re.findall("(100|[0-9][0-9]|[0-9])%", "89%")
This returns only result [89] and I need to return the whole 89%. Any ideas how to do it please?
这只返回结果[89]，我需要返回整个89％。有什么想法可以吗？
3 个解决方案
#1
6
The trivial solution:
琐碎的解决方案：
>>> re.findall("(100%|[0-9][0-9]%|[0-9]%)","89%")
['89%']
More beautiful solution:
更美的解决方案：
>>> re.findall("(100%|[0-9]{1,2}%)","89%")
['89%']
The prettiest solution:
最漂亮的解决方案：
>>> re.findall("(?:100|[0-9]{1,2})%","89%")
['89%']
#2
10
>>> re.findall("(?:100|[0-9][0-9]|[0-9])%", "89%")
['89%']
When there are capture groups findall returns only the captured parts. Use ?: to prevent the parentheses from being a capture group.
当有捕获组时，findall仅返回捕获的部分。使用？：防止括号成为捕获组。
#3
2
Use an outer group, with the inner group a non-capturing group:
使用外部组，内部组为非捕获组：
>>> re.findall("((?:100|[0-9][0-9]|[0-9])%)","89%")
['89%']
#1
6
The trivial solution:
琐碎的解决方案：
>>> re.findall("(100%|[0-9][0-9]%|[0-9]%)","89%")
['89%']
More beautiful solution:
更美的解决方案：
>>> re.findall("(100%|[0-9]{1,2}%)","89%")
['89%']
The prettiest solution:
最漂亮的解决方案：
>>> re.findall("(?:100|[0-9]{1,2})%","89%")
['89%']
#2
10
>>> re.findall("(?:100|[0-9][0-9]|[0-9])%", "89%")
['89%']
When there are capture groups findall returns only the captured parts. Use ?: to prevent the parentheses from being a capture group.
当有捕获组时，findall仅返回捕获的部分。使用？：防止括号成为捕获组。
#3
2
Use an outer group, with the inner group a non-capturing group:
使用外部组，内部组为非捕获组：
>>> re.findall("((?:100|[0-9][0-9]|[0-9])%)","89%")
['89%']