求二维和的最简单方法?(Python)
我需要两个维度求和的帮助 假设我有[[0,1,2],[3,4,5]]作为维度,将这些数字相加将返回int 15求二维和的最简单方法?(Python),python,list,int,sum,dimension,Python,List,Int,Sum,Dimension,我需要两个维度求和的帮助 假设我有[[0,1,2],[3,4,5]]作为维度,将这些数字相加将返回int 15 def sum_dimensions(x): x = [] answer = sum(x) return int(x) 指出我在代码中的错误非常感谢。这就是您要寻找的: data = [[0,1,2],[3,4,5]] sum([sum([item for item in ele]) for ele in data]) 15 正如@f
def sum_dimensions(x):
x = []
answer = sum(x)
return int(x)
指出我在代码中的错误非常感谢。这就是您要寻找的:
data = [[0,1,2],[3,4,5]]
sum([sum([item for item in ele]) for ele in data])
15
正如@filmor所指出的,这可以简化为:
data = [[0,1,2],[3,4,5]]
sum([sum(ele) for ele in data])
15
至于您的代码,如果您将[[0,1,2],[3,4,5]]
作为x
传递给函数:
answer = sum(x)
将抛出一个TypeError
。以下
return int(x)
也没有道理。使代码正常工作的方法是:
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)
这是Python,不是C
ar = [[0,1,2],[3,4,5]]
result = sum ( [sum(block) for block in ar] )
如果你在做其他线性代数,可能值得考虑
numpy
:
import numpy as np
x = np.array( [[0,1,2],[3,4,5]] )
print np.sum(x)
试试这个
>>>sum([sum(i) for i in [[0,1,2],[3,4,5]]])
输出:
15
让我们来讨论一下。我在我的机器上使用了以下数据集:
data = list(list(range(100000)) for i in range(1000))
我有以下结果:
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
然而,对于小数据集,itertools
变量比sum(sum
变量)快2倍。sum(map(sum
似乎正好映射到sum(sum(ele)for ele in data)
构造您可以尝试以下方法:
>>> l = [[0,1,2],[3,4,5]]
>>>
>>> sum(a for v in l for a in v)
15
可以使用,也可以使用:
sum([item for item in ele])
可以写成sum(ele)
。或者sum(j代表i,数据中的j代表i)
更好,是:)这将在大型阵列上消耗额外的资源。当性能可能成为一个问题时,你应该寻找方法用调用math
和itertools
来替换Python循环。具体是什么?M4rtini的解决方案?我更喜欢不使用Numpy。忘了提了!这是关于过早优化的吗?嘿,我想是生成器的总和pression会比这更有效。很高兴知道!@Hyperboreus No。但是这项任务非常简单,并且允许具有相同代码复杂度的多个解决方案。因此,值得研究它们之间的比较,不是吗?例如,有趣的是,看到i结构中的j的数据中的i的j执行得有多糟糕,我没有预料到谢谢你的回复。
a = [[0,1,2],[3,4,5]]
sum(map(sum, a)) #15