How to match a whole word with a regular expression?

I’m having trouble finding the correct regular expression for the scenario below:

Lets say:

a = "this is a sample"

I want to match whole word – for example match "hi" should return False since "hi" is not a word and "is" should return True since there is no alpha character on the left and on the right side.

Answers:

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Method 1

Try

re.search(r'bisb', your_string)

From the docs:

b Matches the empty string, but only at the beginning or end of a word.

Note that the re module uses a naive definition of “word” as a “sequence of alphanumeric or underscore characters”, where “alphanumeric” depends on locale or unicode options.

Also note that without the raw string prefix, b is seen as “backspace” instead of regex word boundary.

Method 2

Try using the “word boundary” character class in the regex module, re:

x="this is a sample"
y="this isis a sample."
regex=re.compile(r"bisb")  # For ignore case: re.compile(r"bisb", re.IGNORECASE)

regex.findall(y)
[]

regex.findall(x)
['is']

From the documentation of re.search().

b matches the empty string, but only at the beginning or end of a word

…

For example r'bfoob' matches 'foo', 'foo.', '(foo)', 'bar foo baz' but not 'foobar' or 'foo3'

Method 3

I think that the behavior desired by the OP was not completely achieved using the answers given. Specifically, the desired output of a boolean was not accomplished. The answers given do help illustrate the concept, and I think they are excellent. Perhaps I can illustrate what I mean by stating that I think that the OP used the examples used because of the following.

The string given was,

a = "this is a sample"

The OP then stated,

I want to match whole word – for example match "hi" should return False since "hi" is not a word …

As I understand, the reference is to the search token, "hi" as it is found in the word, "this". If someone were to search the string, a for the word "hi", they should receive False as the response.

The OP continues,

… and "is" should return True since there is no alpha character on the left and on the right side.

In this case, the reference is to the search token "is" as it is found in the word "is". I hope this helps clarify things as to why we use word boundaries. The other answers have the behavior of “don’t return a word unless that word is found by itself — not inside of other words.” The “word boundary” shorthand character class does this job nicely.

Only the word "is" has been used in examples up to this point. I think that these answers are correct, but I think that there is more of the question’s fundamental meaning that needs to be addressed. The behavior of other search strings should be noted to understand the concept. In other words, we need to generalize the (excellent) answer by @georg using re.match(r"bisb", your_string) The same r"bisb" concept is also used in the answer by @OmPrakash, who started the generalizing discussion by showing

>>> y="this isis a sample."
>>> regex=re.compile(r"bisb")  # For ignore case: re.compile(r"bisb", re.IGNORECASE)
>>> regex.findall(y)
[]

Let’s say the method which should exhibit the behavior I’ve discussed is named

find_only_whole_word(search_string, input_string)

The following behavior should then be expected.

>>> a = "this is a sample"
>>> find_only_whole_word("hi", a)
False
>>> find_only_whole_word("is", a)
True

Once again, this is how I understand the OP’s question. We have a step towards that behavior with the answer from @georg , but it’s a little hard to interpret/implement. to wit

>>> import re
>>> a = "this is a sample"
>>> re.search(r"bisb", a)
<_sre.SRE_Match object; span=(5, 7), match='is'>
>>> re.search(r"bhib", a)
>>>

There is no output from the second command. The useful answer from @OmPrakesh shows output, but not True or False.

Here’s a more complete sampling of the behavior to be expected.

>>> find_only_whole_word("this", a)
True
>>> find_only_whole_word("is", a)
True
>>> find_only_whole_word("a", a)
True
>>> find_only_whole_word("sample", a)
True
# Use "ample", part of the word, "sample": (s)ample
>>> find_only_whole_word("ample", a)
False
# (t)his
>>> find_only_whole_word("his", a)
False
# (sa)mpl(e)
>>> find_only_whole_word("mpl", a)
False
# Any random word
>>> find_only_whole_word("applesauce", a)
False
>>>

This can be accomplished by the following code:

#!/usr/bin/env python3
# -*- coding: utf-8 -*-
#
#@file find_only_whole_word.py

import re

def find_only_whole_word(search_string, input_string):
  # Create a raw string with word boundaries from the user's input_string
  raw_search_string = r"b" + search_string + r"b"

  match_output = re.search(raw_search_string, input_string)
  ##As noted by @OmPrakesh, if you want to ignore case, uncomment
  ##the next two lines
  #match_output = re.search(raw_search_string, input_string, 
  #                         flags=re.IGNORECASE)

  no_match_was_found = ( match_output is None )
  if no_match_was_found:
    return False
  else:
    return True

##endof:  find_only_whole_word(search_string, input_string)

A simple demonstration follows. Run the Python interpreter from the same directory where you saved the file, find_only_whole_word.py.

>>> from find_only_whole_word import find_only_whole_word
>>> a = "this is a sample"
>>> find_only_whole_word("hi", a)
False
>>> find_only_whole_word("is", a)
True
>>> find_only_whole_word("cucumber", a)
False
# The excellent example from @OmPrakash
>>> find_only_whole_word("is", "this isis a sample")
False
>>>

Method 4

The trouble with regex is that if hte string you want to search for in another string has regex characters it gets complicated. any string with brackets will fail.

This code will find a word

 word="is"
    srchedStr="this is a sample"
    if srchedStr.find(" "+word+" ") >=0  or 
       srchedStr.endswith(" "+word):
        <do stuff>

The first part of the conditional searches for the text with a space on each side and the second part catches the end of string situation. Note that the endwith is boolean whereas the find returns an integer

All methods was sourced from stackoverflow.com or stackexchange.com, is licensed under cc by-sa 2.5, cc by-sa 3.0 and cc by-sa 4.0