迭代一个 numpy 数组 [英] Iterating over a numpy array

问题描述

是否有更简洁的替代方法:

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

我想出了这个:

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

这节省了一个缩进，但仍然很丑.

我希望看起来像这个伪代码:

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

这样的东西存在吗?

解决方案

我认为您正在寻找 ndenumerate.

<预><代码>>>>a =numpy.array([[1,2],[3,4],[5,6]])>>>对于(x，y)，numpy.ndenumerate(a)中的值:... 打印 x,y...0 00 11 01 12 02 1

关于性能.它比列表理解慢一点.

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范围(X.shape[2])])1 个循环，最好的 3 个:每个循环 376 毫秒%timeit 列表(np.ndenumerate(X))1 个循环，最好的 3 个:每个循环 570 毫秒

如果您担心性能，您可以通过查看 ndenumerate 的实现来进一步优化，它可以做两件事，转换为数组和循环.如果你知道你有一个数组，你可以调用平面迭代器的 .coords 属性.

a = X.flat%timeit list([(a.coords, x) for x in a.flat])1 个循环，最好的 3 个:每个循环 305 毫秒

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)

I came up with this:

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

Which saves one indentation, but is still pretty ugly.

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

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

Does anything like that exist?

解决方案

I think you're looking for the ndenumerate.

>>> a =numpy.array([[1,2],[3,4],[5,6]])
>>> for (x,y), value in numpy.ndenumerate(a):
...  print x,y
... 
0 0
0 1
1 0
1 1
2 0
2 1

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

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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