Python re.sub 反向引用而不是反向引用 [英] Python re.sub back reference not back referencing
问题描述
我有以下几点:
<text top="52" left="20" width="383" height="15" font="0"><b>test</b></text>
我有以下几点:
fileText = re.sub("(.*?)", "\1", fileText, flags=re.DOTALL)
其中 fileText 是我上面发布的字符串.当我在运行正则表达式替换后打印出 fileText
时,我回来了
而不是预期
<text top="52" left="20" width="383" height="15" font="0">test</text>
现在我相当精通正则表达式,我知道它应该可以工作,事实上我知道它正确匹配,因为我可以在 groups
中看到它,当我进行搜索并打印出groups
但我是 python 新手,我很困惑为什么它不能正确处理反向引用
您需要使用 此处使用原始字符串,以便反斜杠不会作为转义字符处理:
<预><代码>>>>进口重新>>>fileText = '<text top="52" left="20" width="383" height="15" font="0"><b>test</b></text>'>>>fileText = re.sub("<b>(.*?)</b>", r"\1", fileText, flags=re.DOTALL)>>>文件文本'<text top="52" left="20" width="383" height="15" font="0">test</text>'>>>注意如何将 "\1"
更改为 r"\1"
.虽然这是一个很小的变化(一个字符),但它有很大的影响.见下文:
I have the following:
<text top="52" left="20" width="383" height="15" font="0"><b>test</b></text>
and I have the following:
fileText = re.sub("<b>(.*?)</b>", "\1", fileText, flags=re.DOTALL)
In which fileText is the string I posted above. When I print out fileText
after I run the regex replacement I get back
<text top="52" left="20" width="383" height="15" font="0"></text>
instead of the expected
<text top="52" left="20" width="383" height="15" font="0">test</text>
Now I am fairly proficient at regex and I know that it should work, in fact I know that it matches properly because I can see it in the groups
when I do a search and print out the groups
but I am new to python and am confused as to why its not working with back references properly
You need to use a raw-string here so that the backslash isn't processed as an escape character:
>>> import re
>>> fileText = '<text top="52" left="20" width="383" height="15" font="0"><b>test</b></text>'
>>> fileText = re.sub("<b>(.*?)</b>", r"\1", fileText, flags=re.DOTALL)
>>> fileText
'<text top="52" left="20" width="383" height="15" font="0">test</text>'
>>>
Notice how "\1"
was changed to r"\1"
. Though it is a very small change (one character), it has a big effect. See below:
>>> "\1"
'\x01'
>>> r"\1"
'\\1'
>>>
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