我有一个numpy数组,我想重复n次,同时保留行的原始顺序:
>>>a
array([[ 0, 1, 2, 3],
[ 4, 5, 6, 7],
[ 8, 9, 10, 11]])
所需输出(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]])
我发现了一个np.repeat函数,但是它不保留列的原始顺序。还有其他内置函数或技巧可以在保留顺序的同时重复数组吗?
答案 0 :(得分:2)
使用np.repeat
,然后使用np.concatenate
:
np.concatenate(np.repeat(a[None, :], n, axis=0), axis=0)
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]])
另一种选择是使用np.broadcast_to
:
np.broadcast_to(a, (n, *a.shape)).reshape(-1, 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]])
答案 1 :(得分:2)
这是另一种方式。我还添加了一些时间与@coldspeed的解决方案进行比较
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)
冷速解决方案1 ,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)
冷速解决方案2 ,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)
答案 2 :(得分:2)
numpy.repeat
用于逐个元素重复。要重复整个数组,您需要numpy.tile
。
numpy.tile(a, (2, 1))
元组是每个轴上的重复次数。您想要第一个为2,第二个为1,所以元组为(2, 1)
。
答案 3 :(得分:1)
在这种情况下,n
的填充模式很有用:
m
(def whiten(self, x):
shape = x.shape
x = K.batch_flatten(x)
mn = K.mean(x, 0)
std = K.std(x, 0) + K.epsilon()
r = (x - mn) / std
r = K.reshape(x, (-1,shape[1],shape[2],shape[3]))
return r
#
方法是不同的。)
答案 4 :(得分:0)
您可以尝试numpy.tile()。
这是在保存原始订单的同时使用 numpy.tile 重复阵列的方法:
import numpy as np
a = np.array([[ 0, 1, 2, 3],
[ 4, 5, 6, 7],
[ 8, 9, 10, 11]])
n = 5
b = np.tile(a, (n,1))
print b
输出:
[[ 0 1 2 3]
[ 4 5 6 7]
[ 8 9 10 11]
[ 0 1 2 3]
[ 4 5 6 7]
[ 8 9 10 11]
[ 0 1 2 3]
[ 4 5 6 7]
[ 8 9 10 11]
[ 0 1 2 3]
[ 4 5 6 7]
[ 8 9 10 11]
[ 0 1 2 3]
[ 4 5 6 7]
[ 8 9 10 11]]
答案 5 :(得分:0)
您也可以尝试
b=np.append(a,a).reshape(np.shape(a)[0]*2,np.shape(a)[1])
输出
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]])