如何通过 numpy.repeat 创建多维数组 [重复]

Question

我想把这个多维数组写成软代码

adjust = np.array([[0.,1.,0.],
               [0.,1.,0.],
               [0.,1.,0.],
               [0.,1.,0.],
               [0.,1.,0.],
               [0.,1.,0.],
               [0.,1.,0.],
               [0.,1.,0.],
               [0.,1.,0.],   
               [0.,1.,0.]])

我为x = [0.,1.,0.] 尝试了np.repeat(x, 10, axis=0)，它在同一个括号[] 中重复了它们。可以在这里使用 np.repeat 吗？还是其他 numpy？

还有，可以写软代码吗

adjust = np.array([[0.,0.,1.,0.,0.],
               [0.,0.,1.,0.,0.],
               [0.,0.,1.,0.,0.]])

将来我可能需要将左右 0 扩展为各种数字？

Answer 1

您可以在重复之前向数组添加另一个轴（注意我们重复了n = 1000 次）

n = 1000
%timeit adjust = np.repeat(np.array([0., 1., 0.])[None, :], n, axis=0)
# 7.67 µs ± 940 ns per loop (mean ± std. dev. of 7 runs, 100000 loops each)

或者对repeat()的结果进行整形和转置

%timeit adjust = np.repeat([0., 1., 0.], n, axis=0).reshape(3, -1).T
# 22.5 µs ± 1.13 µs per loop (mean ± std. dev. of 7 runs, 100000 loops each)

或使用广播

%timeit adjust = np.array([0., 1., 0.]) * np.ones(n)[:, None]
# 26.8 µs ± 880 ns per loop (mean ± std. dev. of 7 runs, 10000 loops each)

为了性能比较，艾伦的建议：

%timeit adjust = np.asarray([ 0.,  1.,  0.] * n).reshape(n,-1)
# 93.5 µs ± 7.87 µs per loop (mean ± std. dev. of 7 runs, 10000 loops each)

和 Divakars 的建议

%timeit adjust = np.tile(np.array([0., 1., 0.]), (n, 1))
# 11.1 µs ± 686 ns per loop (mean ± std. dev. of 7 runs, 100000 loops each)

结论：np.repeat()加另一个轴后最快。

Answer 2

重复你的列表 n 次，把它放在一个 numpy 数组中，然后将它重新整形为 n 行。

np.asarray([ 0.,  1.,  0.]*10).reshape(10,-1)
Out[139]: 
array([[ 0.,  1.,  0.],
       [ 0.,  1.,  0.],
       [ 0.,  1.,  0.],
       ..., 
       [ 0.,  1.,  0.],
       [ 0.,  1.,  0.],
       [ 0.,  1.,  0.]])

第二个数组也是如此：

np.asarray([0.,0.,1.,0.,0.]*3).reshape(3,-1)
Out[140]: 
array([[ 0.,  0.,  1.,  0.,  0.],
       [ 0.,  0.,  1.,  0.,  0.],
       [ 0.,  0.,  1.,  0.,  0.]])

时间安排：

%timeit np.asarray([ 0.,  1.,  0.]*10).reshape(10,-1)
The slowest run took 14.97 times longer than the fastest. This could mean that an intermediate result is being cached.
100000 loops, best of 3: 4.51 µs per loop

%timeit np.repeat([0., 1., 0.], 10, axis=0).reshape(3, -1).T
The slowest run took 4.44 times longer than the fastest. This could mean that an intermediate result is being cached.
100000 loops, best of 3: 11.3 µs per loop

%timeit np.array([0., 1., 0.]) * np.ones(10)[:, None]
The slowest run took 10.28 times longer than the fastest. This could mean that an intermediate result is being cached.
100000 loops, best of 3: 11.3 µs per loop