在python中使用正则表达式在冒号或括号之前提取字符串

Question

I'm trying to extract the string muscle pain from the following strings. I need to use a regular expression that works for all three cases.

string1 = 'A1 muscle pain: immunotherapy'
string2 = 'A2B_45 muscle pain: topical medicine e.g. ....'
string3 = 'A2_45 muscle pain (pain): topical medicine e.g. ....'

The following code works for string1 and string2 . But it does not work for string3 . What I get is always muscle pain (pain) . Can anyone help me with that. I tried so many times with different expression but could not figure out how.

re.match(r"^[A-Z]+\d*[A-Z]*_?\d*\s(.*)[:\(]", string3).group(1)

Answer 1

You can shorten the expression to:

^A\S+\s([^:(]*)(?=:|\s\()

• ^A Assert position beginning of string.
• \\S+ Any non whitespace characters.
• \\s Whitespace character.
• ([^:(]*) Capture group. Match and capture anything other than a ( bracket or ] bracket.
• (?=:|\\s\\() Positive lookahead for : or whitespace followed by ( .

Try it live here .

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Python snippet:

import re
string1 = 'A1 muscle pain: immunotherapy'
string2 = 'A2B_45 muscle pain: topical medicine e.g. ....'
string3 = 'A2_45 muscle pain (pain): topical medicine e.g. ....'

print(re.match(r'^A\S+\s([^:(]*)(?=:|\s\()',string3).group(1))

Answer 2

Try this pattern: ^[\\dA-Z_]+ ([^\\(:]+) .

It starts with [\\dA-Z_]+ at the beggining (note anchor ^ ), followed by space. Now, start capturing group until one of unwanted characters is met: [^\\(:] . You can add there more "unwanted" characters to alter regex to match differently.

First capturing group is what you want.

Demo

You could try this pattern to remove space after third match: ^[\\dA-Z_]+ ([\\w ]+)(?=(:| \\()) . See demo.