regex 尝试了解findall与finditer [duplicate]的匹配项和结果输出的差异

tv6aics1  于 2023-02-17  发布在  其他
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re.findall behaves weird(3个答案)
6天前关闭。
1.使用findall:

import re

target_string = "please sir, that's obviously a clip-on."

result = re.findall(r"[a-z]+('[a-z])?[a-z]*", target_string)

print(result)

# result: ['', '', "'s", '', '', '', '']

1.使用查找器:

import re

target_string ="please sir, that's obviously a clip-on."

result = re.finditer(r"[a-z]+('[a-z])?[a-z]*", target_string)
matched = []
    
for match_obj in result:
    matched.append(match_obj.group())

print(matched)
    
# result: ['please', 'sir', "that's", 'obviously', 'a', 'clip', 'on']

这两种方法是如何匹配模式的,为什么结果输出有差异。请解释。
尝试阅读文档,但仍然对findall与finditer的工作原理感到困惑

b1zrtrql

b1zrtrql1#

findall的情况下,输出将是捕获组('[a-z])。如果您希望完全匹配,请将您的组转换为非捕获组(?:'[a-z])

target_string = "please sir, that's obviously a clip-on."
result = re.findall(r"[a-z]+(?:'[a-z])?[a-z]*", target_string)
print(result)

输出:

['please', 'sir', "that's", 'obviously', 'a', 'clip', 'on']

请注意,如果有多个捕获组,findall将返回它们的元组:

re.findall(r"([a-z]+('[a-z])?[a-z]*)", target_string)

[('please', ''), ('sir', ''), ("that's", "'s"), ('obviously', ''), ('a', ''), ('clip', ''), ('on', '')]

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