如何沿给定的轴取其索引指定的元素? [英] How to take elements along a given axis, given by their indices?

问题描述

我有一个3D数组，我需要在最后一个轴上挤压"它，以便获得2D数组.我需要按以下方式进行操作.对于前两个维度的每个索引值，我知道第三个维度的索引值应从何处取值.

I have a 3D array and I need to "squeeze" it over the last axis, so that I get a 2D array. I need to do it in the following way. For each values of the indices for the first two dimensions I know the value of the index for the 3rd dimension from where the value should be taken.

例如，我知道如果i1 == 2和i2 == 7然后是i3 == 11.这表示out[2,7] = inp[2,7,11].从前两个维到第三个维的映射在另一个2D数组中给出.换句话说，我有一个数组，其中在位置2,7上我有11作为值.

For example, I know that if i1 == 2 and i2 == 7 then i3 == 11. It means that out[2,7] = inp[2,7,11]. This mapping from first two dimensions into the third one is given in another 2D array. In other words, I have an array in which on the position 2,7 I have 11 as a value.

所以，我的问题是如何组合这两个数组(3D和2D)以获得输出数组(2D).

So, my question is how to combine these two array (3D and 2D) to get the output array (2D).

推荐答案

In [635]: arr = np.arange(24).reshape(2,3,4)
In [636]: idx = np.array([[1,2,3],[0,1,2]])


In [637]: I,J = np.ogrid[:2,:3]
In [638]: arr[I,J,idx]
Out[638]: 
array([[ 1,  6, 11],
       [12, 17, 22]])
In [639]: arr
Out[639]: 
array([[[ 0,  1,  2,  3],   # 1
        [ 4,  5,  6,  7],   # 6
        [ 8,  9, 10, 11]],  # ll

       [[12, 13, 14, 15],
        [16, 17, 18, 19],
        [20, 21, 22, 23]]])

I,J一起广播以选择一组(2,3)值，这些值与idx相匹配:

I,J broadcast together to select a (2,3) set of values, matching idx:

In [640]: I
Out[640]: 
array([[0],
       [1]])
In [641]: J
Out[641]: array([[0, 1, 2]])

---

这是对更简单的2d问题的3d的概括-从每一行中选择一项:

---

This is a generalization to 3d of the easier 2d problem - selecting one item from each row:

In [649]: idx
Out[649]: 
array([[1, 2, 3],
       [0, 1, 2]])
In [650]: idx[np.arange(2), [0,1]]
Out[650]: array([1, 1])

实际上，我们可以将3d问题转换为2d问题:

In fact we could convert the 3d problem into a 2d one:

In [655]: arr.reshape(6,4)[np.arange(6), idx.ravel()]
Out[655]: array([ 1,  6, 11, 12, 17, 22])

---

概括原始案例:

---

Generalizing the original case:

In [55]: arr = np.arange(24).reshape(2,3,4)                                     
In [56]: idx = np.array([[1,2,3],[0,1,2]])                                      
In [57]: IJ = np.ogrid[[slice(i) for i in idx.shape]]                           
In [58]: IJ                                                                     
Out[58]: 
[array([[0],
        [1]]), array([[0, 1, 2]])]
In [59]: (*IJ,idx)                                                              
Out[59]: 
(array([[0],
        [1]]), array([[0, 1, 2]]), array([[1, 2, 3],
        [0, 1, 2]]))
In [60]: arr[_]                                                                 
Out[60]: 
array([[ 1,  6, 11],
       [12, 17, 22]])

关键在于将数组的IJ列表与idx组合在一起以创建新的索引元组.如果idx不是最后一个索引，则构造元组会有点麻烦，但是仍然可能.例如

The key is in combining the IJ list of arrays with the idx to make a new indexing tuple. Constructing the tuple is a little messier if idx isn't the last index, but it's still possible. E.g.

In [61]: (*IJ[:-1],idx,IJ[-1])                                                  
Out[61]: 
(array([[0],
        [1]]), array([[1, 2, 3],
        [0, 1, 2]]), array([[0, 1, 2]]))
In [62]: arr.transpose(0,2,1)[_]                                                
Out[62]: 
array([[ 1,  6, 11],
       [12, 17, 22]])

如果更容易将arr换位为idx维度，则为最后.关键是索引操作采用索引数组的元组，索引数组相互广播以选择特定项. 这就是ogrid的工作，创建与idx一起使用的数组.

Of if it's easier transpose arr to the idx dimension is last. The key is that the index operation takes a tuple of index arrays, arrays which broadcast against each other to select specific items. That's what ogrid is doing, create the arrays that work with idx.

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