在特定维度上选择numpy数组中的索引[重复]

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• Index n dimensional array with (n-1) d array 2 answers

索引n维数组与(n-1)d数组2答案

It is hard to find a clear title but an example will put it clearly. For example, my inputs are:

很难找到明确的标题,但一个例子会清楚地说明。例如,我的输入是:

c = np.full((4, 3, 2), 5)
c[:,:,1] *= 2

ix = np.random.randint(0, 2, (4, 3))

if ix is:

如果ix是:

array([[1, 0, 1],
       [0, 0, 1],
       [0, 0, 1],
       [1, 1, 0]])

if want as a result:

如果想要结果:

array([[10,  5, 10],
       [ 5,  5, 10],
       [ 5,  5, 10],
       [10, 10,  5]])

My c array can be of arbitrary dimensions, as well a the dimension I want to sample in.

我的c数组可以是任意尺寸,也是我想要采样的尺寸。

It sounds like interpolation, but I'm reluctant to construct a be array of indices each time I want to apply this. Is there a way of doing this using some kind of indexing on numpy arrays ? Or do I have to use some interpolation methods... Speed and memory are a concern here because I have to do this many times, and the arrays can be really large.

这听起来像插值,但每次我想应用它时,我都不愿意构造一个索引数组。有没有办法在numpy数组上使用某种索引来做到这一点?或者我必须使用一些插值方法...速度和内存是一个问题,因为我必须多次这样做,并且数组可能非常大。

Thanks for any insight !

感谢您的任何见解!

1 个解决方案

#1

Create the x, y indices with numpy.ogrid, and then use advanced indexing:

使用numpy.ogrid创建x,y索引,然后使用高级索引:

idx, idy = np.ogrid[:c.shape[0], :c.shape[1]]
c[idx, idy, ix]

#array([[10,  5, 10],
#       [ 5,  5, 10],
#       [ 5,  5, 10],
#       [10, 10,  5]])