blocks|key|3559892|text|您的模式只是一个外观，它本身无法与任何东西相匹配。我将使用这种re.findall方法：|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|3559893|string+=+'I:/Etrmtest/PZMALIo4/ETRM841_FX_Deals_Restructuring/FO_PRE\\abo_st_gas_dtd.csv'
filename+=+re.findall(r'\\([%5E.]%2B\.\w%2B)$',+string)[0]
print(filename)++#+abo_st_gas_dtd.csv|code-block|syntax|javascript|3559894|entityMap^0|V|A|0|0^^$0|@$1|2|3|4|5|6|7|M|8|@$9|N|A|O|B|C]]|D|@]|E|$]]|$1|F|3|G|5|H|7|P|8|@]|D|@]|E|$I|J]]|$1|K|3|-4|5|6|7|Q|8|@]|D|@]|E|$]]]|L|$]]

Your pattern is just a lookabehind, which, by itself, can't match anything. I would use this <code>re.findall</code> approach:
<pre class="lang-py prettyprint-override"><code>string = 'I:/Etrmtest/PZMALIo4/ETRM841_FX_Deals_Restructuring/FO_PRE\\abo_st_gas_dtd.csv'
filename = re.findall(r'\\([^.]+\.\w+)$', string)[0]
print(filename) # abo_st_gas_dtd.csv
</code></pre>

blocks|key|3559901|text|有几件事发生了：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|3559902|3559903|您需要使用与模式相同的r'string'表示法来声明字符串定义；现在，您的string只有一个反斜杠，因为第一个反斜杠是转义的。|ordered-list-item|offset|length|style|CODE|3559904|，我不确定您是否正确地使用了*。它的意思是“立即重复前一组”，而不仅仅是“任何字符串”(例如，在通常的shell模式中)。括号中的第一个*在它之前没有任何内容，这意味着正则表达式是无效的。因此，你看到了错误。我认为，您想要的是.*，即重复任何字符0或更多次。此外，在括号中不需要这样做。更正确的regexp是r'(?<=\\\\).*'|3559905|。|3559906|import+re

string+=+r'I:/Etrmtest/PZMALIo4/ETRM841_FX_Deals_Restructuring/FO_PRE\\abo_st_gas_dtd.csv'

pattern+=+r'(?<=\\\\).*'

re.findall(pattern,string)|code-block|syntax|javascript|3559907|entityMap^0|0|0|B|9|11|6|0|E|1|1W|1|35|2|4A|E|0|0|0^^$0|@$1|2|3|4|5|6|7|U|8|@]|9|@]|A|$]]|$1|B|3|-4|5|6|7|V|8|@]|9|@]|A|$]]|$1|C|3|D|5|E|7|W|8|@$F|X|G|Y|H|I]|$F|Z|G|10|H|I]]|9|@]|A|$]]|$1|J|3|K|5|E|7|11|8|@$F|12|G|13|H|I]|$F|14|G|15|H|I]|$F|16|G|17|H|I]|$F|18|G|19|H|I]]|9|@]|A|$]]|$1|L|3|M|5|6|7|1A|8|@]|9|@]|A|$]]|$1|N|3|O|5|P|7|1B|8|@]|9|@]|A|$Q|R]]|$1|S|3|-4|5|6|7|1C|8|@]|9|@]|A|$]]]|T|$]]

There's a couple of things going on:
<ol>
<li>You need to declare your string definition using the same <code>r'string'</code> notation as for the pattern; right now your <code>string</code> only has a single backslash, since the first one of the two is escaped.</li>
<li>I'm not sure you're using <code>*</code> correctly. It means &quot;repeat immediately preceding group&quot;, and not just &quot;any string&quot; (as, e.g., in the usual shell patterns). The first <code>*</code> in parentheses does not have anything preceding it, meaning that the regex is invalid. Hence the error you see. I think, what you want is <code>.*</code>, i.e., repeating any character 0 or more times. Furthermore, it is not needed in the parentheses. A more correct regexp would be <code>r'(?&lt;=\\\\).*'</code>:</li>
</ol>
<pre class="lang-py prettyprint-override"><code>import re

