Python 替换所有匹配正则表达式的出现
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Replace all occurrences that match regular expression
提问by tots_o_tater
I have a regular expression that searches for a string that contains '.00.'or '.11.'as follows:
我有一个正则表达式，用于搜索包含'.00.'或'.11.'如下的字符串：
.*\.(00|11)\..*
What I would like to do is replace all occurrences that match the pattern with 'X00X'or 'X11X'. For example, the string '.00..0..11.'would result in 'X00X.0.X11X'.
我想要做的是用'X00X'or替换所有与模式匹配的事件'X11X'。例如，字符串'.00..0..11.'将导致'X00X.0.X11X'.
I was looking into the Python re.sub method and am unsure of how to do this effectively. The returned match object only matches on the first occurrence and therefore doesn't work well. Any advice? Should I just be using a string replace for this task? Thanks.
我正在研究 Python re.sub 方法，但不确定如何有效地做到这一点。返回的匹配对象仅在第一次出现时匹配，因此效果不佳。有什么建议吗？我应该为此任务使用字符串替换吗？谢谢。
回答by Tim Pietzcker
re.sub()(docs for Python 2and Python 3) does replace all matches it finds, but your use of .*may have caused the regex to match too much (even other occurences of .00.etc.). Simply do:
re.sub()（Python 2和Python 3 的文档）确实替换了它找到的所有匹配项，但是您的使用.*可能会导致正则表达式匹配太多（甚至是其他情况.00.等）。简单地做：
In [2]: re.sub(r"\.(00|11)\.", r"XX", ".00..0..11.")
Out[2]: 'X00X.0.X11X'
Note that patterns cannot overlap:
请注意，模式不能重叠：
In [3]: re.sub(r"\.(00|11)\.", r"XX", ".00.11.")
Out[3]: 'X00X11.'
回答by Siva Prasad Koka
You can try this as well,
你也可以试试这个
data = "otherway-of-try-b-pool"
data_after_regex = re.sub(r'(-[a-z]-pool)', "", data)
out: otherway-of-try (above regEx removed '-b-pool' part)
