That should do for your particular case since you include a string within list. Therefore you need to flatten it:

new_list = [int(item) for sublist in a for item in sublist if item.isdigit()]

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.

Based on @alecxe solution you can also do it without any imports.

If your id numbers are always after id and have a fixed (7) number of digits I would probably just use .split('id ') to separate it and get the 7 digits from the second block onwards.

You can put them together in the desired format by using '; '.join()

Putting everything together:

pattern = ['; '.join([value[:7] for value in valueList.split('id ')[1:]]) for valueList in lst1]

Which prints out:

['3999595; 3999999', '3998895; 5555456; 3998899']

From this question:

l=[[100], [500], [300]]
result=[item for sublist in l for item in sublist]

From wikibooks:

def flatten(seq, list = None):
    """flatten(seq, list = None) -> list

    Return a flat version of the iterator `seq` appended to `list`
    """
    if list == None:
        list = []
    try:                          # Can `seq` be iterated over?
        for item in seq:          # If so then iterate over `seq`
            flatten(item, list)      # and make the same check on each item.
    except TypeError:             # If seq isn't iterable
        list.append(seq)             # append it to the new list.
    return list

Google is your friend ...

list_strings=['1','2','3']
for i in list_strings:
    num_list.append(int(i))
or

list_num = [int(x) for x in list_strings]

Check this sample code once.

You don't have return in your function. Also i[0] is not good in this context, as it will give you the 0th character from the string, so for 90 you'd get only 9 as an int, not 90...

from typing import List
items = ["90", "hello", "55", "Hi", "100"]


def extract_numbers(items: List[str])->List[int]:
    numbers = []
    for item in items:
        try:
            numbers.append(int(item))
        except Exception:
            print("Skipping an item which cannot be casted to an int!")
    return numbers

numbers = extract_numbers(items)
print(numbers)

Side notes:

• give variables (and functions) meaningful names
• use type hints when possible (it will help later on)
• do not use reserved keywords as names, not ever (e.g. list, dict, tuple etc. is always a bad name)
• return from a function if you need its output
• in general do NOT silently pass on exception, do some logging or at least a print

As for "why was the actual output none": By default function with no explicit return statement (or with return, but no variable returned) returns None. Give it a try:

def foo():pass

x = foo()
print(x) # It's gonna be None

def bar():
    return

y = bar()
print(y) # The same here