CODEWITHSUNDEEP
pythonnumpy
mrgloom
mrgloom

Reputation: 21622

Numpy 4d array slicing

Why does slicing a 4d array give me a 3d array? I expected a 4d array with extent 1 in one of the dimensions.

Example:

print X.shape
(1783, 1, 96, 96)

Slice array:

print X[11,:,:,:].shape

or

print X[11,:].shape

gives me (1, 96, 96), but I expected (1, 1, 96, 96)

I can do it by print X[11:12,:].shape, but I wonder why the first method doesn't work as I expect?

Upvotes: 3

Views: 8201

Answers (1)

unutbu
unutbu

Reputation: 879591

Per the docs:

An integer, i, returns the same values as i:i+1 except the dimensionality of the returned object is reduced by 1. In particular, a selection tuple with the p-th element an integer (and all other entries :) returns the corresponding sub-array with dimension N - 1. If N = 1 then the returned object is an array scalar.

Thus, when your index is an integer, the value(s) at that index is(are) returned and the corresponding axis is removed. In one dimension the behavior is as you would expect:

In [6]: a = np.arange(5); a
Out[6]: array([0, 1, 2, 3, 4])

In [7]: a[2]
Out[7]: 2

In [8]: a[2].shape
Out[8]: ()

a is 1-dimensional, a[2] is 0-dimensional.

In higher dimensions, if X is 4-dimensional and of shape (1783,1,96,96), then X[11,:,:,:] returns all the values where the first axis index equals 11 and then that axis is removed. So X[11,:,:,:].shape is (1,96,96).

When the slice specifies a range, such as a[2:3] then all the values within that range are returned and the axis is not removed:

In [9]: a[2:3]
Out[9]: array([2])

In [10]: a[2:3].shape
Out[10]: (1,)

Similarly, X[11:12, :, :, :] has shape (1,1,96,96).

Upvotes: 7

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