子集一个2D numpy数组

Question

我在这里查看了文档和其他问题,但似乎我还没有掌握numpy数组中的子集.

我有一个numpy数组,为了参数,让它定义如下:

import numpy as np
a = np.arange(100)
a.shape = (10,10)
# 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, 27, 28, 29],
#        [30, 31, 32, 33, 34, 35, 36, 37, 38, 39],
#        [40, 41, 42, 43, 44, 45, 46, 47, 48, 49],
#        [50, 51, 52, 53, 54, 55, 56, 57, 58, 59],
#        [60, 61, 62, 63, 64, 65, 66, 67, 68, 69],
#        [70, 71, 72, 73, 74, 75, 76, 77, 78, 79],
#        [80, 81, 82, 83, 84, 85, 86, 87, 88, 89],
#        [90, 91, 92, 93, 94, 95, 96, 97, 98, 99]])

现在我想选择a矢量n1和指定的行和列n2.举个例子:

n1 = range(5)
n2 = range(5)

但是当我使用时:

b = a[n1,n2]
# array([ 0, 11, 22, 33, 44])

然后仅选择前五个对角线元素,而不是整个5x5块.我找到的解决方案就是这样做:

b = a[n1,:]
b = b[:,n2]
# array([[ 0,  1,  2,  3,  4],
#        [10, 11, 12, 13, 14],
#        [20, 21, 22, 23, 24],
#        [30, 31, 32, 33, 34],
#        [40, 41, 42, 43, 44]])

但我确信应该有一种方法可以在一个命令中完成这个简单的任务.

Answer 1

你已经掌握了一些如何做你想做的好例子.但是,理解正在发生的事情以及事情的运作方式也很有用.有一些简单的规则可以帮助您将来.

"花式"索引(即使用列表/序列)和"正常"索引(使用切片)之间存在很大差异.根本原因与阵列是否可以"定期跨越"有关,因此是否需要制作副本.因此,如果我们希望能够创建"视图"而不制作副本,则必须区别对待任意序列.

在你的情况下:

import numpy as np

a = np.arange(100).reshape(10,10)
n1, n2 = np.arange(5), np.arange(5)

# Not what you want
b = a[n1, n2]  # array([ 0, 11, 22, 33, 44])

# What you want, but only for simple sequences
# Note that no copy of *a* is made!! This is a view.
b = a[:5, :5]

# What you want, but probably confusing at first. (Also, makes a copy.)
# np.meshgrid and np.ix_ are basically equivalent to this.
b = a[n1[:,None], n2[None,:]]

使用1D序列进行花式索引基本上等同于将它们压缩在一起并使用结果进行索引.

print "Fancy Indexing:"
print a[n1, n2]

print "Manual indexing:"
for i, j in zip(n1, n2):
    print a[i, j]

但是,如果您要编制索引的序列与您要编制索引的数组的维度相匹配(在本例中为2D),则会对索引进行不同的处理.numpy使用像面具这样的索引,而不是"将两者拼凑在一起".

换句话说,a[[[1, 2, 3]], [[1],[2],[3]]]处理完全不同于a[[1, 2, 3], [1, 2, 3]],因为您传入的序列/数组是二维的.

In [4]: a[[[1, 2, 3]], [[1],[2],[3]]]
Out[4]:
array([[11, 21, 31],
       [12, 22, 32],
       [13, 23, 33]])

In [5]: a[[1, 2, 3], [1, 2, 3]]
Out[5]: array([11, 22, 33])

为了更精确一点,

a[[[1, 2, 3]], [[1],[2],[3]]]

被视为完全像:

i = [[1, 1, 1],
     [2, 2, 2],
     [3, 3, 3]])
j = [[1, 2, 3],
     [1, 2, 3],
     [1, 2, 3]]
a[i, j]

换句话说,输入是否是行/列向量是索引应如何在索引中重复的简写.

np.meshgrid并且np.ix_只是将您的1D序列转换为2D版本以进行索引的简单方法:

In [6]: np.ix_([1, 2, 3], [1, 2, 3])
Out[6]:
(array([[1],
       [2],
       [3]]), array([[1, 2, 3]]))

同样(sparse论证会使它与ix_上面相同):

In [7]: np.meshgrid([1, 2, 3], [1, 2, 3], indexing='ij')
Out[7]:
[array([[1, 1, 1],
       [2, 2, 2],
       [3, 3, 3]]),
 array([[1, 2, 3],
       [1, 2, 3],
       [1, 2, 3]])]

Answer 2

构建所需索引的另一种快速方法是使用以下np.ix_函数:

>>> a[np.ix_(n1, n2)]
array([[ 0,  1,  2,  3,  4],
       [10, 11, 12, 13, 14],
       [20, 21, 22, 23, 24],
       [30, 31, 32, 33, 34],
       [40, 41, 42, 43, 44]])

这提供了从索引序列构造开放网格的便利方式.

Answer 3

您可以使用np.meshgrid给予n1,n2阵列执行所需的索引正确的形状:

In [104]: a[np.meshgrid(n1,n2, sparse=True, indexing='ij')]
Out[104]: 
array([[ 0,  1,  2,  3,  4],
       [10, 11, 12, 13, 14],
       [20, 21, 22, 23, 24],
       [30, 31, 32, 33, 34],
       [40, 41, 42, 43, 44]])

或者,没有meshgrid:

In [117]: a[np.array(n1)[:,np.newaxis], np.array(n2)[np.newaxis,:]]
Out[117]: 
array([[ 0,  1,  2,  3,  4],
       [10, 11, 12, 13, 14],
       [20, 21, 22, 23, 24],
       [30, 31, 32, 33, 34],
       [40, 41, 42, 43, 44]])

有一个类似的例子,解释了这个整数数组索引如何在文档中工作.

另请参阅Cookbook配方挑选行和列.