Regex expression for words with length of even number-CodePudding

I want to write a regex expression for words with even-numbered length.

For example, the output I want from the list containing the words:
{"blue", "ah", "sky", "wow", "neat"} is {"blue", "ah", "neat}.

I know that the expression \w{2} or \w{4} would produce 2-worded or 4-worded words, but what I want is something that could work for all even numbers. I tried using \w{%2==0} but it doesn't work.

CodePudding user response：

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

CodePudding user response：

If you need no extra validation for the strings in your set, you can simply use

words = {"blue", "ah", "sky", "wow", "neat"}
print( list(w for w in words if len(w) % 2 == 0) )
# => ['ah', 'blue', 'neat']

See this Python demo.

If you want to make sure the words you return are made of letters, you can use

import re
words = {"blue", "ah", "sky", "wow", "neat"}
rx = re.compile(r'(?:[^\W\d_]{2}) ')   # For any Unicode letter words
# rx = re.compile(r'(?:[a-zA-Z]{2}) ') # For ASCII only letter words
print( [w for w in words if rx.fullmatch(w)] )
# => ['blue', 'ah', 'neat']

See this Python demo. A (?:[^\W\d_]{2}) pattern matches one or more occurrences of any two Unicode letters. Together with re.fullmatch, it requires a string to consist of an even amount of letters.