假设我有这个 numpy 数组:
[[1, 2, 3, 4],
[5, 6, 7, 8],
[9, 10, 11, 12],
[13, 14, 15, 16]]
我的目标是从每一行中选择两个随机元素并创建一个新的 numpy 数组,它可能如下所示:[[2, 4],
[5, 8],
[9, 10],
[15, 16]]
我可以使用 for 循环轻松地做到这一点。但是,有没有一种方法可以使用广播,比如 np.random.choice
, 以避免必须遍历每一行?
最佳答案
方法#1
基于 this trick
,这是一种矢量化的方式 -
n = 2 # number of elements to select per row
idx = np.random.rand(*a.shape).argsort(1)[:,:n]
out = np.take_along_axis(a, idx, axis=1)
sample 运行 -In [251]: a
Out[251]:
array([[ 1, 2, 3, 4],
[ 5, 6, 7, 8],
[ 9, 10, 11, 12],
[13, 14, 15, 16]])
In [252]: idx = np.random.rand(*a.shape).argsort(1)[:,:2]
In [253]: np.take_along_axis(a, idx, axis=1)
Out[253]:
array([[ 2, 1],
[ 6, 7],
[ 9, 11],
[16, 15]])
方法#2 另一个基于掩码每行选择两个 -
def select_two_per_row(a):
m,n = a.shape
mask = np.zeros((m,n), dtype=bool)
R = np.arange(m)
idx1 = np.random.randint(0,n,m)
mask[R,idx1] = 1
mask2 = np.zeros(m*(n-1), dtype=bool)
idx2 = np.random.randint(0,n-1,m) + np.arange(m)*(n-1)
mask2[idx2] = 1
mask[~mask] = mask2
out = a[mask].reshape(-1,2)
return out
方法#3 另一个基于整数的索引再次选择每行两个 -
def select_two_per_row_v2(a):
m,n = a.shape
idx1 = np.random.randint(0,n,m)
idx2 = np.random.randint(1,n,m)
out = np.take_along_axis(a, np.c_[idx1, idx1 - idx2], axis=1)
return out
时间——
In [209]: a = np.random.rand(100000,10)
# App1 with argsort
In [210]: %%timeit
...: idx = np.random.rand(*a.shape).argsort(1)[:,:2]
...: out = np.take_along_axis(a, idx, axis=1)
23.2 ms ± 137 µs per loop (mean ± std. dev. of 7 runs, 10 loops each)
# App1 with argpartition
In [221]: %%timeit
...: idx = np.random.rand(*a.shape).argpartition(axis=1,kth=1)[:,:2]
...: out = np.take_along_axis(a, idx, axis=1)
18.3 ms ± 115 µs per loop (mean ± std. dev. of 7 runs, 100 loops each)
In [214]: %timeit select_two_per_row(a)
9.89 ms ± 37.7 µs per loop (mean ± std. dev. of 7 runs, 100 loops each)
In [215]: %timeit select_two_per_row_v2(a)
5.78 ms ± 9.19 µs per loop (mean ± std. dev. of 7 runs, 100 loops each)
关于python - 你如何在 numpy 数组的每一行中广播 np.random.choice ?,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/64125560/