RE is probably good for this, but as one RE answer has already been posted, I'll take your non-regex example and modify it:

One example I saw but wouldnt work was - 

I="I=2.7A"
[int(s) for s in I.split() if s.isdigit()]

The good thing is that split() can take arguments. Try this:

extracted = float("".join(i for i in I.split("=")[1] if i.isdigit() or i == "."))

Incidentally, here's a breakdown of the RE you supplied:

"\d+.\d+"
\d+ #match one or more decimal digits
. #match any character -- a lone period is just a wildcard
\d+ #match one or more decimal digits again

One way to do it (correctly) would be:

"\d+\.?\d*"
\d+ #match one or more decimal digits
\.? #match 0 or 1 periods (notice how I escaped the period)
\d* #match 0 or more decimal digits

If you only want to extract only positive integers, try the following:

>>> txt = "h3110 23 cat 444.4 rabbit 11 2 dog"
>>> [int(s) for s in txt.split() if s.isdigit()]
[23, 11, 2]

I would argue that this is better than the regex example because you don't need another module and it's more readable because you don't need to parse (and learn) the regex mini-language.

This will not recognize floats, negative integers, or integers in hexadecimal format. If you can't accept these limitations, jmnas's answer below will do the trick.

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