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

此外，是否可以为

编写软代码

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

我将来可能需要以不同的数字扩展左右0？

Answer 1

您可以在重复之前向阵列添加另一个轴（注意我们重复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)

或重塑并转置repeat()

的结果

%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)

为了进行性能比较，艾伦的建议是：

%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：添加另一个轴后np.repeat()是最快的。

Answer 2

重复你的列表n次，把它放在一个numpy数组中，然后重新整形为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.]])

同样适用于您的第二个阵列：

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.]])

时序：

%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

如何通过numpy.repeat创建多维数组

2 个答案: