blocks|key|4177248|text|假设您的字符串以"ABB_CCCCC“或"A_BB_CCCCC”的形式存在，并且您希望提取A/BB/CCCCC。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|4177249|你可以用：|4177250|import+re
re.findall('(.)_?([%5E_]%2B)_(.%2B)',+your_string)[0]|code-block|syntax|javascript|4177251|NB1。如果您总是有一个字符，两个字符，五个字符，请使用：`‘(.)？(.)(.{5})“”|4177252|NB2。这里假设字符串对一个或另一个公式是有效的，否则就没有匹配和IndexError。|4177253|示例：|4177254|s1+=+'n_tt_j1213'
re.findall('(.)_?([%5E_]%2B)_(.%2B)',+your_string)[0]|4177255|s2+=+'ntt_j1213'
re.findall('(.)_?([%5E_]%2B)_(.%2B)',+your_string)[0]|4177256|这两方面的产出：|4177257|('n',+'tt',+'j1213')|4177258|entityMap^0|0|0|0|0|0|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|Y|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|Z|8|@]|9|@]|A|$]]|$1|D|3|E|5|F|7|10|8|@]|9|@]|A|$G|H]]|$1|I|3|J|5|6|7|11|8|@]|9|@]|A|$]]|$1|K|3|L|5|6|7|12|8|@]|9|@]|A|$]]|$1|M|3|N|5|6|7|13|8|@]|9|@]|A|$]]|$1|O|3|P|5|F|7|14|8|@]|9|@]|A|$G|H]]|$1|Q|3|R|5|F|7|15|8|@]|9|@]|A|$G|H]]|$1|S|3|T|5|6|7|16|8|@]|9|@]|A|$]]|$1|U|3|V|5|F|7|17|8|@]|9|@]|A|$G|H]]|$1|W|3|-4|5|6|7|18|8|@]|9|@]|A|$]]]|X|$]]

Assuming your strings are in the form &quot;ABB_CCCCC&quot; or &quot;A_BB_CCCCC&quot; and you want to extract A/BB/CCCCC.
You could use:
<pre><code>import re
re.findall('(.)_?([^_]+)_(.+)', your_string)[0]
</code></pre>
NB1. If you always have 1 character, 2 characters, 5 characters, use: <code>'(.)_?(..)_(.{5})'</code>
NB2. this assumes here that the strings are valid for one or the other formula, else you will have no match and an IndexError
example:
<pre><code>s1 = 'n_tt_j1213'
re.findall('(.)_?([^_]+)_(.+)', your_string)[0]
</code></pre>
<pre><code>s2 = 'ntt_j1213'
re.findall('(.)_?([^_]+)_(.+)', your_string)[0]
</code></pre>
output for both:
<pre><code>('n', 'tt', 'j1213')
</code></pre>

blocks|key|4177319|text|我就是这样做的：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|4177320|import+re

regex+=+re.compile('%5E([a-z]{3}).*([a-z]{2}).*([a-z]{1}\d{3})$')
matches+=+re.search(regex,+'newtt_j123')
print(matches.groups())|code-block|syntax|javascript|4177321|OUT：('new',+'tt',+'j123')|offset|length|style|CODE|4177322|regex的工作方式如下：%5E([a-z]{3})将匹配定位到字符串的开头，然后匹配3个小写字母，并将其存储在第一个捕获组中。any是一个字符类，它意味着任何小写字母，花括号中的3表示长度必须是3。它在一个捕获组中，因为它在普通括号中(即()+)。|4177323|然后贪婪地将任何内容与.*匹配，直到下一个捕获组，并将任意两个小写字母与([a-z]{2})匹配。|4177324|然后，再一次贪婪地将直到最后一个捕获组的任何内容与([a-z]{1}\d{3})匹配。这将匹配由\d表示的三个数字旁边的一个小写字母。|4177325|然后，当你打印出matches.groups()，你将得到匹配的顺序，即第一个将是三个字母，下一个将是下两个字母，最后一个将是字母后面的三个数字。|4177326|我还使用%5E和$将匹配锚定到字符串的开始和结束。|4177327|entityMap^0|0|0|4|L|0|D|B|0|10|A|0|P|F|1C|2|0|8|G|0|4|1|6|1|0^^$0|@$1|2|3|4|5|6|7|Y|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|Z|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|10|8|@$I|11|J|12|K|L]]|9|@]|A|$]]|$1|M|3|N|5|6|7|13|8|@$I|14|J|15|K|L]]|9|@]|A|$]]|$1|O|3|P|5|6|7|16|8|@$I|17|J|18|K|L]]|9|@]|A|$]]|$1|Q|3|R|5|6|7|19|8|@$I|1A|J|1B|K|L]|$I|1C|J|1D|K|L]]|9|@]|A|$]]|$1|S|3|T|5|6|7|1E|8|@$I|1F|J|1G|K|L]]|9|@]|A|$]]|$1|U|3|V|5|6|7|1H|8|@$I|1I|J|1J|K|L]|$I|1K|J|1L|K|L]]|9|@]|A|$]]|$1|W|3|-4|5|6|7|1M|8|@]|9|@]|A|$]]]|X|$]]

