python - 以运算符作为参数的函数

Question

我刚刚开始编写代码是为了好玩，我正在尝试构建一个使用用户输入的计算器。 2 个号码和 1 个运算符。我对编码真的很陌生，目前仅限于 if 语句和 while/for 循环的非常简单的使用，我刚刚开始研究函数。我一直试图将此代码放入函数中一段时间​​，但我找不到使用字符串“operator”作为函数中实际运算符的解决方案。

一定有办法让这一切变得更短。

if used_op == "+":
    print(">  " + str(number_1) + " + " + str(number_2) + " = " + str(number_1 + number_2) + "  <")
elif used_op == "-":
    print(">  " + str(number_1) + " - " + str(number_2) + " = " + str(number_1 - number_2) + "  <")
elif used_op == "*":
    print(">  " + str(number_1) + " * " + str(number_2) + " = " + str(number_1 * number_2) + "  <")
elif used_op == "/":
    print(">  " + str(number_1) + " / " + str(number_2) + " = " + str(number_1 / number_2) + "  <")
elif used_op == "%":
    print(">  " + str(number_1) + " % " + str(number_2) + " = " + str(number_1 % number_2) + "  <")
elif used_op == "**":
    print(">  " + str(number_1) + " ** " + str(number_2) + " = " + str(number_1 ** number_2) + "  <")
elif used_op == "//":
    print(">  " + str(number_1) + " // " + str(number_2) + " = " + str(number_1 // number_2) + "  <")

我尝试过的是这样的:

def solve(op):
    print(">  " + str(number_1) + op + str(number_2) + " = " + str(
        number_1 + **op** + number_2) + "  <")

solve(used_op)

我尝试在互联网上找到解决方案一段时间，但到目前为止我没有运气。

Answer 1

您可以使用字典和 operator 模块来执行您想要的操作:

import operator

# this will act like a sort of case statement or switch
operations = {
    '>': operator.gt,
    '<': operator.lt,
    '=': operator.eq,
    '+': operator.add,
    '-': operator.sub,
    '/': operator.div,
    '*': operator.mul,
    '**': operator.pow,
    '//': operator.floordiv,
    ... # so on and so forth
}

def calculate(num1, num2, op):
    # operation is a function that is grabbed from the dictionary
    operation = operations.get(op)
    if not operation:
         raise KeyError("Operation %s not supported"%op)

    # Call the operation with your inputs
    num3 = operation(num1, num2)
    print('%3.2f %s %3.2f = %3.2f' % (num1, op, num2, num3))


calculate(1,2, '+')
# 1.00 + 2.00 = 3.00