在Python中如何压缩这个if语句？

我目前正在用Python创建一个计算器，遇到了一个小问题：if-else语句中有28个条件：

if operation0 == "+" and operation1 == "+" and operation2 == "+": # First operation is addition
    print(number0 + number1 + number2 + number3)
elif operation0 == "+" and operation1 == "+" and operation2 == "*":
    print(number0 + number1 + number2 * number3)
elif operation0 == "+" and operation1 == "+" and operation2 == "^":
    print(number0 + number1 + number2 ** number3)
elif operation0 == "+" and operation1 == "*" and operation2 == "+":
    print(number0 + number1 * number2 + number3)
elif operation0 == "+" and operation1 == "*" and operation2 == "*":
    print(number0 + number1 * number2 * number3)
elif operation0 == "+" and operation1 == "*" and operation2 == "^":
    print(number0 + number1 * number2 ** number3)
elif operation0 == "+" and operation1 == "^" and operation2 == "+":
    print(number0 + number1 ** number2 + number3)
elif operation0 == "+" and operation1 == "^" and operation2 == "*":
    print(number0 + number1 ** number2 * number3)
elif operation0 == "+" and operation1 == "^" and operation2 == "^":
    print(number0 + number1 ** number2 ** number3)
elif operation0 == "*" and operation1 == "+" and operation2 == "+": # First operation is multiplication
    print(number0 * number1 + number2 + number3)
elif operation0 == "*" and operation1 == "+" and operation2 == "*":
    print(number0 * number1 + number2 * number3)
elif operation0 == "*" and operation1 == "+" and operation2 == "^":
    print(number0 * number1 + number2 ** number3)
elif operation0 == "*" and operation1 == "*" and operation2 == "+":
    print(number0 * number1 * number2 + number3)
elif operation0 == "*" and operation1 == "*" and operation2 == "*":
    print(number0 * number1 * number2 * number3)
elif operation0 == "*" and operation1 == "*" and operation2 == "^":
    print(number0 * number1 * number2 ** number3)
elif operation0 == "*" and operation1 == "^" and operation2 == "+":
    print(number0 * number1 ** number2 + number3)
elif operation0 == "*" and operation1 == "^" and operation2 == "*":
    print(number0 * number1 ** number2 * number3)
elif operation0 == "*" and operation1 == "^" and operation2 == "^":
    print(number0 * number1 ** number2 ** number3)
elif operation0 == "^" and operation1 == "+" and operation2 == "+": # First operation is exponentiation
    print(number0 ** number1 + number2 + number3)
elif operation0 == "^" and operation1 == "+" and operation2 == "*":
    print(number0 ** number1 + number2 * number3)
elif operation0 == "^" and operation1 == "+" and operation2 == "^":
    print(number0 ** number1 + number2 ** number3)
elif operation0 == "^" and operation1 == "*" and operation2 == "+":
    print(number0 ** number1 * number2 + number3)
elif operation0 == "^" and operation1 == "*" and operation2 == "*":
    print(number0 ** number1 * number2 * number3)
elif operation0 == "^" and operation1 == "*" and operation2 == "^":
    print(number0 ** number1 * number2 ** number3)
elif operation0 == "^" and operation1 == "^" and operation2 == "+":
    print(number0 ** number1 ** number2 + number3)
elif operation0 == "^" and operation1 == "^" and operation2 == "*":
    print(number0 ** number1 ** number2 * number3)
elif operation0 == "^" and operation1 == "^" and operation2 == "^":
    print(number0 ** number1 ** number2 ** number3)
else:
    print("Error")

---

我试过的东西都不管用，我在这个网站上也找不到任何东西来帮助我压缩代码。如此多的重复ELIF是不合理的，任何数量的减少都是值得赞赏的。

您可以使用操作符库并尝试类似的操作

from operator import pow, truediv, mul, add, sub  

operators = {
  '+': add,
  '-': sub,
  '*': mul,
  '/': truediv
}

def calculate(s):
    if s.isdigit():
        return float(s)
    for c in operators.keys():
        left, operator, right = s.partition(c)
        if operator in operators:
            return operators[operator](calculate(left), calculate(right))

calc = input("Type calculation:\n")
print("Answer: " + str(calculate(calc)))

