Remove a pattern from list element and return another list in Python

Question

Let's assume I have a list like this

List=["Face123","Body234","Face565"]

I would like to obtain as output a list without character/substring described in another list.

NonDesideredPattern["Face","Body"]
Output=[123,234,565]. 

Answer 1

I am not sure, this is 100% efficient, but you could do something like this:

def eval_list(og_list):
    list_parts = []
    list_nums = []
    for element in og_list:
        part = ""
        num = ""
        for char in element:
            if char.isalpha():
                part += char
            else:
                num += char
        list_parts.append(part)
        list_nums.append(num)
    return list_parts, list_nums

(if you are always working with alphabetical syntax and then a number)

Answer 2

Create a function which returns a string without the undesired patterns. Then use this function in a comprehension list:

import re

def remove_pattern(string, patterns):
    result = string
    for p in patterns:
        result = re.sub(p, '', result)
    return result

inputs = ["Face123", "Body234", "Face565"]
undesired_patterns = ["Face", "Body"]
outputs = [remove_pattern(e, undesired_patterns) for e in inputs]

Answer 3

Use re.compile and re.sub

import re
lst = ["Face123", "Body234", "Face565"]
lst_no_desired_pattern = ["Face","Body"]
pattern = re.compile("|".join(lst_no_desired_pattern))
lst_output = [re.sub(pattern, "", word) for word in lst]

Result:

['123', '234', '565']