python中的计算器 [英] Calculator in python

问题描述

我正在尝试制作一个可以使用基本4个运算符来求解表达式的计算器，例如1 + 2 * 3-4 / 5，但是它不起作用，我也不知道出了什么问题。请检查我的代码。
运行它时，我在8中得到无限个错误。line return ret（parts [0]）* ret（parts [2]）
这是代码

I am trying to make calculator that can solve expressions with basic 4 operators, like 1+2*3-4/5, however it does not work and I do not know what is wrong. Please check my code. When I run it, I am getting infinte number of errors in 8. line return ret(parts[0]) * ret(parts[2]) Here is code

def ret(s):
    s = str(s)
    if s.isdigit():
        return float(s)
    for c in ('*','/','+','-'):
        parts = s.partition(c)
        if c == '*':
            return ret(parts[0]) * ret(parts[2])
        elif c == '/':
            return ret(parts[0]) / ret(parts[2])
        elif c == '+':
            return ret(parts[0]) + ret(parts[2])
        elif c == '-':
            return ret(parts[0]) - ret(parts[2])
print(ret('1+2'))

错误回溯结束于：

  File "C:\Python33\calc.py", line 8, in ret
    return ret(parts[0]) * ret(parts[2])
  File "C:\Python33\calc.py", line 2, in ret
    s = str(s)
RuntimeError: maximum recursion depth exceeded while calling a Python object

推荐答案

无论如何，都对输入字符串进行分区，从不检查如果操作员在那里 .partition（）如果输入中不存在分区字符，则返回空字符串：

You partition the input string regardless, never checking if the operator is even there. .partition() returns empty strings if the partition character is not present in the input:

 >>> '1+1'.partition('*')
 ('1+1', '', '')

所以您将调用 s.partition（'*'），但不要检查是否有 这样的运算符当前，导致无条件调用 ret（）。您将始终调用 ret（parts [0]）* ret（parts [2]），而不管 * 是否出现在 s 中。

So you'll call s.partition('*') but never check if there is any such operator present, resulting in unconditional calls to ret(). You'll always call ret(parts[0]) * ret(parts[2]) regardless of wether * is present in s or not.

解决方案是测试运算符 first 或检查 .partition（）的返回值。后者可能是最简单的：

The solution is to either test for the operator first or to check the return value of .partition(). The latter is probably easiest:

for c in ('+','-','*','/'):
    parts = s.partition(c)
    if parts[1] == '*':
        return ret(parts[0]) * ret(parts[2])
    elif parts[1] == '/':
        return ret(parts[0]) / ret(parts[2])
    elif parts[1] == '+':
        return ret(parts[0]) + ret(parts[2])
    elif parts[1] == '-':
        return ret(parts[0]) - ret(parts[2])

请注意，我颠倒了操作员订单；是的，在加法和减法之前需要应用乘法和除法，但是您正在此处进行 reverse ；将表达式拆分为较小的部分，然后在处理子表达式后应用操作。

Note that I reversed the operator order; yes, multiplication and division need to be applied before addition and subtraction, but you are working in reverse here; splitting up the expression into smaller parts, and the operations are then applied when the sub-expression has been processed.

您可以使用赋值拆包来赋值3的返回值。 .partition（）以更简单的名称：

You could use assignment unpacking to assign the 3 return values of .partition() to easier names:

for c in ('+','-','*','/'):
    left, operator, right = s.partition(c)
    if operator == '*':
        return ret(left) * ret(right)
    elif operator == '/':
        return ret(left) / ret(right)
    elif operator == '+':
        return ret(left) + ret(right)
    elif operator == '-':
        return ret(left) - ret(right)

接下来，您可以使用 operator 模块，其功能与算术运算的功能相同。地图应该执行以下操作：

Next you can simplify all this by using the operator module, which has functions that perform the same operations as your arithmetic operations. A map should do:

import operator
ops = {'*': operator.mul, '/': operator.div, '+': operator.add, '-': operator.sub}

for c in ('+','-','*','/'):
    left, operator, right = s.partition(c)
    if operator in ops:
        return ops[operator](ret(left), ret(right))

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