Python 正则表达式 - 删除所有“.”和除小数点以外的特殊字符

Question

我有些句子有多个“.”。

如何删除所有特殊字符和“.”数据中除小数点外?

输入示例为

What? The Census Says It’s Counted 99.9 Percent of Households. Don’t Be Fooled.

我想删除所有“.” s 和除小数点以外的特殊字符'.'

输出应该像

What The Census Says Its Counted 99.9 Percent of Households Dont Be Fooled

我试过了，

regex = re.compile('[^ (\w+\.\w+)0-9a-zA-Z]+')
regex.sub('', test)

但输出是

What The Census Says Its Counted 99.9 Percent of Households. Dont Be Fooled.

Answer 1

使用捕获组只捕获十进制数字，同时匹配特殊字符（即不是 space 和单词字符）。

替换后，只需引用捕获组，以便仅使用捕获的字符。 IE。整个匹配项将被删除并替换为十进制数（如果存在）。

s = 'What? The Census Says It’s Counted 99.9 Percent of Households. Don’t Be Fooled.'
import re
rgx = re.compile(r'(\d\.\d)|[^\s\w]')
rgx.sub(lambda x: x.group(1), s)
# 'What The Census Says Its Counted 99.9 Percent of Households Dont Be Fooled'

或

匹配除数字之间存在的点以外的所有点和除特殊字符外的所有字符，然后最后将这些匹配字符替换为空字符串。

re.sub(r'(?!<\d)\.(?!\d)|[^\s\w.]', '', s)
# 'What The Census Says Its Counted 99.9 Percent of Households Dont Be Fooled'

Answer 2

您需要以下正则表达式：

[^ 0-9a-zA-Z](?!(?<=\d\.)\d)

或者，如果您需要一个完全支持 Unicode 的正则表达式：

(?:_|[^\s\w])(?!(?<=\d\.)\d)

参见regex demo。详情:

[^ 0-9a-zA-Z] - 除了 space、ASCII 字母或数字
(?:_|[^\s\w]) - _ 字符或除 whitespace 和单词 char
(?!(?<=\d\.)\d) - 如果在当前位置右侧紧邻数字和一个点之前有一个数字，则匹配失败的否定前瞻。

看到一个Python demo:

import re
s = 'What? The Census Says It’s Counted 99.9 Percent of Households. Don’t Be Fooled.'
print(re.sub(r'[^ 0-9a-zA-Z](?!(?<=\d\.)\d)', '', s))
# => What The Census Says Its Counted 99.9 Percent of Households Dont Be Fooled
print(re.sub(r'(?:_|[^\s\w])(?!(?<=\d\.)\d)', '', s))
# => What The Census Says Its Counted 99.9 Percent of Households Dont Be Fooled

Python 正则表达式 - 删除所有“.”和除小数点以外的特殊字符

Python Regex - remove all "." and special characters EXCEPT the decimal point

python-3.x

regex

regex-group

regexp-replace

nsregularexpression