Question

是否可以从结果中打印出某个重新匹配对象？

Python

sentence = ''' 
Tue, 20 August 2019
17:30 - 21:00 CEST 
'''

pattern = re.compile(r'\d\d[:]\d\d')
matches = pattern.finditer(sentence)

for match in matches:
    print(match)
    print(match.group(0))

输出-> print(match)

<re.Match object; span=(20, 25), match='17:30'>
<re.Match object; span=(28, 33), match='21:00'>

输出-> print(match.group(0))

17:30
21:00

是否只能print进行某些匹配？

print(first re.Match object in list) = 17:30

print(second re.Match object in list) = 21:00

尝试->我为解决此问题而做的尝试

for match in matches:
    print(matches[0].group(0))
    print(match.group(0))
    print(match[0])
    print(match)
    print(match.__getitem__(0))
    print(match.group(0))
    print(match.groupdict(0))
    print(match.start(0))
    print(match.groups)
    print(match.group(1))
    print(re.match(r'<img.*?>', matches))
    print(re.match(r'<img.*?>', match).group(0))

Answer 1

for match in matches ==您为所有匹配项进行打印。

执行以下操作：

sentence = ''' 
Tue, 20 August 2019
17:30 - 21:00 CEST 
'''

pattern = re.compile(r'\d\d[:]\d\d')
matches = pattern.findall(sentence) # returns a list
print(matches[0].group(0)) # get first element from the list (0-indexed) and get its first group

Answer 2

如果需要，可以添加一个计数器，在for循环中增加它，并检查它是否是您需要打印的匹配项。

在线查看Python demo：

import re

sentence = ''' 
Tue, 20 August 2019
17:30 - 21:00 CEST 
'''
pattern = re.compile(r'\d\d:\d\d')
cnt = 0          # Initialize the counter
wanted = [1,2]   # Defines the 1-based IDs of the matches you want to display

for match in pattern.finditer(sentence):
    cnt += 1                  # Increment the counter
    if cnt in wanted:         # If the ID is in wanted IDs
        print(match.group())  # Print it

如何打印特定的re.Match对象

2 个答案: