在Python中加上偶数，减去不均匀数

Question

我有一个创建for循环的任务，以添加偶数并从数字列表中减去不均匀数字。 像这样的东西： list = [(6,(+8),(-95),(+2),(+12),(+152),(+4),(+78),(-621),(-45))]我不确定从哪里开始，但我已经走了这么远：

list = [6,8,95,2,12,152,4,78,621,45]
sum = 0

for x in list:
    if list (0 % 2) == 0:
        sum = sum + list[x]
    elif list (0 % 2) != 0:
        sum = sum - list[x]
    return sum

虽然不确定如何遍历列表...

Answer 1

不要将list用作变量名，因为它会隐藏list数据类型！

lst = [6, 8, 95, 2, 12, 152, 4, 78, 621, 45]

total = 0
for x in lst:
    if x % 2:    # odd
        total -= x
    else:        # even
        total += x

编辑：只是为了好玩，您也可以尝试

total = sum(x * (1 - (x % 2 * 2)) for x in lst)

弄清楚它是如何工作的奖励积分;-)

Answer 2

您可以将条件表达式与for循环一起使用，以在运行的合计/ sm中添加偶数并从中减去奇数：

lst = [6,8,95,2,12,152,4,78,621,45]
sm = 0
for ele in lst:
    sm = sm + ele if not ele % 2 else sm - ele

if not ele % 2对于偶数将为True，因为0是一个假值。

您还可以检查最低有效位（ if ele & 1 （如果为True表示数字是奇数），或者if ele & 0以查找偶数：

sm = 0
for ele in lst:
    sm = sm - ele if ele & 1 else sm + ele

print(sm)

所有这些都可以放在生成器表达式中 ：

print(sum(-ele if ele & 1 else ele for ele in lst))

sum是一个内置函数，与list因此请尽量避免将其中任一个用作变量名。

只是出于兴趣一些时间：

In [8]: timeit sum(-x if x % 2 else x for x in lst)
1000000 loops, best of 3: 1.44 µs per loop

In [9]: %%timeit                                   
sm = 0
for ele in lst:
    sm = sm + ele if not ele % 2 else sm - ele
   ...: 
1000000 loops, best of 3: 1.12 µs per loop

In [11]: timeit sum(-ele if ele & 1 else ele for ele in lst)
1000000 loops, best of 3: 1.27 µs per loop

In [13]: %%timeit                                           
sm = 0
for ele in lst:
    sm = sm + ele if not ele % 2 else sm - ele
   ....: 
1000000 loops, best of 3: 1.11 µs per loop

In [14]: %%timeit                                           
sm = 0
for ele in lst:
    sm = sm - ele if ele & 1 else sm + ele
   ....: 
1000000 loops, best of 3: 875 ns per loop

In [15]: %%timeit
   ....: total = 0
   ....: for x in lst:
   ....:     if x % 2:    # odd
   ....:         total -= x
   ....:     else:        # even
   ....:         total += x 
1000000 loops, best of 3: 1.02 µs per loop

In [16]: timeit sum(x * (1 - (x % 2 * 2)) for x in lst)
100000 loops, best of 3: 2.2 µs per loop

Answer 3

使用+=和-=将总和相减。

num_list = [6,8,95,2,12,152,4,78,621,45]
sum = 0

for x in num_list:
    if x % 2 == 0:
        sum += x
    else:
        sum -= x

Answer 4

您实际上可以在一行中完成此操作，作为列表理解：

print sum([j * [1, -1][j % 2] for j in data])