---

其他一些解决方案也可以在

操作顺序中找到，这很困难，但并非不可能：

#执行操作的辅助函数
def doop（num1、op、num2）：
如果op==“^”：
返回num1**num2
如果op==“+”：
返回num1+num2
如果op==“-”：
返回num1-num2
如果op==“*”：
返回num1*num2
如果op==“/”：
返回num1/num2
#如果我们无法识别该操作，则会出现错误
引发异常（“无效操作”）
#具有运算符优先级的集合列表
优先权=[
{"^"},
{"*", "/"},
{"+", "-"}
]
#测试用硬编码值
nums=[1,2,3,4]
ops=[“+”、“-”、“*”]
#这表示1+2-3*4
#从最高到最低优先级开始
对于优先的prec：
#我们必须使用while循环来手动控制'i'`
i=0
而我

免责声明：解决方案可能会使用各种相关概念，但这些概念与此处的解释不太相关

要获得优先级，必须正确解析输入。有几种方法可以做到这一点；一个常见的是Dijkstra算法。
由于这只是一个示例，我们将放弃良好实践（错误处理、类）以获得更简洁的代码

首先，我们需要列出先例。更高的优先级意味着操作符具有更高的优先级，因此它执行得更早

先例={
'^': 3,
'*': 2, '/': 2,
'+': 1, '-': 1
}

然后，我们需要将操作与每个操作符关联起来。为此，我们可以使用

操作员

模块中的功能：

从操作员导入pow、truediv、mul、add、sub
do={
"(二):战俘,，
“*”：mul“/”：truediv，
“+”：添加“-”：子
}

最后，我们需要实现决定何时计算表达式的逻辑。首先，我们比操作多了一个数字，所以我们将其放在堆栈上：

def分流（编号、操作）：
堆栈=[numbers.pop（0）]
运算符=[]

然后，我们处理每个运算符对

while len（数字）：
运算符=operations.pop（0）
优先级=优先级[运算符]

当一个操作的优先级（执行时间早于当前操作）高于当前操作时（使用

do

dictionary）对其进行求值。我们从堆栈中删除左操作数和右操作数，并从

do

中查找适当的函数以计算结果：

while运算符和优先级[运算符[-1]]>=优先级：
[左，右]，堆栈[-2:]=堆栈[-2:]，[]
stack.append（do[operators.pop（）]（左、右））

然后，我们将当前运算符和编号附加到堆栈：

operators.append（运算符）
stack.append（numbers.pop（0））

所有数字和运算符用完后，我们评估其余的运算符，直到只剩下一个数字，然后返回：

而len（stack）>1：
[左，右]，堆栈[-2:]=堆栈[-2:]，[]
stack.append（do[operators.pop（）]（左、右））
返回堆栈[0]

把它们放在一起（）：

从操作员导入pow、truediv、mul、add、sub
先例={
'^': 3,
'*': 2, '/': 2,
'+': 1, '-': 1
}
do={
"(二):战俘,，
“*”：mul“/”：truediv，
“+”：添加“-”：子
}
def分流（数量、操作）：
堆栈=[numbers.pop（0）]
运算符=[]
而len（数字）：
运算符=operations.pop（0）
优先级=优先级[运算符]
而运算符和优先级[运算符[-1]]>=优先级：
[左，右]，堆栈[-2:]=堆栈[-2:]，[]
stack.append（do[operators.pop（）]（左、右））
运算符。追加（运算符）
stack.append（numbers.pop（0））
而len（stack）>1：
[左，右]，堆栈[-2:]=堆栈[-2:]，[]
stack.append（do[operators.pop（）]（左、右））
返回堆栈[0]

对多个值执行公共处理时，应使用列表而不是单个变量。这允许您将通用逻辑应用于元素，并基于可存储在变量中的索引对其进行操作：

def compute(*operNums):
    operNums   = list(operNums)
    numbers    = operNums[::2]  # [number0,number1,number2,number3]
    operations = operNums[1::2] # [operation0,operation1,operation2]
    while operations:           # process list until all operations performed
        for op in ("^","**","*","+"):          # find highest priority operation
            if op not in operations: continue
            i = operations.index(op)           # index of operation
            operations.pop(i)                  # remove completed operation
            if   op in ["^","**"]: numbers[i] **= numbers.pop(i+1) # remove right side
            elif op == "*":        numbers[i]  *= numbers.pop(i+1) # left side number
            elif op == "+":        numbers[i]  += numbers.pop(i+1) # is replaced by result   
            else: return "ERROR"
    return numbers[0]

print(compute(1,"+",2,"**",3,"*",5)) # 41

---

就像karl的回答一样，不处理先例我认为最好的是实现.aplet的解释在哪里我认为更专业的异常会更好，例如，

算术错误

。