除括号外的所有逗号的索引

Question

如何排除小括号（少于20个字符）中的逗号

Get index of this comma, but (not this , comma). Get other commas like, or ,or, 1,1 2 ,2. (not this ,) BUT (get index of this comma, if more than 20 characters are inside the parentheses)

this example所有逗号索引的预期输出： [23, 71, 76, 79, 82, 87, 132]

Answer 1

使用PyPi正则表达式：

,(?![^()]*\))|(?<=\((?=[^()]{20})[^()]*),

见proof

Python code：

import regex
text = r"Get index of this comma, but (not this , comma). Get other commas like, or ,or, 1,1 2 ,2. (not this ,) BUT (get index of this comma, if more than 20 characters are inside the parentheses)"
reg_expression = r',(?![^()]*\))|(?<=\((?=[^()]{20})[^()]*),'
print(regex.sub(reg_expression, '<COMMA>\g<0></COMMA>', text))
# Get index of this comma<COMMA>,</COMMA> but (not this , comma). Get other commas like<COMMA>,</COMMA> or <COMMA>,</COMMA>or<COMMA>,</COMMA> 1<COMMA>,</COMMA>1 2 <COMMA>,</COMMA>2. (not this ,) BUT (get index of this comma<COMMA>,</COMMA> if more than 20 characters are inside the parentheses)
indices = [x.start() for x in regex.finditer(reg_expression, text)]
print(indices)
# [23, 70, 75, 78, 81, 86, 131]

表达解释

--------------------------------------------------------------------------------
  ,                        ','
--------------------------------------------------------------------------------
  (?!                      look ahead to see if there is not:
--------------------------------------------------------------------------------
    [^()]*                   any character except: '(', ')' (0 or
                             more times (matching the most amount
                             possible))
--------------------------------------------------------------------------------
    \)                       ')'
--------------------------------------------------------------------------------
  )                        end of look-ahead
--------------------------------------------------------------------------------
 |                        OR
--------------------------------------------------------------------------------
  (?<=                     look behind to see if there is:
--------------------------------------------------------------------------------
    \(                       '('
--------------------------------------------------------------------------------
    (?=                      look ahead to see if there is:
--------------------------------------------------------------------------------
      [^()]{20}                any character except: '(', ')' (20
                               times)
--------------------------------------------------------------------------------
    )                        end of look-ahead
--------------------------------------------------------------------------------
    [^()]*                   any character except: '(', ')' (0 or
                             more times (matching the most amount
                             possible))
--------------------------------------------------------------------------------
  )                        end of look-behind
--------------------------------------------------------------------------------
  ,                        ','

Answer 2

您还可以使用PyPi regex module和SKIP FAIL来匹配和排除匹配结果中不需要的字符

在这种情况下，可以在不应匹配逗号的括号之间匹配1-20

\([^()]{1,20}\)(*SKIP)(*FAIL)|,

解释

• \(匹配(
• [^()]{1,20}匹配除(或)之外的任何字符的1-20倍
• \)匹配)
• (*SKIP)(*FAIL)从匹配结果中排除字符
• |或
• ,匹配一个逗号

Regex demoPython demo

示例代码

import regex

s = """Get index of this comma, but (not this , comma). Get other commas like , or ,or, 1,1 2 ,2.
(not this ,) BUT (get index of this comma, if more than 20 characters are inside the parentheses)"""
pattern = r"\([^()]{1,20}\)(*SKIP)(*FAIL)|,"
indices = [m.start(0) for m in regex.finditer(pattern, s)]
print(indices)

输出

[23, 71, 76, 79, 82, 87, 132]

Answer 3

正则表达式模式：(,)|(\([^()]{0,20}\))

这种模式背后的直觉：

• (,)查找所有逗号。这些存储在捕获组1中

• (\([^()]{0,20}\))查找中间最多20个字符的所有括号。这些存储在捕获组2中

然后，我们可以找到组1中的所有匹配项，只排除长度为20的括号内的逗号

现在要查找这些匹配项的索引，请使用re.finditer()与Match.start()和Match.group()组合使用，以查找组1中每个匹配项的起始索引：

import re

string = """Get index of this comma, but (not this , comma). Get other commas like , or ,or, 1,1 2 ,2.
(not this ,) BUT (get index of this comma, if more than 20 characters are inside the parentheses)"""

indices = [m.start(1) for m in re.finditer('(,)|(\([^()]{0,20}\))', string) if m.group(1)]

print(indices)
# > [23, 71, 76, 79, 82, 87, 132]
print([string[index] for index in indices])
# > [',', ',', ',', ',', ',', ',', ',']

m.start(1)返回组1匹配的起始索引。由于re.finditer()返回来自所有捕获组的匹配项，因此添加if m.group(1)需要为组1找到匹配项（来自其他组的匹配项为None）

编辑：这将忽略内部包含20个或更少字符的括号，这与第一条语句不一致，但与示例中解释的内容一致。如果希望小于20，只需使用{0,19}