英文:
Numpy mask : dimension loss
问题
我遇到了关于mask和numpy的问题。我有一个3D张量，尝试仅选择其中的特定单元。
更具体地说，我创建了一个3D张量：
array = np.repeat(np.arange(15).reshape(3,5)[None,:],3, axis = 0)
# array([[[ 0, 1, 2, 3, 4],
[ 5, 6, 7, 8, 9],
[10, 11, 12, 13, 14]],
[[ 0, 1, 2, 3, 4],
[ 5, 6, 7, 8, 9],
[10, 11, 12, 13, 14]],
[[ 0, 1, 2, 3, 4],
[ 5, 6, 7, 8, 9],
[10, 11, 12, 13, 14]]])
我想要移除第一个矩阵的第一行，第二个矩阵的第二行，依此类推，以便得到：
# array([[[ 5, 6, 7, 8, 9],
[10, 11, 12, 13, 14]],
[[ 0, 1, 2, 3, 4],
[10, 11, 12, 13, 14]],
[[ 0, 1, 2, 3, 4],
[ 5, 6, 7, 8, 9]]])
我要说明的是，我希望使用以下形式的代码来实现：
array[mask]
（我不想重新整形或选择索引和mask.array以获得更好的性能）。
我创建了一个掩码：
mask = [[False, True, True],
[True, False, True],
[False, True, True]]
但它丢失了一个维度：
array[mask]
array([[ 5, 6, 7, 8, 9],
[10, 11, 12, 13, 14],
[ 0, 1, 2, 3, 4],
[10, 11, 12, 13, 14],
[ 5, 6, 7, 8, 9],
[10, 11, 12, 13, 14]])
非常感谢您的帮助！
英文:
I struggle with mask and numpy. I have a 3D-tensor and I try to select only some specific cells.
More preciseley I create a 3D tensor :
array = np.repeat(np.arange(15).reshape(3,5)[None,:],3, axis = 0)
# array([[[ 0, 1, 2, 3, 4],
[ 5, 6, 7, 8, 9],
[10, 11, 12, 13, 14]],
[[ 0, 1, 2, 3, 4],
[ 5, 6, 7, 8, 9],
[10, 11, 12, 13, 14]],
[[ 0, 1, 2, 3, 4],
[ 5, 6, 7, 8, 9],
[10, 11, 12, 13, 14]]])
and I want to have remove the first row of the first matrix, the second row of the second matrix etc... So in order to have :
# array([[[ 5, 6, 7, 8, 9],
[10, 11, 12, 13, 14]],
[[ 0, 1, 2, 3, 4],
[10, 11, 12, 13, 14]],
[[ 0, 1, 2, 3, 4],
[ 5, 6, 7, 8, 9]]])
I precise that I want to do it with a code of this form :
array[mask]
(I don't want to reshape or select indexes and mask.array for its performance).
I created a mask :
mask = [[False ,True ,True],
[True ,False ,True],
[False, True, True]]
But it lost a dimension :
array[mask]
array([[ 5, 6, 7, 8, 9],
[10, 11, 12, 13, 14],
[ 0, 1, 2, 3, 4],
[10, 11, 12, 13, 14],
[ 5, 6, 7, 8, 9],
[10, 11, 12, 13, 14]])
Thank you very much for the help !
答案1
得分: 1
抱歉，我无法识别你要翻译的代码部分。以下是已经翻译好的非代码内容：
不幸的是，我认为使用布尔掩码无法实现你想要的目标。你所看到的行为在numpy索引指南中有描述。特别是：
> 一般来说，如果索引包括布尔数组，结果将等同于在相同位置插入 obj.nonzero() 并使用上面描述的整数数组索引机制。x[ind_1, boolean_array, ind_2] 等同于 x[(ind_1,) + boolean_array.nonzero() + (ind_2,)]。
在你的情况下，np.nonzero(mask) 会产生两个一维数组，这意味着你的掩码索引了array的前两个维度，这两个维度被替换为一个单一维度（因为np.nonzero(mask) 的结果会广播到一个一维数组）。
你可以通过使用两个二维数组索引数据array的前两个维度来实现你想要的效果，但我不确定这是否符合你对“选择索引”的定义。以下是一个示例：
import numpy as np
array = np.repeat(np.arange(15).reshape(3,5)[None,:],3, axis = 0)
# 元组，包含两个形状为(3, 1)和(3, 2)的数组，这两个数组会广播到(3, 2)的子空间
idx = [[0], [1], [2]], [[1, 2], [0, 2], [0, 1]]
out = array[idx]
输出结果：
array([[[ 5, 6, 7, 8, 9],
[10, 11, 12, 13, 14]],
[[ 0, 1, 2, 3, 4],
[10, 11, 12, 13, 14]],
[[ 0, 1, 2, 3, 4],
[ 5, 6, 7, 8, 9]]])
英文:
Unfortunately, I don't think it's possible to achieve what you want using boolean masks. The behavior you're seeing is described in numpy's indexing guide. In particular:
> In general if an index includes a Boolean array, the result will be identical to inserting obj.nonzero() into the same position and using the integer array indexing mechanism described above. x[ind_1, boolean_array, ind_2] is equivalent to x[(ind_1,) + boolean_array.nonzero() + (ind_2,)].
In your case, np.nonzero(mask) would produce two 1 dimensional arrays, which means your mask indexes the first two dimensions of array which is replaced by a single dimension (since the np.nonzero(mask) results broadcast to a 1-D array).
You could achieve what you want by indexing the first two dimensions of the data array with two 2-d arrays, but I'm not sure whether that falls under your definition of not wanting to "select indexes". Here is an example:
import numpy as np
array = np.repeat(np.arange(15).reshape(3,5)[None,:],3, axis = 0)
# Tuple of (3, 1) and (3, 2) shaped array which broadcasts to (3, 2) subspace
idx = [[0], [1], [2]], [[1, 2], [0, 2], [0, 1]]
out = array[idx]
out:
array([[[ 5, 6, 7, 8, 9],
[10, 11, 12, 13, 14]],
[[ 0, 1, 2, 3, 4],
[10, 11, 12, 13, 14]],
[[ 0, 1, 2, 3, 4],
[ 5, 6, 7, 8, 9]]])
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