Can this Python postfix notation (reverse polish n

问题:

Here is a Python postfix notation interpreter which utilizes a stack to evaluate the expressions. Is it possible to make this function more efficient and accurate?

#!/usr/bin/env python


import operator
import doctest


class Stack:
    """A stack is a collection, meaning that it is a data structure that 
    contains multiple elements.

    """

    def __init__(self):
        """Initialize a new empty stack."""
        self.items = []       

    def push(self, item):
        """Add a new item to the stack."""
        self.items.append(item)

    def pop(self):
        """Remove and return an item from the stack. The item 
        that is returned is always the last one that was added.

        """
        return self.items.pop()

    def is_empty(self):
        """Check whether the stack is empty."""
        return (self.items == [])


# Map supported arithmetic operators to their functions
ARITHMETIC_OPERATORS = {"+":"add", "-":"sub", "*":"mul", "/":"div", 
                        "%":"mod", "**":"pow", "//":"floordiv"}


def postfix(expression, stack=Stack(), operators=ARITHMETIC_OPERATORS):
    """Postfix is a mathematical notation wherein every operator follows all 
    of its operands. This function accepts a string as a postfix mathematical 
    notation and evaluates the expressions. 

    1. Starting at the beginning of the expression, get one term 
       (operator or operand) at a time.
       * If the term is an operand, push it on the stack.
       * If the term is an operator, pop two operands off the stack, 
         perform the operation on them, and push the result back on 
         the stack.

    2. When you get to the end of the expression, there should be exactly 
       one operand left on the stack. That operand is the result.

    See http://en.wikipedia.org/wiki/Reverse_Polish_notation

    >>> expression = "1 2 +"
    >>> postfix(expression)
    3
    >>> expression = "5 4 3 + *"
    >>> postfix(expression)
    35
    >>> expression = "3 4 5 * -"
    >>> postfix(expression)
    -17
    >>> expression = "5 1 2 + 4 * + 3 -"
    >>> postfix(expression, Stack(), ARITHMETIC_OPERATORS)
    14

    """
    if not isinstance(expression, str):
        return
    for val in expression.split(" "):
        if operators.has_key(val):
            method = getattr(operator, operators.get(val))
            # The user has not input sufficient values in the expression
            if len(stack.items) < 2:
                return
            first_out_one = stack.pop()
            first_out_two = stack.pop()
            operand = method(first_out_two, first_out_one)
            stack.push(operand)
        else:
            # Type check and force int
            try:
                operand = int(val)
                stack.push(operand)
            except ValueError:
                continue
    return stack.pop()


if __name__ == '__main__':
    doctest.testmod()

回答1:

General suggestions:

• Avoid unnecessary type checks, and rely on default exception behavior.
• has_key() has long been deprecated in favor of the in operator: use that instead.
• Profile your program, before attempting any performance optimization. For a zero-effort profiling run of any given code, just run: python -m cProfile -s cumulative foo.py

Specific points:

• list makes a good stack out of the box. In particular, it allows you to use slice notation (tutorial) to replace the pop/pop/append dance with a single step.
• ARITHMETIC_OPERATORS can refer to operator implementations directly, without the getattr indirection.

Putting all this together:

ARITHMETIC_OPERATORS = {
    '+':  operator.add, '-':  operator.sub,
    '*':  operator.mul, '/':  operator.div, '%':  operator.mod,
    '**': operator.pow, '//': operator.floordiv,
}

def postfix(expression, operators=ARITHMETIC_OPERATORS):
    stack = []
    for val in expression.split():
        if val in operators:
            f = operators[val]
            stack[-2:] = [f(*stack[-2:])]
        else:
            stack.append(int(val))
    return stack.pop()

回答2:

Lists can be used directly as stacks:

>>> stack = []
>>> stack.append(3) # push
>>> stack.append(2)
>>> stack
[3, 2]
>>> stack.pop() # pop
2
>>> stack
[3]

You can also put the operator functions directly into your ARITHMETIC_OPERATORS dict:

ARITHMETIC_OPERATORS = {"+":operator.add,
                        "-":operator.sub,
                        "*":operator.mul,
                        "/":operator.div, 
                        "%":operator.mod,
                        "**":operator.pow,
                        "//":operator.floordiv}

then

if operators.has_key(val):
    method = operators[val]

The goal of these is not to make things more efficient (though it may have that effect) but to make them more obvious to the reader. Get rid of unnecessary levels of indirection and wrappers. That will tend to make your code less obfuscated. It will also provide (trivial) improvements in performance, but don't believe that unless you measure it.

Incidentally, using lists as stacks is fairly common idiomatic Python.

回答3:

You can directly map the operators: {"+": operator.add, "-": operator.sub, ...}. This is simpler, doesn't need the unnecessary getattr and also allows adding additional functions (without hacking the operator module). You could also drop a few temporary variables that are only used once anyway:

rhs, lhs = stack.pop(), stack.pop()
stack.push(operators[val](lhs, rhs)).    

Also (less of a performance and more of a style issue, also subjective), I would propably don't do error handling at all in the loop and wrap it in one try block with an except KeyError block ("Unknown operand"), an except IndexError block (empty stack), ...

But accurate? Does it give wrong results?