用python替换不同长度的数字（re.sub）

Question

corpus = """In the US 555-0198 and 1-206-5705-0100 are examples fictitious numbers.
            In the UK, 044-113-496-1834 is a fictitious number.
            In Ireland, the number 353-020-917-1234 is fictitious.
            And in Australia, 061-970-654-321 is a fictitious number.
            311 is a joke."""

我是python的新手，正在研究正则表达式，试图将所有7,11,12和13位数字都更改为零。我希望它仍然看起来像一个电话号码。例如将555-0198更改为000-0000，是否有一种方法可以将311保持原样而不变为零？以下是我能够提出的内容

起初我尝试过，但是使所有数字都为零

    for word in corpus.split():
        nums = re.sub("(\d)", "0",word)
        print(nums)

然后我尝试过，但是我意识到用这种方式对11位和13位数字不正确

    def sub_nums():
        for word in corpus.split():
           nums = re.sub("(\d{1,4})-+(\d{1,4})", "000-0000",word)
           print(nums)
    sub_nums()

Answer 1

我使用的正则表达式是：

r'(?<!\S)(?:(?=(-*\d-*){7}(\s|\Z))[\d-]+|(?=(-*\d-*){11}(\s|\Z))[\d-]+|(?=(-*\d-*){12}(\s|\Z))[\d-]+|(?=(-*\d-*){13}(\s|\Z))[\d-]+)'

[7、11、12和13位电话号码有重复的“主题”或模式，所以我只解释7位电话号码的模式：

1. (?!\S)这是negative lookbehind，适用于所有模式，并说电话号码必须not后面带有not空格字符。这是一个双重否定，并且positive lookbehind (?=\s|\A)，它表示电话号码必须以空格或字符串开头。但是，这是可变长度的回溯，Python随附的regex引擎不支持该变量（但PyPi存储库的regex程序包支持）。(?=(-*\d-*){7}(\s|\Z)) 7位电话号码的
2. lookahead
要求说，下一个字符必须由数字和连字符组成，后跟空格或字符串末尾and数字必须正好是7。[\d-]+进行输入中下一个数字和连字符的实际匹配。
3. See Regex Demo

import re


corpus = """In the US 555-0198 and 1-206-5705-0100 are examples fictitious numbers.
            In the UK, 044-113-496-1834 is a fictitious number.
            In Ireland, the number 353-020-917-1234 is fictitious.
            And in Australia, 061-970-654-321 is a fictitious number.
            311 is a joke."""

regex = r'(?<!\S)(?:(?=(-*\d-*){7}(\s|\Z))[\d-]+|(?=(-*\d-*){11}(\s|\Z))[\d-]+|(?=(-*\d-*){12}(\s|\Z))[\d-]+|(?=(-*\d-*){13}(\s|\Z))[\d-]+)'
new_corpus = re.sub(regex, lambda m: re.sub(r'\d', '0', m[0]), corpus)
print(new_corpus)

打印：

In the US 000-0000 and 0-000-0000-0000 are examples fictitious numbers.
            In the UK, 000-000-000-0000 is a fictitious number.
            In Ireland, the number 000-000-000-0000 is fictitious.
            And in Australia, 000-000-000-000 is a fictitious number.
            311 is a joke.

Answer 2

[我认为使用模式先匹配由连字符分隔的数字，然后检查匹配项中的位数是否等于7,11,12或13，可能会更容易。