在python re中使用findall时，未指定长度/字符的区域的语法是什么？

Question

我正在尝试使用正则表达式在一些数据中捕获蛋白质名称及其对应的氨基酸序列。以下是我的代码的精简版本：

import re
line=">sp|A0A385XJ53|INSA9_ECOLI Insertion element OS=Escherichia coli (strain K12) PE=3 SV=1 MASVSISCPSCSATDGVVRNGKSTAGHQRYLCSHCRKTWQLQFTYTASQPGTHQKIIDMA >sp|A0A385XJE6|INH21_ECOLI Transposase InsH for insertion sequence element OS=Escherichia coli (strain K12) PE=3 SV=1 MFVIWSHRTGFIMSHQLTFADSEFSSKRRQTRKEIFLSRMEQILPWQNMVEVIEPFYPKA >sp|A0A385XJL4|INSB9_ECOLI Insertion element IS1 9 protein OS=Escherichia coli (strain K12) PE=3 SV=2 MPGNSPHYGRWPQHDFTSLKKLRPQSVTSRIQPGSDVIVCAEMDEQWGYVGAKSRQRWLF"

result1=re.findall(r'SV=\d\s([A-Z]+)', line)
result2=re.findall(r'>sp\|(\w+)\|', line)
result3=re.findall(r'>sp\|(\w+)\|\.\SV=\d\s([A-Z]+)', line)
for item1 in result1:
    print(item1)
for item2 in result2:
    print(item2)
for item3 in result3:
    print(item3)

Result1输出：

MASVSISCPSCSATDGVVRNGKSTAGHQRYLCSHCRKTWQLQFTYTASQPGTHQKIIDMA
MFVIWSHRTGFIMSHQLTFADSEFSSKRRQTRKEIFLSRMEQILPWQNMVEVIEPFYPKA
MPGNSPHYGRWPQHDFTSLKKLRPQSVTSRIQPGSDVIVCAEMDEQWGYVGAKSRQRWLF

和result2输出：

A0A385XJ53
A0A385XJE6
A0A385XJL4

但是，result3不输出任何内容。我的印象是“。在使用正则表达式时，可用于未指定字符的序列。什么语法可以用于一系列没有设定长度的未指定字符？实际上，我希望python查找>sp|(\w+)|的匹配项，直到找到SV=\d\s(A-Z+)。在这一点上，它将重置为查找>sp|(\w+)|的匹配。我如何实现这一点?我希望它输出如下内容：

A0A385XJ53 MASVSISCPSCSATDGVVRNGKSTAGHQRYLCSHCRKTWQLQFTYTASQPGTHQKIIDMA
A0A385XJE6 MFVIWSHRTGFIMSHQLTFADSEFSSKRRQTRKEIFLSRMEQILPWQNMVEVIEPFYPKA
A0A385XJL4 MPGNSPHYGRWPQHDFTSLKKLRPQSVTSRIQPGSDVIVCAEMDEQWGYVGAKSRQRWLF

我尝试过一些不同的东西，我想也许我不理解".“的用法。由于我的代码已将所有蛋白质转换为单个字符串，因此我认为可以使用"\b+“或"\b*”来代替它，因为没有新的行。我得到了下面的两个错误代码。

error                                     Traceback (most recent call last)
<ipython-input-76-f43b57fdde31> in <module>()
      8 result1=re.findall(r'SV=\d\s([A-Z]+)', line)
      9 result2=re.findall(r'>sp\|(\w+)\|', line)
---> 10 result3=re.findall(r'>sp\|(\w+)\|\b*\SV=\d\s([A-Z]+)', line)
     11 for item1 in result1:
     12     print(item1)

~\OneDrive\Documents\Python stuff\Pythonstuff\lib\re.py in findall(pattern, string, flags)
    220 
    221     Empty matches are included in the result."""
--> 222     return _compile(pattern, flags).findall(string)
    223 
    224 def finditer(pattern, string, flags=0):

~\OneDrive\Documents\Python stuff\Pythonstuff\lib\re.py in _compile(pattern, flags)
    299     if not sre_compile.isstring(pattern):
    300         raise TypeError("first argument must be string or compiled pattern")
--> 301     p = sre_compile.compile(pattern, flags)
    302     if not (flags & DEBUG):
    303         if len(_cache) >= _MAXCACHE:

~\OneDrive\Documents\Python stuff\Pythonstuff\lib\sre_compile.py in compile(p, flags)
    560     if isstring(p):
    561         pattern = p
--> 562         p = sre_parse.parse(p, flags)
    563     else:
    564         pattern = None

~\OneDrive\Documents\Python stuff\Pythonstuff\lib\sre_parse.py in parse(str, flags, pattern)
    853 
    854     try:
--> 855         p = _parse_sub(source, pattern, flags & SRE_FLAG_VERBOSE, 0)
    856     except Verbose:
    857         # the VERBOSE flag was switched on inside the pattern.  to be

~\OneDrive\Documents\Python stuff\Pythonstuff\lib\sre_parse.py in _parse_sub(source, state, verbose, nested)
    414     while True:
    415         itemsappend(_parse(source, state, verbose, nested + 1,
--> 416                            not nested and not items))
    417         if not sourcematch("|"):
    418             break

~\OneDrive\Documents\Python stuff\Pythonstuff\lib\sre_parse.py in _parse(source, state, verbose, nested, first)
    614             if not item or (_len(item) == 1 and item[0][0] is AT):
    615                 raise source.error("nothing to repeat",
--> 616                                    source.tell() - here + len(this))
    617             if item[0][0] in _REPEATCODES:
    618                 raise source.error("multiple repeat",

error: nothing to repeat at position 14

Answer 1

在第三个模式中，对\S进行转义，这意味着匹配一个非空格字符，而不是逐字匹配S。(Is确实与S本身匹配)

当您转义点\.时，它从字面上匹配一个点，这在示例数据中不存在。

在问题I essentially want python to look for a match to >sp\|(\w+)\| and continue until it finds SV=\d\s([A-Z]+). At which point, it will reset to looking for >sp\|(\w+)\|'s match.中阅读此内容

我认为你想使用一个非贪婪的点.+?来匹配两个模式之间的内容，使它至少匹配一个字符。

>sp\|(\w+)\|.+?SV=\d\s([A-Z]+)

Regex demo