对列表中的数字求和，但遇到零后更改其符号

我需要对列表中的所有数字求和。如果发生0，则开始减法，直到另一个0，开始加法。

例如：

[1, 2, 0, 3, 0, 4] -> 1 + 2 - 3 + 4 = 4
[0, 2, 1, 0, 1, 0, 2] -> -2 - 1 + 1 - 2 = -4
[1, 2] -> 1 + 2 = 3
[4, 0, 2, 3] = 4 - 2 - 3 = -1

这是我尝试过的：

sss = 0

for num in numbers:
    if 0 == num:
        sss = -num
    else:
        sss += num
return sss

当列表的元素等于0时更改符号。

result = 0
current_sign = 1
for element in your_list:
    if element == 0:
       current_sign *= -1
    result += current_sign*element

这是一个解决方案，只要列表中的值为零，就可以在两个运算符（加减法）之间循环：

from operator import add, sub
from itertools import cycle

cycler = cycle([add, sub])
current_operator = next(cycler)

result = 0
my_list = [1, 2, 0, 3, 0, 4]

for number in my_list:
    if number == 0:
        current_op = next(cycler)
    else:
        result = current_operator(result, number)

尝试这个：

d = [1, 2, 0, 3, 0, 4]

sum = 0
sign = False
for i in d:
    if i == 0:
        if sign == False:
            sign = True
        else:
            sign = False
    else:
        if sign == False:
            sum += i
        else:
            sum -= i
print(sum)

operator模块和按位取反的另一种变化~：

import operator

def accum_on_zero(lst):
    res = 0
    ops, pos = (operator.add, operator.sub), 0
    for i in lst:
        if i == 0:
            pos = ~pos
        res = ops[pos](res, i)
    return res


print(accum_on_zero([1, 2, 0, 3, 0, 4]))     # 4
print(accum_on_zero([0, 2, 1, 0, 1, 0, 2]))  # -4 

lst = [1, 2, 0, 3, 0, 4]
print(sum([v * (-1) ** lst[0:i].count(0) for i, v in enumerate(lst)]))