Evaluate expression with infix operators in Python

infix.py

from functools import reduce

def inifix_operator(expression, variables=None):

    opr = set(['/','+','-','*'])
    ex = expression.strip().split()
    num_count = len(opr.difference(ex))
    num_opr = len(opr.intersection(ex))
    valid = num_count - num_opr

    if valid != 1:
        print("nothing")

    stack = []

    for i in ex[::-1]:
        print(i)
        # if i is a number
        if i.isdigit():
            stack.append(i)
            print(stack)
        else:
            # operand found
            if i == '+':
                stack = stack[:-2] + [sum([int(x) for x in stack[-2:][::-1]])]
                print(stack)
            elif i == '*':
                stack = stack[:-2] + [reduce(lambda x,y: x*y, [int(x) for x in stack[-2:][::-1]])]
                print(stack)
            elif i == '-':
                stack = stack[:-2] + [reduce(lambda x,y: x-y, [int(x) for x in stack[-2:][::-1]])]
                print(stack)
            elif i == '/':
                stack = stack[:-2] + [reduce(lambda x,y: x//y, [int(x) for x in stack[-2:][::-1]])]
                print(stack)

    print(stack)



if __name__ == '__main__':
    sample1 = "+ 6 * - 4 + 2 3 8"
    sample2 = "+ 3 8"

    inifix_operator(d_small)
    inifix_operator(d)

Note: The complexity of this solution is O(n). It works the following way:

1. Reverses the input string.
2. Append numbers to the stack until a operator is found.
3. As soon as the operator is found, it takes the last 2 numbers from the last, reverses them and makes computation. For example:
   [2,3,4,'+'] will become [2, 7] as the new list.
4. Repeat this logic for all operators.