python - 重复二维数组的行

Question

我有一个 numpy 数组，我想在保留行的原始顺序的同时重复它 n 次:

>>>a
array([[ 0,  1,  2,  3],
       [ 4,  5,  6,  7],
       [ 8,  9, 10, 11]])

期望的输出(对于 n =2):

>>>a
array([[ 0,  1,  2,  3],
       [ 4,  5,  6,  7],
       [ 8,  9, 10, 11],
       [ 0,  1,  2,  3],
       [ 4,  5,  6,  7],
       [ 8,  9, 10, 11]])

我找到了一个 np.repeat 函数，但是它没有保留列的原始顺序。是否有任何其他内置函数或技巧可以在保留顺序的同时重复数组？

Answer 1

这是另一种方法。我还添加了一些与@coldspeed 解决方案的时间比较

n = 2
a_new = np.tile(a.flatten(), n) 
a_new.reshape((n*a.shape[0], a.shape[1]))
# array([[ 0,  1,  2,  3],
#        [ 4,  5,  6,  7],
#        [ 8,  9, 10, 11],
#        [ 0,  1,  2,  3],
#        [ 4,  5,  6,  7],
#        [ 8,  9, 10, 11]])

与coldspeed解决方案的性能比较

我的方法 n = 10000

a = np.array([[ 0,  1,  2,  3],
       [ 4,  5,  6,  7],
       [ 8,  9, 10, 11]])
n = 10000

def tile_flatten(a, n):
    a_new = np.tile(a.flatten(), n).reshape((n*a.shape[0], a.shape[1])) 
    return a_new

%timeit tile_flatten(a,n)
# 149 µs ± 20.2 µs per loop (mean ± std. dev. of 7 runs, 10000 loops each)   

coldspeed 的解决方案 1 对于 n = 10000

a = np.array([[ 0,  1,  2,  3],
   [ 4,  5,  6,  7],
   [ 8,  9, 10, 11]])
n = 10000

def concatenate_repeat(a, n):
    a_new =  np.concatenate(np.repeat(a[None, :], n, axis=0), axis=0)
    return a_new

%timeit concatenate_repeat(a,n)
# 7.61 ms ± 1.37 ms per loop (mean ± std. dev. of 7 runs, 100 loops each)

coldspeed 的解决方案 2 对于 n = 10000

a = np.array([[ 0,  1,  2,  3],
       [ 4,  5,  6,  7],
       [ 8,  9, 10, 11]])
n = 10000

def broadcast_reshape(a, n):
    a_new =  np.broadcast_to(a, (n, *a.shape)).reshape(-1, a.shape[1])
    return a_new

%timeit broadcast_reshape(a,n)
# 162 µs ± 29.8 µs per loop (mean ± std. dev. of 7 runs, 10000 loops each)

@user2357112的解决方案

def tile_only(a, n):
    a_new = np.tile(a, (n, 1))
    return a_new

%timeit tile_only(a,n)
# 142 µs ± 21.8 µs per loop (mean ± std. dev. of 7 runs, 10000 loops each)