Python Regex return words containing double character

user_16590479 0
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I have a big string "a" and I want to extract all words with double characters from there. I need words as output (not lines or characters). I tried simple things like this:

re.findall(r'\b.*([A-Za-z])\1.*\b', a)

which gives a list of letters.

I also tried this:

pat='(\w*)\1'
def f(a, pat):
   pat = r'(\w*%s\w*)' % pat
   return re.findall(pat, a)
print (f(a,pat))

It gave me empty list. So I have two problems: 1) How to return words? 2) What regex to use to select words with double letters (like 'look', 'settle' etc).

python regex
3 Answers

You can try this:

import re


def find_words(text):
    words = re.findall(r'(\w*(\w)\2\w*)', text, re.IGNORECASE)
    return words

text = 'look book cat battle settle america google facebook gmail ball'
words = find_words(text=text)

for word, char in words:
    print(f'Word: {word} - Double Char: {char}')

Output:

Word: look - Double Char: o
Word: book - Double Char: o
Word: battle - Double Char: t
Word: settle - Double Char: t
Word: google - Double Char: o
Word: facebook - Double Char: o
Word: ball - Double Char: l
user_165552790

You can do this:

import re

def f(a):
   return [x[0] for x in re.findall(r'\b([a-z]*([a-z]+)\2[a-z]*)\b', a, re.IGNORECASE)]

print(f("We Choose To Go To The Moon"))

Which will output:

['Choose', 'Moon']

This works when there's multiple repeated letters (success) or when the repeated letters are different cases (Eel). Returns an empty array if there are no words with repeating letters.

user_165847780

1). The pattern you tried \b.*([A-Za-z])\1.*\b will:

  • match the first word boundary
  • then match 2 of the same chars from ranges a-z A-Z in the string
  • then match the last word boundary because .* can match any character so the whole match is not limited to a "word"

Then using re.findall will only return the capture group value, which is a single character.

2). In the pattern (\w*)\1 the quantifier for the word characters is 0 or more. It will for example match all positions as an empty string will also match, and could possible also match 2 times abc like abcabc

You can use a single capture group, and to get all the matches use re.finditer instead.

\w*([a-zA-Z])\1\w*

The pattern matches:

  • \w* Match optional word characters
  • ([a-zA-Z])\1 Match a single char from ranges a-z A-Z, followed by matching the exact same char using a backreference \1
  • \w* Match optional word characters

Regex demo | Python demo

import re

pattern = r"\w*([a-zA-Z])\1\w*"
s = 'settle, look, test'
for m in re.finditer(pattern, s):
    print(m.group())

Output

settle
look
user_54249880
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