string = r'I:/Etrmtest/PZMALIo4/ETRM841_FX_Deals_Restructuring/FO_PRE\\abo_st_gas_dtd.csv'

pattern = r'(?&lt;=\\\\).*'

re.findall(pattern,string)
</code></pre>

blocks|key|670333|text|files+=+'I:E\\trm.csvest/PZMALIo4\ETRM841_FX_.csvDeals_Restructuring/FO_PRE\\abo_st_gas_dtd.csv'
counter+=+-1
my_files+=+[]
for+f+in+files:
++++counter+%2B=+1
++++if+ord(f)+==+92:#'\'
++++++++temp+=+files[counter%2B1:len(files)]
++++++++temp_file+=+""
++++++++for+f1+in+temp:
++++++++++++temp_file+%2B=+f1
++++++++++++#+[0-len(temp_file)]+=>+if+[char+after+.+to+num+index+of+type+file]==+csv
++++++++++++if+f1+==+'.'+and+temp[len(temp_file):len(temp_file)%2B3]+==+"csv":
++++++++++++++++my_files.append(temp_file+%2B+"csv")
++++++++++++++++break
print(my_files)#['trm.csv',+'ETRM841_FX_.csv',+'abo_st_gas_dtd.csv']|type|code-block|depth|inlineStyleRanges|entityRanges|data|syntax|javascript|670334|unstyled|entityMap^0|0^^$0|@$1|2|3|4|5|6|7|G|8|@]|9|@]|A|$B|C]]|$1|D|3|-4|5|E|7|H|8|@]|9|@]|A|$]]]|F|$]]

<pre><code>files = 'I:E\\trm.csvest/PZMALIo4\ETRM841_FX_.csvDeals_Restructuring/FO_PRE\\abo_st_gas_dtd.csv'
counter = -1
my_files = []
for f in files:
 counter += 1
 if ord(f) == 92:#'\'
 temp = files[counter+1:len(files)]
 temp_file = &quot;&quot;
 for f1 in temp:
 temp_file += f1
 # [0-len(temp_file)] =&gt; if [char after . to num index of type file]== csv
 if f1 == '.' and temp[len(temp_file):len(temp_file)+3] == &quot;csv&quot;:
 my_files.append(temp_file + &quot;csv&quot;)
 break
print(my_files)#['trm.csv', 'ETRM841_FX_.csv', 'abo_st_gas_dtd.csv']


</code></pre>

I have the following code that tries to retrieve the name of a file from a directory based on a double \ character:
<pre><code>import re

string = 'I:/Etrmtest/PZMALIo4/ETRM841_FX_Deals_Restructuring/FO_PRE\\abo_st_gas_dtd.csv'
pattern = r'(?&lt;=*\\\\)*'
re.findall(pattern,string)
</code></pre>
The reasoning behind is that the name of the file is always after a double \ , so I try to look any string which is preceeded by any text that finishes with \ .
Neverthless, when I apply this code I get the following error:
<pre><code>error: nothing to repeat at position 4
</code></pre>
What am I doing wrong?
Edit: The concrete output I am looking for is getting the string <code>'abo_st_gas_dtd_csv'</code> as a match.

regex: pattern fails to match what I am looking for

翻译质量差，导致语言生硬或混乱。

没有提供实际的解决方法或示例。

解答不清晰，无法理解或解决问题。

页面排版不美观，阅读体验差。

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下面的代码试图根据双字符从目录中检索文件名：import restring = 'I:/Etrmtest/PZMALIo4/ETRM841_FX_Deals_Restructuring/FO_PRE\\abo_st_gas_dtd.csv'pattern = r'(?<=*\\\\)*'re.findall(patte...

问regex:模式无法匹配我正在寻找的
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