沿特定轴将另一个数组排序为一个numpy数组 [英] Sort a numpy array by another array, along a particular axis
问题描述
类似于此答案，我有一对3D numpy数组a和b，我想按a的值对b的条目进行排序.与此答案不同，我想进行排序仅沿数组的一个轴.
Similar to this answer, I have a pair of 3D numpy arrays, a and b, and I want to sort the entries of b by the values of a. Unlike this answer, I want to sort only along one axis of the arrays.
我对numpy.argsort()文档的幼稚阅读:
Returns
-------
index_array : ndarray, int
Array of indices that sort `a` along the specified axis.
In other words, ``a[index_array]`` yields a sorted `a`.
让我相信我可以使用以下代码进行排序:
led me to believe that I could do my sort with the following code:
import numpy
a = numpy.zeros((3, 3, 3))
a += numpy.array((1, 3, 2)).reshape((3, 1, 1))
print "a"
print a
"""
[[[ 1. 1. 1.]
[ 1. 1. 1.]
[ 1. 1. 1.]]
[[ 3. 3. 3.]
[ 3. 3. 3.]
[ 3. 3. 3.]]
[[ 2. 2. 2.]
[ 2. 2. 2.]
[ 2. 2. 2.]]]
"""
b = numpy.arange(3*3*3).reshape((3, 3, 3))
print "b"
print b
"""
[[[ 0 1 2]
[ 3 4 5]
[ 6 7 8]]
[[ 9 10 11]
[12 13 14]
[15 16 17]]
[[18 19 20]
[21 22 23]
[24 25 26]]]
"""
print "a, sorted"
print numpy.sort(a, axis=0)
"""
[[[ 1. 1. 1.]
[ 1. 1. 1.]
[ 1. 1. 1.]]
[[ 2. 2. 2.]
[ 2. 2. 2.]
[ 2. 2. 2.]]
[[ 3. 3. 3.]
[ 3. 3. 3.]
[ 3. 3. 3.]]]
"""
##This isnt' working how I'd like
sort_indices = numpy.argsort(a, axis=0)
c = b[sort_indices]
"""
Desired output:
[[[ 0 1 2]
[ 3 4 5]
[ 6 7 8]]
[[18 19 20]
[21 22 23]
[24 25 26]]
[[ 9 10 11]
[12 13 14]
[15 16 17]]]
"""
print "Desired shape of b[sort_indices]: (3, 3, 3)."
print "Actual shape of b[sort_indices]:"
print c.shape
"""
(3, 3, 3, 3, 3)
"""
什么是正确的方法?
推荐答案
您仍然必须为其他两个维度提供索引，以使其正常工作.
You still have to supply indices for the other two dimensions for this to work correctly.
>>> a = numpy.zeros((3, 3, 3))
>>> a += numpy.array((1, 3, 2)).reshape((3, 1, 1))
>>> b = numpy.arange(3*3*3).reshape((3, 3, 3))
>>> sort_indices = numpy.argsort(a, axis=0)
>>> static_indices = numpy.indices((3, 3, 3))
>>> b[sort_indices, static_indices[1], static_indices[2]]
array([[[ 0, 1, 2],
[ 3, 4, 5],
[ 6, 7, 8]],
[[18, 19, 20],
[21, 22, 23],
[24, 25, 26]],
[[ 9, 10, 11],
[12, 13, 14],
[15, 16, 17]]])
numpy.indices 计算当通过其他两个轴(或n-1轴，其中n =轴的总数)展平"时，数组每个轴的索引.换句话说，这(对冗长的帖子表示歉意):
numpy.indices calculates the indices of each axis of the array when "flattened" through the other two axes (or n - 1 axes where n = total number of axes). In other words, this (apologies for the long post):
>>> static_indices
array([[[[0, 0, 0],
[0, 0, 0],
[0, 0, 0]],
[[1, 1, 1],
[1, 1, 1],
[1, 1, 1]],
[[2, 2, 2],
[2, 2, 2],
[2, 2, 2]]],
[[[0, 0, 0],
[1, 1, 1],
[2, 2, 2]],
[[0, 0, 0],
[1, 1, 1],
[2, 2, 2]],
[[0, 0, 0],
[1, 1, 1],
[2, 2, 2]]],
[[[0, 1, 2],
[0, 1, 2],
[0, 1, 2]],
[[0, 1, 2],
[0, 1, 2],
[0, 1, 2]],
[[0, 1, 2],
[0, 1, 2],
[0, 1, 2]]]])
这些是每个轴的标识索引；当用于索引b时，它们将重新创建b.
These are the identity indices for each axis; when used to index b, they recreate b.
>>> b[static_indices[0], static_indices[1], static_indices[2]]
array([[[ 0, 1, 2],
[ 3, 4, 5],
[ 6, 7, 8]],
[[ 9, 10, 11],
[12, 13, 14],
[15, 16, 17]],
[[18, 19, 20],
[21, 22, 23],
[24, 25, 26]]])
作为numpy.indices的替代方法，您可以使用uncbu建议的numpy.ogrid.由于ogrid生成的对象较小，因此，出于一致性考虑，我将创建所有三个轴，但是请注意unutbu的注释，以通过仅生成两个轴来实现此目的.
As an alternative to numpy.indices, you could use numpy.ogrid, as unutbu suggests. Since the object generated by ogrid is smaller, I'll create all three axes, just for consistency sake, but note unutbu's comment for a way to do this by generating only two.
>>> static_indices = numpy.ogrid[0:a.shape[0], 0:a.shape[1], 0:a.shape[2]]
>>> a[sort_indices, static_indices[1], static_indices[2]]
array([[[ 1., 1., 1.],
[ 1., 1., 1.],
[ 1., 1., 1.]],
[[ 2., 2., 2.],
[ 2., 2., 2.],
[ 2., 2., 2.]],
[[ 3., 3., 3.],
[ 3., 3., 3.],
[ 3., 3., 3.]]])
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