在不使用多个函数的情况下使用python评估数学表达式的算法

Question

I am trying to evaluate infix expressions using Stacks in python.

I want to know if I have the following infix expressions below:

'2 ^ ( 1 + 3 ^ 2 )'
'( 3 * 5 ) - ( 1 > 2 > 3 < 4 )'
'4 ^ ( 10 < 2 ) / 5 + 100'

how I would go on about evaluating this using 2 stacks and not calling upon multiple other functions, so that the output would be:

1024
12
103

I'm just wondering what the algorithm would be to do this. I know the standard way is to change the infix to postfix and then solve, but is there a quicker way to do this in one single function. Thank you very much.

Answer 1

You might want to search for the "shunting yard algorithm" . It's a pretty fundamental expression evaluation algorithm, and it uses two stacks (although one is called the 'output queue').

Basically, you filter numbers from operators. Numbers go onto the output stack, operators go onto the "holding" stack. When an operator comes along that has lower precedence than what's on the holding stack, you move the contents of the holding stack to the output stack until either the holding stack is empty, or the item on the holding stack has lower precedence than the input operator.

Precedence

Remember that a + b * c - d is evaluated as (a + (b*c)) - d . Remember also that exponentiation has higher precedence than multiplication, so a * b ^ c is going to be a * (b ^ c) .

EDIT:

Here's some code that doesn't work. I don't know what your operators '>' and '<' are. Apparently (10 < 2) should be 2, from the third expression?

I just implemented them as C-style booleans (1 for true, 0 false).

There's a heap of redundant code in there, because only one function. Feel free to clean that up. It modifies the Shunting Yard algorithm to perform the RPN computations on the fly. I've left in a bunch of print statements that illustrate the contents of the lists that I'm using as stacks. Feel free to do a proper job with your stack classes. In my version, pop() is pop, and append() is push.

tests = [
    '2 ^ ( 1 + 3 ^ 2 )',
    '( 3 * 5 ) - ( 1 > 2 > 3 < 4 )',
    '4 ^ ( 10 < 2 ) / 5 + 100',
]

expected = [
    1024,
    12,
    103,
]

def expr_eval(s):
    print("EXPR:", s)
    tokens = s.split()
    output_stk = []
    operator_stk = []

    precedence = {
        '(': 0,
        '+':1, '-':1, '<':1, '>':1,
        '*':2, '/':2,
        '^':3,
    }

    for t in tokens:
        print("OUT:", output_stk, "OP:", operator_stk)
        print("Tok: ",t)
        if t.isdigit():
            output_stk.append(int(t))
            continue
        elif t == '(':
            operator_stk.append(t)
            continue
        elif t == ')':
            # End of subexpression. Do math until we find opening (
            while operator_stk:
                op = operator_stk.pop()
                if op == '(':
                    break

                b = output_stk.pop()
                a = output_stk.pop()
                if op == '-': output_stk.append(a - b)
                elif op == '+': output_stk.append(a + b)
                elif op == '<': output_stk.append(1 if a < b else 0)
                elif op == '>': output_stk.append(1 if a > b else 0)
                elif op == '*': output_stk.append(a * b)
                elif op == '/': output_stk.append(a // b)
                elif op == '^': output_stk.append(a ** b)
                else:
                    raise Exception("Unknown operator: %s" % op)
                print("OUT:", output_stk, "OP:", operator_stk)
            continue

        # Not a number - check operator precedence
        prec_t = precedence[t]
        while operator_stk and prec_t <= precedence[operator_stk[-1]]:
            print("OUT:", output_stk, "OP:", operator_stk)
            op = operator_stk.pop()
            b = output_stk.pop()
            a = output_stk.pop() # 'a' went on first!
            if op == '-': output_stk.append(a - b)
            elif op == '+': output_stk.append(a + b)
            elif op == '<': output_stk.append(1 if a < b else 0)
            elif op == '>': output_stk.append(1 if a > b else 0)
            elif op == '*': output_stk.append(a * b)
            elif op == '/': output_stk.append(a // b)
            elif op == '^': output_stk.append(a ** b)
            else:
                raise Exception("Unknown operator: %s" % op)
        operator_stk.append(t)

    print("OUT:", output_stk, "OP:", operator_stk)

    while operator_stk:
        op = operator_stk.pop()
        if op == '(':
            raise Exception('Mismatched opening parenthesis!')

        b = output_stk.pop()
        a = output_stk.pop() # 'a' went on first!
        if op == '-':   output_stk.append(a - b)
        elif op == '+': output_stk.append(a + b)
        elif op == '<': output_stk.append(1 if a < b else 0)
        elif op == '>': output_stk.append(1 if a > b else 0)
        elif op == '*': output_stk.append(a * b)
        elif op == '/': output_stk.append(a // b)
        elif op == '^': output_stk.append(a ** b)
        else:
            raise Exception("Unknown operator: %s" % op)
        print("OUT:", output_stk, "OP:", operator_stk)

    return output_stk.pop()

for i in range(len(tests)):
    r = expr_eval(tests[i])
    if r == expected[i]:
        print("PASS: %d = %s" % (r, tests[i]))
    else:
        print("FAIL: %d != %d = %s" % (r, expected[i], tests[i]))