Python regex, how to match a string for n times of occurrences

Question

I want to match the line that contains word 100.00% for 3 occurrences.

e.g

some string 100.00% foo 100.00% 100.00%
some string  99.91%  foo 100.00% 99.91%
some string 100.00%100.00%100.00%

So line 1 and 3 should all be matched.

However, my following regex seems only to match the case that has 100.00% repeated for 3 times. (Only match line 3)

re.search(r"([1][0][0]\.[0][0][%]){3}",string)

I wonder how should I do to match line 1 as well?

Thanks everybody!

Answer 1

Halemur Ali's answer is much cleaner than this one which I leave only for completeness.

The notation {3} is merely a shortcut, you can always replace it by expanding the repeated sequence. To match line where the sequence 100.00% appears exactly 3 times, you can use:

"^(?:(?!100\.00%).)*(?:100\.00%(?:(?!100\.00%).)*){3}$"

This uses negative-lookaheads.

It can be read this way:

• ^ start of the line
• (?:(?!100\.00%).)* any character (zero or more), until a 100.00% sequence
• (?:100\.00% the 100.00% sequence
• (?!100\.00%).)* followed by any character (zero or more) until a 100.00% sequence
• {3} repeated 3 times
• $ end of line

Consider taking a look at python's documentation of the re module.

Note: brackets are not necessary to match a single character.

Answer 2

an alternative that requires a simpler regular expression is to find all substrings that match 100.00% and test if the count == 3.

example

import re

p = re.compile(r'100\.00%')
texts = ['some string 100.00% foo 100.00% 100.00%',
         'some string  99.91%  foo 100.00% 99.91%',
         'some string 100.00%100.00%100.00%']

matches = [i for i, t in enumerate(texts)
           if len(re.findall(p, t)) == 3]
# matches = [0, 2]