How to convert string from list into integer, and then substract?

Question

I want to substract the two numbers found in each string values from the list array and convert into an integer.

current df:

                     pbp
0   ['0-0', '0-2', '0-4']
1   ['0-0', '2-0', '4-2']
2   ['0-0', '0-1', '0-2']

expected df:

            pbp
0   [0, -2, -4]
1   [0, 2, 2]
2   [0, -1, -2]

tried: df[['pbp']].applymap(lambda x: x.strip()).applymap(ast.literal_eval)

               pbp
0   [0-0, 0-2, 0-4]
1   [0-0, 2-0, 4-2]
2   [0-0, 0-1, 0-2]

I just don't know how to substract them

Answer 1

Here's a quick and naive way to achieve what you want:

def subtract_list(my_list):
    result=[]
    for val in my_list:
        numbers = val.split('-')
        res = int(numbers[0] )- int(numbers[1])
        result.append(res)
    print(result)
    return result

Calling the method would return:

subtract_list(['0-0', '0-2', '0-4'])

[0, -2, -4]

Answer 2

While not stated in the question this answer can cover multiple math operations using a pattern and re.split .

import re

def compute(data):
    output = []
    for entry in data:
        equation = re.split('([\/|\-|\+|\*\*|\*|%])', entry)

        result = {
           '*' : lambda e: int(e[0]) * int(e[2]),
           '+' : lambda e: int(e[0]) + int(e[2]),
           '-' : lambda e: int(e[0]) - int(e[2]),
           '\/': lambda e: int(e[0]) / int(e[2]),
           '%' : lambda e: int(e[0]) % int(e[2]),
        }[equation[1]](equation)

        output.append(result)
    return output