меня есть дата в формате 22/09/1992, я хочу преобразовать эту дату в двадцать два сентября ninteen nighty two в Python, как мне это сделать.

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

ожидаемый результат:

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]