Question

我有一个数组A（m X n）和另一个W（m X n'），其中n'<= n。 W可以包含nan个条目。

我想用W设置A的前n'列。但是，在W有In [1]: import numpy as np In [2]: A = np.random.rand(10, 4) In [3]: A Out[3]: array([[ 0.60879191, 0.13556633, 0.26043647, 0.69874723], [ 0.23775325, 0.5744113 , 0.76446265, 0.82177711], [ 0.44320766, 0.43535148, 0.33745034, 0.63270876], [ 0.81903997, 0.15170996, 0.9847644 , 0.77856538], [ 0.89771006, 0.11521106, 0.92180393, 0.82296974], [ 0.57292616, 0.47236245, 0.643821 , 0.39011254], [ 0.1478904 , 0.1597816 , 0.61934679, 0.87229069], [ 0.6395053 , 0.40529633, 0.02389057, 0.0144438 ], [ 0.25381449, 0.28045816, 0.98475699, 0.03257699], [ 0.91827695, 0.85382925, 0.94231584, 0.5303724 ]]) In [4]: W = np.random.rand(10, 2) In [5]: W Out[5]: array([[ 0.85731947, 0.02603337], [ 0.46941828, 0.12485814], [ 0.607665 , 0.61742206], [ 0.67579577, 0.44169912], [ 0.77938579, 0.04609614], [ 0.55431149, 0.12936694], [ 0.05191665, 0.94768561], [ 0.9494111 , 0.21739947], [ 0.77785379, 0.35316779], [ 0.72959474, 0.72603156]]) In [6]: W[2, 1] = np.NAN In [7]: W[4, 0] = np.NAN In [8]: W Out[8]: array([[ 0.85731947, 0.02603337], [ 0.46941828, 0.12485814], [ 0.607665 , nan], [ 0.67579577, 0.44169912], [ nan, 0.04609614], [ 0.55431149, 0.12936694], [ 0.05191665, 0.94768561], [ 0.9494111 , 0.21739947], [ 0.77785379, 0.35316779], [ 0.72959474, 0.72603156]]) In [9]: W_non_nan = ~np.isnan(W) In [10]: W_non_nan Out[10]: array([[ True, True], [ True, True], [ True, False], [ True, True], [False, True], [ True, True], [ True, True], [ True, True], [ True, True], [ True, True]], dtype=bool) In [11]: A[W_non_nan] Out[11]: array([ 0.60879191, 0.13556633, 0.23775325, 0.5744113 , 0.44320766, 0.81903997, 0.15170996, 0.11521106, 0.57292616, 0.47236245, 0.1478904 , 0.1597816 , 0.6395053 , 0.40529633, 0.25381449, 0.28045816, 0.91827695, 0.85382925])的位置，我希望保留原始A条目。

以下是示例代码。

@types/filesystem

禁止A [2,1]和A [4,0]; A的前两列应该用W代替。最干净的方法是什么？

Answer 1

您可以使用masking/boolean-indexing将非NaN的掩码切成第一个n'列，并使用该掩码选择W之外的元素并分配到{的切片部分{1}}，就像这样 -

示例运行 -

mask = ~np.isnan(W)
A[:,:mask.shape[1]][mask] = W[mask]

Answer 2

另一种解决方案：

A = np.array([[ 68.,  54.,  54.,  15.],
       [ 50.,  86.,  84.,  74.],
       [ 27.,  39.,  43.,  24.],
       [ 41.,  45.,  42.,  40.],
       [ 32.,  90.,  93.,  19.]])

W = np.array([[  3.,   4.],
       [  3.,   1.],
       [  4.,  np.nan],
       [  6.,   8.],
       [ np.nan,   6.]])   
#replace first two cols of A with W except the positions where W has nan.    
A[:,0:2] = np.where(np.isnan(W),A[:,0:2],W)
print(A)

[[  3.   4.  54.  15.]
 [  3.   1.  84.  74.]
 [  4.  39.  43.  24.]
 [  6.   8.  42.  40.]
 [ 32.   6.  93.  19.]]

NumPy：在非NaN位置设置numpy数组的子集

2 个答案: