正则表达式从 python 中的字符串中提取 substring

Question

How do we get the following substring from a string using re in python.

string1 = "fgdshdfgsLooking: 3j #123"
substring = "Looking: 3j #123"

string2 = "Looking: avb456j #13fgfddg"
substring = "Looking: avb456j #13"

tried:

re.search(r'Looking: (.*#\d+)$', string1)

Answer 1

You need to remove the $ from the regex:

 re.search(r'Looking: (.*#\d+)', string1)

If you also want re to return Looking , you'll have to wrap it in parens:

 re.search(r'(Looking: (.*#\d+))', string1)

Answer 2

Try,

re.search(r'Looking: (.)*#(\d)+', string1)

1. It will match "Looking: "
2. After that it will look for 0 or more any character
3. After that a "#"
4. and 1 or more digits

Answer 3

Your regex is mostly correct, you just need to remove EOL(End of Line) $ as in some case like string2 the pattern does not end with a EOL, and have some extra string after the pattern ends.

import re

string1 = 'fgdshdfgsLooking: 3j #123'
string2 = 'Looking: avb456j #13fgfddg'

pattern = r'Looking: (.*?#\d+)'

match1 = re.search(pattern, string1)
match2 = re.search(pattern, string2)

print('String1:', string1, '|| Substring1:', match1.group(0))
print('String2:', string2, '|| Substring2:', match2.group(0))

Output:

String1: fgdshdfgsLooking: 3j #123 || Substring1: Looking: 3j #123
String2: Looking: avb456j #13fgfddg || Substring2: Looking: avb456j #13

should work, also I've matched everything before # lazily by using ? to match as few times as possible, expanding as needed, that is to avoid matching everything upto second # , in case there is a second # followed by few digits in the string somewhere further down.

Live Demo

Answer 4

try this:

re.search("[A-Z]\w+:\s?\w+\s#\d+",string1)