CODEWITHSUNDEEP
pythonregex
Amith
Amith

Reputation: 401

Regular expression: Match everything after a particular word

I am using Python and would like to match all the words after test till a period (full-stop) or space is encountered.

text = "test : match this."

At the moment, I am using :

import re
re.match('(?<=test :).*',text)

The above code doesn't match anything. I need match this as my output.

Upvotes: 23

Views: 108303

Answers (4)

Punnerud
Punnerud

Reputation: 8081

Everything after test, including test

test.*

Everything after test, without test

(?<=test).*

Example here on regexr.com

Upvotes: 56

Wiktor Stribiżew
Wiktor Stribiżew

Reputation: 627537

In a general case, as the title mentions, you may capture with (.*) pattern any 0 or more chars other than newline after any pattern(s) you want:

import re
p = re.compile(r'test\s*:\s*(.*)')
s = "test : match this."
m = p.search(s)           # Run a regex search anywhere inside a string
if m:                     # If there is a match
    print(m.group(1))     # Print Group 1 value

If you want . to match across multiple lines, compile the regex with re.DOTALL or re.S flag (or add (?s) before the pattern):

p = re.compile(r'test\s*:\s*(.*)', re.DOTALL)
p = re.compile(r'(?s)test\s*:\s*(.*)')

However, it will retrun match this.. See also a regex demo.

You can add \. pattern after (.*) to make the regex engine stop before the last . on that line:

test\s*:\s*(.*)\.

Watch out for re.match() since it will only look for a match at the beginning of the string (Avinash aleady pointed that out, but it is a very important note!)

See the regex demo and a sample Python code snippet:

import re
p = re.compile(r'test\s*:\s*(.*)\.')
s = "test : match this."
m = p.search(s)           # Run a regex search anywhere inside a string
if m:                     # If there is a match
    print(m.group(1))     # Print Group 1 value

If you want to make sure test is matched as a whole word, add \b before it (do not remove the r prefix from the string literal, or '\b' will match a BACKSPACE char!) - r'\btest\s*:\s*(.*)\.'.

Upvotes: 6

NDevox
NDevox

Reputation: 4086

I don't see why you want to use regex if you're just getting a subset from a string.

This works the same way:

if line.startswith('test:'):
    print(line[5:line.find('.')])

example:

>>> line = "test: match this."
>>> print(line[5:line.find('.')])
 match this

Regex is slow, it is awkward to design, and difficult to debug. There are definitely occassions to use it, but if you just want to extract the text between test: and ., then I don't think is one of those occasions.

See: https://softwareengineering.stackexchange.com/questions/113237/when-you-should-not-use-regular-expressions

For more flexibility (for example if you are looping through a list of strings you want to find at the beginning of a string and then index out) replace 5 (the length of 'test:') in the index with len(str_you_looked_for).

Upvotes: 5

Avinash Raj
Avinash Raj

Reputation: 174874

You need to use re.search since re.match tries to match from the beging of the string. To match until a space or period is encountered.

re.search(r'(?<=test :)[^.\s]*',text)

To match all the chars until a period is encountered,

re.search(r'(?<=test :)[^.]*',text)

Upvotes: 19

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