如何在Python中将数字转换为单词

Question

我需要将数字从1-99变成单词.这是我到目前为止所得到的:

num2words1 = {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'}
num2words2 = ['Twenty', 'Thirty', 'Forty', 'Fifty', 'Sixty', 'Seventy', 'Eighty', 'Ninety']

def number(Number):

    if (Number > 1) or (Number < 19):
        return (num2words1[Number])
    elif (Number > 20) or (Number < 99):
        return (num2words2[Number])
    else:
        print("Number Out Of Range")
        main()

def main():
    num = eval(input("Please enter a number between 0 and 99: "))
    number(num)
main()

现在,我到目前为止最大的问题是if,elif和else语句似乎不起作用.只运行第一个if语句.

第二个问题是从20-99创建数字的字符串版本....

请提前帮助,谢谢.

PS是的,我知道num2word库,但我不允许使用它.

Answer 1

您的第一个语句逻辑不正确.除非Number是1或更小,否则该语句始终为 True; 200也大于1.

请and改用,并包含1在可接受的值中:

if (Number >= 1) and (Number < 19):

你也可以使用链接:

if 1 <= Number < 19:

对于20或更大的数字,用于divmod()获得数十和余数:

tens, below_ten = divmod(Number, 10)

演示:

>>> divmod(42, 10)
(4, 2)

然后使用这些值从部分构建您的数字:

return num2words2[tens - 2] + '-' + num2words1[below_ten]

全部放在一起:

def number(Number):
    if 1 <= Number < 19:
        return num2words1[Number]
    elif 20 <= Number <= 99:
        tens, below_ten = divmod(Number, 10)
        return num2words2[tens - 2] + '-' + num2words1[below_ten]
    else:
        print("Number out of range")

Answer 2

你可以通过使用一个字典和一个try/except子句来简化这个:

num2words = {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', 0: 'Zero'}

>>> def n2w(n):
        try:
            print num2words[n]
        except KeyError:
            try:
                print num2words[n-n%10] + num2words[n%10].lower()
            except KeyError:
                print 'Number out of range'

>>> n2w(0)
Zero
>>> n2w(13)
Thirteen        
>>> n2w(91)
Ninetyone
>>> n2w(21)
Twentyone
>>> n2w(33)
Thirtythree

Answer 3

代码 2 和 3：

ones = {
    0: '', 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'}
tens = {
    2: 'twenty', 3: 'thirty', 4: 'forty', 5: 'fifty', 6: 'sixty',
    7: 'seventy', 8: 'eighty', 9: 'ninety'}
illions = {
    1: 'thousand', 2: 'million', 3: 'billion', 4: 'trillion', 5: 'quadrillion',
    6: 'quintillion', 7: 'sextillion', 8: 'septillion', 9: 'octillion',
    10: 'nonillion', 11: 'decillion'}



def say_number(i):
    """
    Convert an integer in to it's word representation.

    say_number(i: integer) -> string
    """
    if i < 0:
        return _join('negative', _say_number_pos(-i))
    if i == 0:
        return 'zero'
    return _say_number_pos(i)


def _say_number_pos(i):
    if i < 20:
        return ones[i]
    if i < 100:
        return _join(tens[i // 10], ones[i % 10])
    if i < 1000:
        return _divide(i, 100, 'hundred')
    for illions_number, illions_name in illions.items():
        if i < 1000**(illions_number + 1):
            break
    return _divide(i, 1000**illions_number, illions_name)


def _divide(dividend, divisor, magnitude):
    return _join(
        _say_number_pos(dividend // divisor),
        magnitude,
        _say_number_pos(dividend % divisor),
    )


def _join(*args):
    return ' '.join(filter(bool, args))

测试：

def test_say_number(data, expected_output):
    """Test cases for say_number(i)."""
    output = say_number(data)
    assert output == expected_output, \
        "\n    for:      {}\n    expected: {}\n    got:      {}".format(
            data, expected_output, output)


test_say_number(0, 'zero')
test_say_number(1, 'one')
test_say_number(-1, 'negative one')
test_say_number(10, 'ten')
test_say_number(11, 'eleven')
test_say_number(99, 'ninety nine')
test_say_number(100, 'one hundred')
test_say_number(111, 'one hundred eleven')
test_say_number(999, 'nine hundred ninety nine')
test_say_number(1119, 'one thousand one hundred nineteen')
test_say_number(999999,
                'nine hundred ninety nine thousand nine hundred ninety nine')
test_say_number(9876543210,
                'nine billion eight hundred seventy six million '
                'five hundred forty three thousand two hundred ten')
test_say_number(1000**1, 'one thousand')
test_say_number(1000**2, 'one million')
test_say_number(1000**3, 'one billion')
test_say_number(1000**4, 'one trillion')
test_say_number(1000**5, 'one quadrillion')
test_say_number(1000**6, 'one quintillion')
test_say_number(1000**7, 'one sextillion')
test_say_number(1000**8, 'one septillion')
test_say_number(1000**9, 'one octillion')
test_say_number(1000**10, 'one nonillion')
test_say_number(1000**11, 'one decillion')
test_say_number(1000**12, 'one thousand decillion')
test_say_number(
    1-1000**12,
    'negative nine hundred ninety nine decillion nine hundred ninety nine '
    'nonillion nine hundred ninety nine octillion nine hundred ninety nine '
    'septillion nine hundred ninety nine sextillion nine hundred ninety nine '
    'quintillion nine hundred ninety nine quadrillion nine hundred ninety '
    'nine trillion nine hundred ninety nine billion nine hundred ninety nine'
    ' million nine hundred ninety nine thousand nine hundred ninety nine')

Answer 4

使用名为num2words Link - > HERE的 python库

Answer 5

您被允许使用其他软件包吗？这一次的作品真的很适合我： 活用。它对于自然语言生成很有用，并且具有将数字转换为英文文本的方法。

我安装了

$ pip install inflect

然后在您的Python会话中

>>> import inflect
>>> p = inflect.engine()
>>> p.number_to_words(1234567)
'one million, two hundred and thirty-four thousand, five hundred and sixty-seven'

>>> p.number_to_words(22)
'twenty-two'