Numpy 4D数组切片 [英] Numpy 4d array slicing

问题描述

为什么切片4d数组会给我3d数组?我期望其中一个维度的范围为1的4d数组.

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

示例:

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

切片数组:

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

或

print X[11,:].shape

给了我(1, 96, 96)，但是我期望(1, 1, 96, 96)

我可以通过print X[11:12,:].shape做到这一点，但是我想知道为什么第一种方法不能按我预期的那样工作?

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

推荐答案

每文档:

整数i返回与i:i+1 相同的值，除了返回的对象的维数减小了1.特别是，第p个元素为整数(以及所有其他条目:)的选择元组将返回尺寸为N - 1的相应子数组.如果为N = 1，则返回的对象为数组标量.

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.

---

因此，当索引为整数时，将返回该索引处的值并删除相应的轴.在一个维度上，行为符合您的预期:

---

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是1维的，a[2]是0维的.

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

在较大的尺寸中，如果X是4维且形状为(1783,1,96,96)，则 X[11,:,:,:]返回所有值，其中第一个轴索引等于11，然后删除该轴.所以X[11,:,:,:].shape是(1,96,96).

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).

当切片指定一个范围(例如a[2:3])时，将返回该范围内的所有值，并且不删除轴:

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,)

类似地，X[11:12, :, :, :]的形状为(1,1,96,96).

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