'A program to evaluate arithmetic expression

Here's an interesting problem I haven't managed to deal with yet.

Given an arithmetic expression in Reverse Polish Notation, write a program to evaluate it.

The expression is given as a list of numbers and operands. For example [5, 3, '+'] should return 5 + 3 = 8.

For example,

[15, 7, 1, 1, '+', '-', '/', 3, '*', 2, 1, 1, '+', '+', '-']

should return 5 since it is equivalent to ((15 / (7 - (1 + 1))) * 3) - (2 + (1 + 1)) = 5.

pythonalgorithmarithmetic-expressions


Solution 1:[1]

This code would do the job:

ops = {
    "+": (lambda a, b: a + b),
    "-": (lambda a, b: a - b),
    "*": (lambda a, b: a * b),
    "/": (lambda a, b: a / b)
}


def pol(tokens):
    stack = []

    for token in tokens:
        # Check if the current element is an operator
        if token in ops:
            # Take the last two elements from the list
            arg2 = stack.pop()
            arg1 = stack.pop()
            # Execute an operation based on the current operator
            result = ops[token](arg1, arg2)
            # Append the result to the list in order to keep working with it
            stack.append(result)
        else:
            # If it is a number, just append in to the list
            stack.append(int(token))

    return stack.pop()


print(pol([15, 7, 1, 1, '+', '-', '/', 3, '*', 2, 1, 1, '+', '+', '-']))

Solution 2:[2]

Tried to solve it with the string approach.


import re

d = [15, 7, 1, 1, '+', '-', '/', 3, '*', 2, 1, 1, '+', '+', '-']

ops = {
    "+": (lambda a, b: a + b),
    "-": (lambda a, b: a - b),
    "*": (lambda a, b: a * b),
    "/": (lambda a, b: a / b)
}

while len(d) > 1:
    s_d = ' '.join(map(str, d))
    l_d = re.findall(r'\d+ \d+ \W', s_d)
    # print(l_d)
    for l1 in l_d:
        l = l1.split(' ')
        a = int(ops[l[-1]](int(l[0]),int(l[1])))
        s_d = s_d.replace(l1, str(int(a)))

    d = s_d.split()

    print(d)
print('----------------------')
print(d[0])

Sources

This article follows the attribution requirements of Stack Overflow and is licensed under CC BY-SA 3.0.

Source: Stack Overflow

Solution Source
Solution 1 enbermudas
Solution 2 Tanmay Jagtap

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