This is how I would approach this:
<pre><code>import re

regex = re.compile('^([a-z]{3}).*([a-z]{2}).*([a-z]{1}\d{3})$')
matches = re.search(regex, 'newtt_j123')
print(matches.groups())
</code></pre>
OUT:
<code>('new', 'tt', 'j123')</code>
The regex works like this:
<code>^([a-z]{3})</code> anchors the match to the start of the string and then matches 3 lowercase letters and stores this in the first capture group. The [a-z] is a character class which means any lowercase letters, the 3 in curly brackets mean it has to be of length 3. It's in a capture group because it's in normal brackets (i.e. () ).
Then greedily match anything with .* up to the next capture group and<code>([a-z]{2})</code> any two lowercase letters.
Then again greedily match anything up to to the last capture group with <code>([a-z]{1}\d{3})</code>. This matches one lowercase letter next to three numbers, which is denoted by <code>\d</code>.
Then when you print out <code>matches.groups()</code> you will get the matches in that order, i.e. the first one will be the three letters, the next one will be the next two letters, and the last one will be the letter followed by three numbers.
I also used <code>^</code> and <code>$</code> to anchor the matches to the start and end of the string.

blocks|key|2073167|text|据我所知，你可以不用regex就能做到。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|2073168|首先，删除所有_+(如果有s.replace('_',+'')+)，然后使用括号提取如下所示：|offset|length|style|CODE|2073169|s+=+"newtt_j1213"
s+=+s.replace('_',+'')
ss+=+[+s[0:3],+s[3:5],+s[5:]+]
print(ss)|code-block|syntax|javascript|2073170|输出：['new',+'tt',+'j1213']|2073171|entityMap^0|0|7|1|D|I|0|0|3|M|0^^$0|@$1|2|3|4|5|6|7|Q|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|R|8|@$D|S|E|T|F|G]|$D|U|E|V|F|G]]|9|@]|A|$]]|$1|H|3|I|5|J|7|W|8|@]|9|@]|A|$K|L]]|$1|M|3|N|5|6|7|X|8|@$D|Y|E|Z|F|G]]|9|@]|A|$]]|$1|O|3|-4|5|6|7|10|8|@]|9|@]|A|$]]]|P|$]]

From what i understand, you can do it without regex.
First remove all <code>_</code> if present with <code>s.replace('_', '')</code>
Next use bracket to extract like this :
<pre><code>s = &quot;newtt_j1213&quot;
s = s.replace('_', '')
ss = [ s[0:3], s[3:5], s[5:] ]
print(ss)
</code></pre>
Output:
<code>['new', 'tt', 'j1213']</code>

I have a string that looks like this: new_tt_j1213.
I need to split it into 'new', 'tt', 'j1213'.
string rules:
<ol>
<li>The first three characters are always letters
</li>
<li>after the first '_' (if it is present) there are 2 letters only
</li>
<li>after the second '_' (if it is present) there are EITHER t1524 OR t014 formats (the first symbol is always a letter and there can be 3 or 4 digits)
</li>
<li>all letters are lower case
</li>
<li>'_' can be missing due to corruption
</li>
</ol>
It is very easy with the split() method BUT because data is corrupted sometimes the first underscore is missing: 'newtt_j1213'
Since I am very new to Regex, can you help me to adjust my code below so the split works as described even without the first '_'. (or even without the 2nd underscore)?
<pre><code> p1 =''
 p2 =''
 p3 =''

str_t = 'newtt_j123'
str_t1 = 'new_tt_j1213'
str_t2 = 'newttj1213'
str3 = 'new_ttj1213'

# Test for data Corruption
 tr_lst = re.split('_', str_t) 
 if len(tr_lst) &lt;3: print('DATA CORRUPTION - ',tr_lst) 
 
 p1, p2, p3 = re.split('_', str_t) # THIS LINE NEEDS ADJUSTMENT (REGEX?)
 print(p1, p2, p3, str_t)
</code></pre>
Thank you!!

Regex use to split a string with an OPTIONAL SEPARATOR (char)

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。

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我有一个看起来像这样的字符串: new_tt_j1213。我需要把它分成'new'，'tt'，‘j 1213’。字符串规则：前三个字符总是字母在第一个'_‘之后(如果它存在的话)，只有两个字母只有在第二个'_‘之后(如果它存在的话)，有t1524或t014格式(第一个符号总是一个字母，可以有3或4个数字)所有字母都是小...

问Regex用于使用可选分隔符(char)拆分字符串。
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问Regex用于使用可选分隔符(char)拆分字符串。EN