finding substring within a list in Python

import re

a = ['aaa/aa/aa/a001','aaa/aa/aa/a002','aaa/aa/aa/a003']

for elem in a:
    print (re.sub(r"[^0-9]","",elem))

Output:

001
002
003

Or, with list comprehension:

numbers = [re.sub(r"[^0-9]","",x) for x in a]

print (numbers)

Output:

['001', '002', '003']

You could try regex:

import re
a = ['aaa/aa/aa/a001','aaa/aa/aa/a002','aaa/aa/aa/a003']
print([re.sub('[^\d+]', '', i) for i in a])

Output:

['001', '002', '003']

Or you could try using str.join:

print([''.join(x for x in i if x.isdigit()) for i in a])

Output:

['001', '002', '003']

Using a list comprehension with re.sub:

a = ['aaa/aa/aa/a001','aaa/aa/aa/a002','aaa/aa/aa/a003']
out = [re.sub(r'^.*?(\d+)$', r'\1', x) for x in a]
print(out)

This prints:

['001', '002', '003']

The strategy here is to match and capture the digits at the end of each string in the list. Note that this approach is robust even if digits might also appear earlier in the string.