在Python中加上偶數，減去不均勻數

Question

我有一個創建for循環的任務，以添加偶數並從數字列表中減去不均勻數字。 像這樣的東西： list = [(6,(+8),(-95),(+2),(+12),(+152),(+4),(+78),(-621),(-45))]我不確定從哪里開始，但我已經走了這么遠：

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

雖然不確定如何遍歷列表...

Answer 1

不要將list用作變量名，因為它會隱藏list數據類型！

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

編輯：只是為了好玩，您也可以嘗試

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

弄清楚它是如何工作的獎勵積分;-)

Answer 2

您可以將條件表達式與for循環一起使用，以在運行的合計/ sm中添加偶數並從中減去奇數：

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對於偶數將為True，因為0是一個假值。

您還可以檢查最低有效位（ if ele & 1 （如果為True表示數字是奇數），或者if ele & 0以查找偶數：

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

print(sm)

所有這些都可以放在生成器表達式中 ：

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

sum是一個內置函數，與list因此請盡量避免將其中任一個用作變量名。

只是出於興趣一些時間：

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-=將總和相減。

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

您實際上可以在一行中完成此操作，作為列表理解：

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