[英]Slicing numpy.ndarray by column index
有没有一种方法可以切片下面的数组而不必定义行索引,即不必写range(len(X))
?
X = np.arange(10*2).reshape((10,2))
L = np.random.randint(0,2,10)
Xs = X[range(len(X)),L]
我以为可以用X[:,L]
切片,但看起来好像不是。
您可能正在寻找np.choose :
In [25]: X = np.arange(10*2).reshape((10,2)); X
Out[25]:
array([[ 0, 1],
[ 2, 3],
[ 4, 5],
[ 6, 7],
[ 8, 9],
[10, 11],
[12, 13],
[14, 15],
[16, 17],
[18, 19]])
In [26]: L = np.random.randint(0,2,10); L
Out[26]: array([1, 1, 1, 1, 1, 0, 0, 0, 0, 1])
In [27]: L.choose(X.T)
Out[27]: array([ 1, 3, 5, 7, 9, 10, 12, 14, 16, 19])
In [28]: # or otherwise
In [29]: np.choose(L, X.T)
Out[29]: array([ 1, 3, 5, 7, 9, 10, 12, 14, 16, 19])
性能说明 :尽管此解决方案是对问题的直接答案,但随着len(X)
增加,它很快就变得不是最佳的。 从numpy 1.9.0开始, np.arange
方法更快:
In [17]: %timeit X[range(len(X)), L]
1000 loops, best of 3: 629 µs per loop
In [18]: %timeit X[np.arange(len(X)), L]
10000 loops, best of 3: 78.8 µs per loop
In [19]: %timeit L.choose(X.T)
10000 loops, best of 3: 146 µs per loop
In [20]: X.shape, L.shape
Out[20]: ((10000, 2), (10000,))
注意
In [9]: X[:, L]
Out[9]:
array([[ 1, 1, 0, 0, 1, 0, 1, 0, 1, 0],
[ 3, 3, 2, 2, 3, 2, 3, 2, 3, 2],
[ 5, 5, 4, 4, 5, 4, 5, 4, 5, 4],
[ 7, 7, 6, 6, 7, 6, 7, 6, 7, 6],
[ 9, 9, 8, 8, 9, 8, 9, 8, 9, 8],
[11, 11, 10, 10, 11, 10, 11, 10, 11, 10],
[13, 13, 12, 12, 13, 12, 13, 12, 13, 12],
[15, 15, 14, 14, 15, 14, 15, 14, 15, 14],
[17, 17, 16, 16, 17, 16, 17, 16, 17, 16],
[19, 19, 18, 18, 19, 18, 19, 18, 19, 18]])
您想要对角线元素:
所以做:
In [14]: X[:, L].diagonal()
Out[14]: array([ 1, 3, 4, 6, 9, 10, 13, 14, 17, 18])
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