Question

我有一个字符串。我想将字符串剪切成子字符串，其中包含一个包含数字的单词，其中包含两边的（最多）4个单词。如果子串重叠，则它们应该结合起来。

Sampletext = "by the way I know 54 how to take praise for 65 excellent questions 34 thank you for asking appreciated."
re.findall('(\s[*\s]){1,4}\d(\s[*\s]){1,4}', Sampletext)
desired output = ['the way I know 54 how to take praise', 'to take praise for 65 excellent questions 34 thank you for asking']

Answer 1

重叠匹配：使用前瞻

这样做：

subject = "by the way I know 54 how to take praise for 65 excellent questions 34 thank you for asking appreciated."
for match in re.finditer(r"(?=((?:\b\w+\b ){4}\d+(?: \b\w+\b){4}))", subject):
    print(match.group(1))

什么是Word？

输出取决于您对单词的定义。总而言之，我允许数字。这会产生以下输出。

输出（允许使用单词中的数字）

the way I know 54 how to take praise
to take praise for 65 excellent questions 34 thank
for 65 excellent questions 34 thank you for asking

选项2：单词中没有数字

subject = "by the way I know 54 how to take praise for 65 excellent questions 34 thank you for asking appreciated."    
for match in re.finditer(r"(?=((?:\b[a-z]+\b ){4}\d+(?: \b[a-z]+\b){4}))", subject, re.IGNORECASE):
    print(match.group(1))

输出2

the way I know 54 how to take praise

选项3：扩展为四个不间断的非数字字

根据您的评论，此选项将扩展到数据透视表的左侧和右侧，直到匹配四个不间断的非数字字词。逗号被忽略了。

subject = "by the way I know 54 how to take praise for 65 excellent questions 34 thank you for asking appreciated. One Two Three Four 55 Extend 66 a b c d AA BB CC DD 71 HH DD, JJ FF"
for match in re.finditer(r"(?=((?:\b[a-z]+[ ,]+){4}(?:\d+ (?:[a-z]+ ){1,3}?)*?\d+.*?(?:[ ,]+[a-z]+){4}))", subject, re.IGNORECASE):
    print(match.group(1))

输出3

the way I know 54 how to take praise
to take praise for 65 excellent questions 34 thank you for asking
One Two Three Four 55 Extend 66 a b c d
AA BB CC DD 71 HH DD, JJ FF

python正则表达式查找以数字为中心的子串

1 个答案: