用运算符计算数字表[闭合]

按照@roganjosh的建议,创建一个dict并执行操作

import operator
ops = { "plus": operator.add, "minus": operator.sub,'multiply':operator.mul, 'divide':operator.div } 
a=[2025, 'minus', 5, 'plus',3] 
try:
    val=int(a[0])
    stack=[]
    error=False

    for i in range(1,len(a)):
        if isinstance(a[i],str):
            stack.append(a[i])
        if isinstance(a[i],int):
            temp_operator =stack.pop()
            operation=ops.get(temp_operator)
            val=operation(val,a[i])
except Exception:
    print('Invalid input')
    error=True
if(stack):
    print('Invalid input')
    error=True
if(not error):
    print(val)

2023

import operator

string_operator = {
    "plus" : operator.add,
    "minus" : operator.sub,
    "multiply" : operator.mul,
    "divide" : operator.truediv}

problem = [2025, "minus", 5, "plus", 3]

for i in range(len(problem)):
    if problem[i] in string_operator.keys():
        problem[i] = string_operator[problem[i]]
        solution = problem[i](problem[i-1],problem[i +1])
        problem[i+1] = solution

print(solution)

输出

(xenial)vash@localhost:~/python$ python3 helping.py  
2023

problem = [2, "multiply", 13] :

(xenial)vash@localhost:~/python$ python3 helping.py 
26

评论

这将遵循代码并按操作符的显示顺序处理操作符,不确定是否要遵循操作顺序,没有提及。

truediv floordiv ).

如果 problem 是运营商之一。然后将字符串转换为相应的运算符( problem[i] = string_operator[problem[i]] 它将采用之前的值( i-1 )以及( i+1

( solution = problem[i](problem[i-1], problem[i+1])

为了继续计算,我将输出存储在所述操作符后面的项中( 我+1 ),通过设置,它将是下一个运算符之前的项,这将允许进程继续。

problem = [26, "multiply", 100, "divide", 2, "plus", 40, "minus", 3]

(xenial)vash@localhost:~/python$ python3 helping.py 
1337

:)