Match any part of the word

asked Jul 19, 2011 at 12:42

/^p(r(e(v(i(o(u(s)?)?)?)?)?)?)?$/

And just to double check that it works:

for (qw/p pr pre previous prevalent previse/) {
    $result = (/^p(r(e(v(i(o(u(s)?)?)?)?)?)?)?$/? "yes" : "no");
    print "Checking $_: $resultn";
}

Produces:

Checking p: yes
Checking pr: yes
Checking pre: yes
Checking previous: yes
Checking prevalent: no
Checking previse: no

answered Jul 19, 2011 at 12:45

Chankey PathakChankey Pathak

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I suggest bookmarking the MSDN Regular Expression Quick Reference

you want to achieve a case insensitive match for the word «rocket» surrounded by non-alphanumeric characters. A regex that would work would be:

W*((?i)rocket(?-i))W*

What it will do is look for zero or more (*) non-alphanumeric (W) characters, followed by a case insensitive version of rocket ( (?i)rocket(?-i) ), followed again by zero or more (*) non-alphanumeric characters (W). The extra parentheses around the rocket-matching term assigns the match to a separate group. The word rocket will thus be in match group 1.

UPDATE 1:
Matt said in the comment that this regex is to be used in python. Python has a slightly different syntax. To achieve the same result in python, use this regex and pass the re.IGNORECASE option to the compile or match function.

W*(rocket)W*

On Regex101 this can be simulated by entering «i» in the textbox next to the regex input.

UPDATE 2 Ismael has mentioned, that the regex is not quite correct, as it might match «1rocket1». He posted a much better solution, namely

(?:^|W)rocket(?:$|W)

answered Apr 18, 2015 at 17:32

I think the look-aheads are overkill in this case, and you would be better off using word boundaries with the ignorecase option,

brocketb

In other words, in python:

>>> x="rocket's"
>>> y="rocket1."
>>> c=re.compile(r"brocketb",re.I)  # with the ignorecase option
>>> c.findall(y)
[]
>>> c.findall(x)
['rocket']

answered Apr 19, 2015 at 6:17

Use the Search for whole words only option.

As far as punctuations, you can’t answer it till you know the flavour/flavor.

It’s a very old thread, so posted for someone who might visit with a need, later. Ones who originated the thread might have moved to something else… No?

answered Nov 23, 2019 at 10:06

answered Mar 4, 2021 at 10:14

KishorKishor

1012 bronze badges

I don’t have enough reputation to comment, so I have to make a post to share why I think the user beroe’s solution is the best way to do this problem.
Take for example this string of text from the codewars challenge Most frequently used words in a text:

a a a b c c d d d d e e e e e

The goal of this challenge is to count the occurrences of words in the text.
If we go with the most popular solution:

(?:^|W)rocket(?:$|W)

in our string of text if we search for ‘a’ instead of ‘rocket’ using re.findall for python it will only return two matches (the first and last a), since the W capture overlaps the middle a from matching. Using b for the word barrier on the other hand returns all 3 a’s as matches

brocketb

Agian, credit to user beroe’s solution above

answered May 13, 2022 at 4:16