Question

我的任务是创建一个for循环来添加偶数并从数字列表中减去不均匀的数字。这样的事情：list = [(6,(+8),(-95),(+2),(+12),(+152),(+4),(+78),(-621),(-45))] 我不确定从哪里开始，但我到目前为止：

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用作变量名称，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)

确定其工作原理的奖励点; - ）

Answer 2

你可以使用conditional expression和for循环添加偶数并将奇数减去运行总数/ sm：

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

对于偶数，

if not ele % 2将为True，因为0是一个假值。

您还可以查看least significant bit, if ele & 1，如果True表示数字是奇数，或if ele & 0找到偶数数字：

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

print(sm)

可以将所有内容都放在generator expression中：

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

sum是内置函数list，因此请尽量避免将其用作变量名。

只是为了感兴趣一些时间：

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-=添加/减去总和。

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

你可以在一行中实现这一点，作为列表理解：

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

添加偶数，在Python中减去不均匀的数字

4 个答案: