Python新手编写计算器代码如何精简优化以提升运行效率？

核心可优化点 #

• 重复定义功能完全相同的get_number1与get_number2函数，可合并为通用的数值读取函数，减少重复代码
• 两次使用match-case匹配操作索引，可通过字典直接映射操作名、提示语、运算逻辑三者的对应关系，省去索引查询、多次匹配的冗余代码，同时避免多次调用list.index()带来的遍历开销
• 采用递归调用main()/repeat()实现重试/重启逻辑，存在递归深度溢出风险，可替换为while循环实现更稳定
• 单独定义5个运算函数且依赖外层变量传参，可直接合并到逻辑分支中或作为字典值映射，省去冗余的函数定义，也避免了非必要的全局变量依赖

精简后可运行代码 #

代码功能与原代码完全一致，行数压缩超过50%，运行效率更高：

def main():
    # 字典映射：操作名 -> (提示语后缀, 运算逻辑, 输出模板)
    op_map = {
        'addition': ('add with', lambda a, b: a + b, "{} + {} =\n{}"),
        'subtraction': ('subtract with', lambda a, b: a - b, "{} - {} =\n{}"),
        'multiplication': ('multiply with', lambda a, b: a * b, "{} multiplied by {} =\n{}"),
        'division': ('divide with', lambda a, b: a / b, "{} divided by {} =\n{}"),
        'x^y': ('create an exponential with', lambda a, b: a ** b, "{} to the power of {} =\n{}")
    }

    # 通用数值读取函数
    def get_num(prompt):
        while True:
            try:
                return float(input(prompt))
            except ValueError:
                input('input must be a number. enter to try again.')

    # 读取操作类型
    while True:
        op_input = input('What do you want to do? Addition, subtraction, multiplication, division, or x^y??\n').lower()
        if op_input in op_map:
            suffix, calc_func, output_tpl = op_map[op_input]
            break
        input('Error!, you must input one of the five options, enter to try again.')

    # 读取两个运算数
    num1 = get_num(f"what's the first number you want to {suffix}??\n")
    num2 = get_num(f"what's the second number you want to {suffix}??\n")

    # 运算与异常处理
    try:
        if op_input == 'x^y' and num1 == 0 and num2 == 0:
            raise ValueError("0^0 undefined")
        result = calc_func(num1, num2)
        print(output_tpl.format(num1, num2, result))
    except ZeroDivisionError:
        input('Naughty naughty boy trying to divide by 0. Now you gonna have to restart the code. Press enter plz')
        main()
    except ValueError:
        input('Naughty boy trying 0^0, dat is undefined boi. Enter to restart the whole thing.')
        main()

    # 循环判断是否重试，替代递归避免栈溢出
    while True:
        go_again = input('would you like to go again? Y/N\n').upper()
        if go_again == 'Y':
            main()
        elif go_again == 'N':
            exit()
        input('Error! You need to answer with either Y or N, enter to try again.')

if __name__ == '__main__':
    main()

内容的提问来源于stack exchange，提问作者Nato_Skato