尝试了解 findall 与 finditer 的匹配内容和结果输出的差异[重复]

Question

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", '', '', '', '']

2. 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']

这两种方法如何匹配模式以及为什么结果输出存在差异。请解释一下。

尝试阅读文档，但仍然对 findall 与 finditer 的工作原理感到困惑

Answer 1

在

findall

情况下，输出将是捕获组

('[a-z])

。 如果您想要完整的比赛，请将您的团队转变为非捕获团队

(?:'[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，如果您有多个捕获组，

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', '')]