遍历一个numpy数组 [英] Iterating over a numpy array

问题描述

有没有那么冗长的替代方法呢?

Is there a less verbose alternative to this:

for x in xrange(array.shape[0]):
    for y in xrange(array.shape[1]):
        do_stuff(x, y)

我想到了这个

for x, y in itertools.product(map(xrange, array.shape)):
    do_stuff(x, y)

保存了一个缩进，但仍然很丑陋.

Which saves one indentation, but is still pretty ugly.

我希望看起来像下面的伪代码:

I'm hoping for something that looks like this pseudocode:

for x, y in array.indices:
    do_stuff(x, y)

有没有类似的东西存在?

Does anything like that exist?

推荐答案

我认为您正在寻找关于性能.它比列表理解要慢一些.

Regarding the performance. It is a bit slower than a list comprehension.

X = np.zeros((100, 100, 100))

%timeit list([((i,j,k), X[i,j,k]) for i in range(X.shape[0]) for j in range(X.shape[1]) for k in range(X.shape[2])])
1 loop, best of 3: 376 ms per loop

%timeit list(np.ndenumerate(X))
1 loop, best of 3: 570 ms per loop

如果您担心性能，可以通过查看ndenumerate的实现来进一步优化，该实现可以完成两件事，转换为数组并循环执行.如果知道有数组，则可以调用平面迭代器的.coords属性.

If you are worried about the performance you could optimise a bit further by looking at the implementation of ndenumerate, which does 2 things, converting to an array and looping. If you know you have an array, you can call the .coords attribute of the flat iterator.

a = X.flat
%timeit list([(a.coords, x) for x in a.flat])
1 loop, best of 3: 305 ms per loop

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