根据特定值的索引筛选numpy数组

huangapple
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英文:

Filtering numpy arrays based on the index of certain value

问题

我有一个类似的numpy数组：

  1. array([[ 1, 17, 33, ..., 28,  9, 22],
  2.        [ 3, 11,  1, ..., 25, 45, 14],
  3.        [ 3, 11,  1, ..., 21, 23,  5],
  4.        ...,
  5.        [20,  6, 27, ..., 43, 15, 14],
  6.        [27,  6, 39, ..., 37, 17,  2],
  7.        [ 3, 11,  8, ..., 27, 35, 32]], dtype=int32)

从这里，我想要筛选出在索引10或之前出现值4的行。

例如：
[1, 2, 3, 4, ..., 34, 35] - 筛选，因为值4在索引3之前出现，即在索引10之前。

例如：
[35, 34, 33, 32..., 4, 3, 2, 1] - 保留，因为值4在索引10之后出现。

使用numpy掩码，可以实现这种筛选的方式是什么？

英文:

I have a numpy array like:

  1. array([[ 1, 17, 33, ..., 28,  9, 22],
  2.        [ 3, 11,  1, ..., 25, 45, 14],
  3.        [ 3, 11,  1, ..., 21, 23,  5],
  4.        ...,
  5.        [20,  6, 27, ..., 43, 15, 14],
  6.        [27,  6, 39, ..., 37, 17,  2],
  7.        [ 3, 11,  8, ..., 27, 35, 32]], dtype=int32)

From here, I would like to filter out rows where value 4 is occurring at index of 10 or before.

e.g.
[1, 2, 3, 4, ..., 34, 35] - filter, as value 4 is occurring at index3, which is before index 10

e.g.
[35, 34, 33, 32..., 4, 3, 2, 1] - keep, as value 4 is occurring after index10.

what would be the way to achieve this filtering using numpy masking?

答案1

得分: 2

这是您要翻译的代码部分：

  1. arr = np.array(
  2.     [[1, 2, 3, 4, 34, 35, 2, 1],
  3.      [1, 2, 3, 5, 6, 7, 8, 9],
  4.      [1, 2, 4, 5, 6, 7, 8, 9],
  5.      [4, 2, 3, 5, 6, 7, 8, 9],
  6.      [35, 34, 33, 32, 4, 3, 2, 1]]
  7. )
  9. V, T = 4, 3 # <-- change the threshold to 10
  11. m = np.any(arr[:, :T+1] == V, axis=1)
  13. out = arr[~m]
  15. Output :
  17. print(out)
  19. array([[ 1,  2,  3,  5,  6,  7,  8,  9],
  20.        [35, 34, 33, 32,  4,  3,  2,  1]])

Intermediates :

  1. >>> arr[:, :T+1]
  2. array([[ 1,  2,  3,  4],
  3.        [ 1,  2,  3,  5],
  4.        [ 1,  2,  4,  5],
  5.        [ 4,  2,  3,  5],
  6.        [35, 34, 33, 32]])
  7.     
  8. >>> arr[:, :T+1] == V
  9. array([[False, False, False,  True],
  10.        [False, False, False, False],
  11.        [False, False,  True, False],
  12.        [ True, False, False, False],
  13.        [False, False, False, False]])
  14.     
  15. >>> np.any(arr[:, :T+1] == V, axis=1)
  16. array([ True, False,  True,  True, False])

请注意，代码部分已经被复制到翻译中。

英文:

You can try this :

  1. arr = np.array(
  2.     [[1, 2, 3, 4, 34, 35, 2, 1],
  3.      [1, 2, 3, 5, 6, 7, 8, 9],
  4.      [1, 2, 4, 5, 6, 7, 8, 9],
  5.      [4, 2, 3, 5, 6, 7, 8, 9],
  6.      [35, 34, 33, 32, 4, 3, 2, 1]]
  7. )
  9. V, T = 4, 3 # <-- change the threshold to 10
  11. m = np.any(arr[:, :T+1] == V, axis=1)
  13. out = arr[~m]

Output :

  1. print(out)
  3. array([[ 1,  2,  3,  5,  6,  7,  8,  9],
  4.        [35, 34, 33, 32,  4,  3,  2,  1]])

Intermediates :

  1. >>> arr[:, :T+1]
  2. array([[ 1,  2,  3,  4],
  3.        [ 1,  2,  3,  5],
  4.        [ 1,  2,  4,  5],
  5.        [ 4,  2,  3,  5],
  6.        [35, 34, 33, 32]])
  8. >>> arr[:, :T+1] == V
  9. array([[False, False, False,  True],
  10.        [False, False, False, False],
  11.        [False, False,  True, False],
  12.        [ True, False, False, False],
  13.        [False, False, False, False]])
  15. >>> np.any(arr[:, :T+1] == V, axis=1)
  16. array([ True, False,  True,  True, False])

答案2

得分: -1

以下是翻译后的代码部分：

  1. import numpy as np
  3. np.random.seed(0)
  5. vals = np.random.choice(50, size=(10, 100))
  7. print(vals[:,:10])
  8. print(vals.shape)
  9. vals = vals[[4 not in i for i in vals[:,:10]]]
  10. print(vals.shape)
  11. print(vals[:,:10])

应该输出：

  1. [[44 47  0  3  3 39  9 19 21 36]
  2.  [ 5 41 35  0 31  5 30  0 49 36]
  3.  [24 15 41 18 40 15 11 38 47 29]
  4.  [ 2  5 37 12 44  2 47 27 21 39]
  5.  [42 48 30 16 26 35 49 42  9 44]
  6.  [ 3 34 40 33 28  4 26 32 45  9]
  7.  [41 38 43 18  7 28  1 41  2 28]
  8.  [36 49 24 33 18 33 14 49  7 43]
  9.  [ 6 27 35  6 19 34 38 20 43  0]
  10.  [ 9 20 37 48 17  9 44 15 38 14]]
  11. (10, 100)
  12. (9, 100)
  13. [[44 47  0  3  3 39  9 19 21 36]
  14.  [ 5 41 35  0 31  5 30  0 49 36]
  15.  [24 15 41 18 40 15 11 38 47 29]
  16.  [ 2  5 37 12 44  2 47 27 21 39]
  17.  [42 48 30 16 26 35 49 42  9 44]
  18.  [41 38 43 18  7 28  1 41  2 28]
  19.  [36 49 24 33 18 33 14 49  7 43]
  20.  [ 6 27 35  6 19 34 38 20 43  0]
  21.  [ 9 20 37 48 17  9 44 15 38 14]]
英文:

Here's one way to do it, the key bit is [4 not in i for i in vals[:,:10]] which creates a mask by iterating through a view of the first 10 elements of each row.

  1. import numpy as np
  2. np.random.seed(0)
  3. vals = np.random.choice(50, size=(10, 100))
  4. print(vals[:,:10])
  5. print(vals.shape)
  6. vals = vals[[4 not in i for i in vals[:,:10]]]
  7. print(vals.shape)
  8. print(vals[:,:10])

which should yield:

  1. [[44 47  0  3  3 39  9 19 21 36]
  2. [ 5 41 35  0 31  5 30  0 49 36]
  3. [24 15 41 18 40 15 11 38 47 29]
  4. [ 2  5 37 12 44  2 47 27 21 39]
  5. [42 48 30 16 26 35 49 42  9 44]
  6. [ 3 34 40 33 28  4 26 32 45  9]
  7. [41 38 43 18  7 28  1 41  2 28]
  8. [36 49 24 33 18 33 14 49  7 43]
  9. [ 6 27 35  6 19 34 38 20 43  0]
  10. [ 9 20 37 48 17  9 44 15 38 14]]
  11. (10, 100)
  12. (9, 100)
  13. [[44 47  0  3  3 39  9 19 21 36]
  14. [ 5 41 35  0 31  5 30  0 49 36]
  15. [24 15 41 18 40 15 11 38 47 29]
  16. [ 2  5 37 12 44  2 47 27 21 39]
  17. [42 48 30 16 26 35 49 42  9 44]
  18. [41 38 43 18  7 28  1 41  2 28]
  19. [36 49 24 33 18 33 14 49  7 43]
  20. [ 6 27 35  6 19 34 38 20 43  0]
  21. [ 9 20 37 48 17  9 44 15 38 14]]

huangapple
  • 本文由 发表于 2023年6月19日 00:56:44
  • 转载请务必保留本文链接：https://go.coder-hub.com/76501649.html
  • arrays
  • filter
  • masking
  • numpy
  • python
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