Extract digits from a string within a word

Question

I want a regular expression, which returns only digits, which are within a word, but I can only find expressions, which returns all digits in a string.

I’ve used this example:
text = 'I need this number inside my wor5d, but also this word3 and this 4word, but not this 1 and not this 555.'

The following code returns all digits, but I am only interested in [‘5’, ‘3’, ‘4’]
import re print(re.findall(r'd+', text))

Any suggestions?

Asked By: Kiri

||

Source

Answer 1

You can use

re.findall(r'(?<=[a-zA-Z])d+|d+(?=[a-zA-Z])', text)

This regex will extract all one or more digit chunks that are immediately preceded or followed with an ASCII letter.

A fully Unicode version for Python re would look like

(?<=[^Wd_])d+|d+(?=[^Wd_])

where [^Wd_] matches any Unicode letter.

See the regex demo for reference.

Answered By: Wiktor Stribiżew

Answer 2

An approach with str.translate, without the use of regex or re module:

from string import ascii_letters

delete_dict = {sp_character: '' for sp_character in ascii_letters}
table = str.maketrans(delete_dict)

text = 'I 77! need 1:5 this number inside my wor5d, but also this word3 and this 4word, but not this 1 and not this 555.'

print([res for s in text.rstrip('.').split()
       if not (s2 := s.rstrip(',')).isnumeric() and (res := s2.translate(table)) and res.isnumeric()])

Out:

['5', '3', '4']

Performance

I was curious so I did some benchmark tests to compare performance against other approaches. Looks like str.translate is faster even than the regex implementation.

Here is my benchmark code with timeit:

import re
from string import ascii_letters
from timeit import timeit


_NUM_RE = re.compile(r'(?<=[a-zA-Z])d+|d+(?=[a-zA-Z])')

delete_dict = {sp_character: '' for sp_character in ascii_letters}
_TABLE = str.maketrans(delete_dict)

text = 'I need this number inside my wor5d, but also this word3 and this 4word, but not this 1 and not this 555.'


def main():
    n = 100_000

    print('regex:         ', timeit("re.findall(r'(?<=[a-zA-Z])d+|d+(?=[a-zA-Z])', text)",
                 globals=globals(), number=n))

    print('regex (opt):   ', (timeit("_NUM_RE.findall(text)",
                 globals=globals(), number=n)))

    print('iter_char:     ', timeit("""
k=set()
for x in range(1,len(text)-1):
    if text[x-1].isdigit() and text[x].isalpha():
        k.add(text[x-1])
    if text[x].isdigit() and text[x+1].isalpha():
        k.add(text[x])
    if text[x-1].isalpha() and text[x].isdigit() and text[x+1].isalpha():
        k.add(text[x])
    if text[x-1].isalpha() and text[x].isdigit():
        k.add(text[x])
    """, globals=globals(), number=n))

    print('str.translate: ', timeit("""
[
    res for s in text.rstrip('.').split()
    if not (s2 := s.rstrip(',')).isnumeric() and (res := s2.translate(_TABLE)) and res.isnumeric()
]
    """, globals=globals(), number=n))


if __name__ == '__main__':
    main()

Results (Mac OS X – M1):

regex:          0.5315765410050517
regex (opt):    0.5069837079936406
iter_char:      2.5037198749923846
str.translate:  0.37348733299586456

Answered By: rv.kvetch

Extract digits from a string within a word

Question:

Answers:

Performance