最簡單的方法來總結兩個維度？ （Python）

Question

我需要總結兩個維度的幫助。

比方說，我有[0,1,2]，[3,4,5]作爲我的尺寸，總結了這些數字將返回INT 15

def sum_dimensions(x): 
     x = [] 
     answer = sum(x) 
     return int(x) 

指出我的錯誤（ s）在代碼中表示讚賞。

Answer 1

這是你在找什麼：

data = [[0,1,2],[3,4,5]] 
sum([sum([item for item in ele]) for ele in data]) 
15 

正如@filmor指出，這可以簡化爲：

data = [[0,1,2],[3,4,5]] 
sum([sum(ele) for ele in data]) 
15 

至於爲您的代碼而言，如果你是經過[[0,1,2],[3,4,5]]作爲x的功能：

answer = sum(x) 

會THR ow一個TypeError。以下

return int(x) 

也沒有意義。使你的代碼的工作方式是：

def sum_dimensions(x): 
    answer = sum(x) 
    return answer 

data = [[0,1,2],[3,4,5]] 
total_sum = 0 
for sub_list in data: 
    total_sum += sum_dimensions(sub_list) 

Answer 2

它的Python，而不是C.

ar = [[0,1,2],[3,4,5]] 
result = sum ([sum(block) for block in ar]) 

Answer 3

如果你正在做其他線性代數，它可能是值得考慮numpy：

import numpy as np 

x = np.array([[0,1,2],[3,4,5]]) 
print np.sum(x) 

Answer 4

試試這個，

>>>sum([sum(i) for i in [[0,1,2],[3,4,5]]]) 

輸出：

15 

Answer 5

讓我們來看看這個。我用下面的數據我的機器上設置：

data = list(list(range(100000)) for i in range(1000)) 

我有以下結果：

In [13]: %%timeit            
sum(sum(ele) for ele in data) 
    ....: 
1 loops, best of 3: 1.15 s per loop 

In [14]: %%timeit            
sum([sum([item for item in ele]) for ele in data]) 
    ....: 
1 loops, best of 3: 3.78 s per loop 


In [15]: %%timeit            
sum(j for i in data for j in i) 
    ....: 
1 loops, best of 3: 4.92 s per loop 

In [16]: %%timeit            
sum(itertools.chain.from_iterable(data)) 
    ....: 
1 loops, best of 3: 1.61 s per loop 

In [18]: %%timeit 
sum(map(sum, data)) 
    ....: 
1 loops, best of 3: 1.16 s per loop 

不過，對於小數據集itertools變種比sum(sum變快2倍。 sum(map(sum似乎準確地映射到sum(sum(ele) for ele in data)構建

Answer 6

你可以試試這個：

>>> l = [[0,1,2],[3,4,5]] 
>>> 
>>> sum(a for v in l for a in v) 
15 

Answer 7

它可以通過map()和sum()也可以做：

a = [[0,1,2],[3,4,5]] 
sum(map(sum, a)) #15