Question

How to convert amount to word in Indian?

我正在使用N2words 图书馆,但该图书馆在提出“ la”和“ores”。

例如:

num2words(903614.55, lang= en-IN ) Its printing nine hundred and three thousand, six hundred and fourteen point five five

但是,印度实际数额的列报方式应为:nine lakhs 3 000, 640 14 and five paisa。

随后,我尝试了以下法典:

def num2words(num):
    under_20 = [ Zero , One , Two , Three , Four , Five , Six , Seven , Eight , Nine , Ten , Eleven , Twelve , Thirteen , Fourteen , Fifteen , Sixteen , Seventeen , Eighteen , Nineteen ]
    tens = [ Twenty , Thirty , Forty , Fifty , Sixty , Seventy , Eighty , Ninety ]
    above_100 = {100:  Hundred ,1000: Thousand , 100000: Lakhs , 10000000: Crores }

    if num < 20:
         return under_20[num]

    if num < 100:
        return tens[(int)(num/10)-2] + (   if num%10==0 else     + under_20[num%10])

    # find the appropriate pivot -  Million  in 3,603,550, or  Thousand  in 603,550
    pivot = max([key for key in above_100.keys() if key <= num])

    return num2words((int)(num/pivot)) +     + above_100[pivot] + (   if num%pivot==0 else     + num2words(num%pivot))

但现在出现错误

类型:清单指标必须是分类账或斜体,而不是小数。 Decimal

我的问题是人数少,Integer正在做罚款。

Answer 1

I guess you could just cast it on the first line if you re not handling decimals. You can also use // to do integer division.

import decimal    

def num2words(num):
    num = decimal.Decimal(num)
    decimal_part = num - int(num)
    num = int(num)

    if decimal_part:
        return num2words(num) + " point " + (" ".join(num2words(i) for i in str(decimal_part)[2:]))

    under_20 = [ Zero ,  One ,  Two ,  Three ,  Four ,  Five ,  Six ,  Seven ,  Eight ,  Nine ,  Ten ,  Eleven ,  Twelve ,  Thirteen ,  Fourteen ,  Fifteen ,  Sixteen ,  Seventeen ,  Eighteen ,  Nineteen ]
    tens = [ Twenty ,  Thirty ,  Forty ,  Fifty ,  Sixty ,  Seventy ,  Eighty ,  Ninety ]
    above_100 = {100:  Hundred , 1000:  Thousand , 100000:  Lakhs , 10000000:  Crores }

    if num < 20:
        return under_20[num]

    if num < 100:
        return tens[num // 10 - 2] + (   if num % 10 == 0 else     + under_20[num % 10])

    # find the appropriate pivot -  Million  in 3,603,550, or  Thousand  in 603,550
    pivot = max([key for key in above_100.keys() if key <= num])

    return num2words(num // pivot) +     + above_100[pivot] + (   if num % pivot==0 else     + num2words(num % pivot))


print(num2words(decimal.Decimal("650958.32")))
# Six Lakhs Fifty Thousand Nine Hundred Fifty Eight point Three Two

Answer 2

解决办法简单,错误在后瞻中显而易见。

仅以“IN”代替“N2words”。没有必要制定任何习惯法。我也被烧毁。

num2words.num2words(123456, lang= en_IN )

回返

one lakh, twenty-three thousand, four hundred and fifty-six

Answer 3

您可以分立精细和分数部分,可以把南2字数的两倍换成单数,而其他则分部分。

import math
def num2words(num):
    under_20 = [ Zero , One , Two , Three , Four , Five , Six , Seven , Eight , Nine , Ten , Eleven , Twelve , Thirteen , Fourteen , Fifteen , Sixteen , Seventeen , Eighteen , Nineteen ]
    tens = [ Twenty , Thirty , Forty , Fifty , Sixty , Seventy , Eighty , Ninety ]
    above_100 = {100:  Hundred ,1000: Thousand , 100000: Lakhs , 10000000: Crores }

    if num < 20:
         return under_20[(int)(num)]

    if num < 100:
        return tens[(int)(num/10)-2] + (   if num%10==0 else     + under_20[(int)(num%10)])

    # find the appropriate pivot -  Million  in 3,603,550, or  Thousand  in 603,550
    pivot = max([key for key in above_100.keys() if key <= num])

    return num2words((int)(num/pivot)) +     + above_100[pivot] + (   if num%pivot==0 else     + num2words(num%pivot))

num="5.12"
print(num2words(int(num.split(".")[0])))
print(num2words(int(num.split(".")[1])))

https://ide.geeksforgeeks.org/J7zsZyIT6m

Answer 4

如果你不实写明的货币数额,就会尝试:

num2words(amount, lang= en_IN , to= currency , currency= USD )

它使用N2字(人数的精华译员)。你只是提供数额(如50.75美元)和任择货币(如美元“美元”),法典将写给你。 “美元和七十五美分”

友情链接