Regex to Match Words of a Certain Length

Regex to match words of a certain length

I think you want \b\w{1,10}\b. The \b matches a word boundary.

Of course, you could also replace the \b and do ^\w{1,10}$. This will match a word of at most 10 characters as long as its the only contents of the string. I think this is what you were doing before.

Since it's Java, you'll actually have to escape the backslashes: "\\b\\w{1,10}\\b". You probably knew this already, but it's gotten me before.

find words of length 4 using regular expression

Use word boundaries \b. When you add anchors in your regex like ^[a-zA-Z]{4}$, this would match the lines which have only four alphabets. It won't check for each individual words. ^ asserts that we are at the start and $ asserts that we are at the end. \b matches between a word character and a non-word character(vice versa). So it matches the start (zero width) of a word or end (zero width) of a word.

>>> s = "here we are having fun these days"
>>> re.findall(r'\b[a-zA-Z]{4}\b', s)
['here', 'days']

Regular expression: matching a string of a certain length that contains specific letters?

^(?=.*p)(?=.*o)(?=.*m)(?=.*s)(?=.*t)(?=.*h)(?=.*u).{10}$

Try something like this.This will work.

See demo.

http://regex101.com/r/yR3mM3/22

Match all words with a specific length

You're close! Just add a \b boundary to the front as well:

\b\w{1,2}\b

Matches:

I ABCDE FGH IG KLMNOPQ RS T
~           ~~         ~~ ~          

Regex matching word by exact word length

For the pattern that you tried you are using an anchor ^ which asserts the start of the string and $ which asserts the end $. If you don't want to match spaces you could use \S instead of .

What you could do instead is match 11 times a non whitespace char and check if it is not surrounded by non whitespace chars.

(?<!\S)\S{11}(?!\S)

Regex demo | Php demo

For example

$re = '~(?<!\S)\S{11}(?!\S)~';
$str = 'Hello match this C90.083635. 11 character long word not C90.083635.73G because it exceed 11 character and not C90.08363 because is not 11 character long.';

preg_match_all($re, $str, $matches);
print_r($matches[0]);

Output

Array
(
    [0] => C90.083635.
)

One option for the second part could be making use of word boundaries \b and use \S+ to match 1+ times a non whitespace char.

\bV\S+\.00\b

Regex demo

Regex expression for words with length of even number

You can repeat 2 word characters as a group between anchors ^ to assert the start and $ to assert the end of the string, or between word boundaries \b

^(?:\w{2})+$

See a regex demo.

import re

strings = [
    "blue",
    "ah",
    "sky",
    "wow",
    "neat"
]

for s in strings:
    m = re.match(r"(?:\w{2})+$", s)
    if m:
        print(m.group())

Output

blue
ah
neat

Regex: How to find a pattern on words of a certain length

If I understood you correctly, this should work:

(?<=\w{3})(s|es)\b /i

Be advised that last /i is not part of regex, it's just case-insensitive flag. You also may want to add g and m flags to read entire string. Here is the breakdown:

• (?<=\w{3}) - positive lookbehind, checking that there are 3 characters preceding the following pattern
• (s|es) - a capture group, looking for characters s or es
• \b - checking that end of a word follows right after the pattern.

Also be advised that this pattern does not differentiate words which ends with s in singular form (like proteus), and i'm very doubtful this task can be properly done by regular expression only.

Match word of specific length

It's matching because it found a string of 5 lowercase letters within the string of longer length. You need to adapt your regex so that the "word" match is surrounded by white space. Don't forget to also address the start/end of the string in the "word" boundary.

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