我相信你可以这样使用.take
:
In [185]: X
Out[185]:
array([[ 1, 2, 3, 4],
[ 5, 6, 7, 8],
[ 9, 10, 11, 12]])
In [186]: idx
Out[186]:
array([[1, 3],
[0, 1],
[1, 2]])
In [187]: X.take(idx + (np.arange(X.shape[0]) * X.shape[1]).reshape(-1, 1))
Out[187]:
array([[ 2, 4],
[ 5, 6],
[10, 11]])
如果你的数组维度非常大,使用以下方式可能更快,尽管不太美观:
idx+np.arange(0, X.shape[0]*X.shape[1], X.shape[1])[:,None]
只是为了好玩,看看下面的表现如何:
np.fromiter((X[i, j] for i, row in enumerate(ixs) for j in row), dtype=X.dtype, count=ixs.size).reshape(ixs.shape)
编辑以添加时间
In [15]: X = np.arange(1000*10000, dtype=np.int32).reshape(1000,-1)
In [16]: ixs = np.random.randint(0, 10000, (1000, 2))
In [17]: ixs.sort(axis=1)
In [18]: ixs
Out[18]:
array([[2738, 3511],
[3600, 7414],
[7426, 9851],
...,
[1654, 8252],
[2194, 8200],
[5497, 8900]])
In [19]: %timeit np.array([row[ix] for row, ix in zip(X, ixs)])
928 µs ± 23.8 µs per loop (mean ± std. dev. of 7 runs, 1000 loops each)
In [20]: %timeit X[np.arange(len(ixs)), ixs.T].T
23.6 µs ± 491 ns per loop (mean ± std. dev. of 7 runs, 10000 loops each)
In [21]: %timeit X.take(idx+np.arange(0, X.shape[0]*X.shape[1], X.shape[1])[:,None])
20.6 µs ± 530 ns per loop (mean ± std. dev. of 7 runs, 10000 loops each)
In [22]: %timeit np.fromiter((X[i, j] for i, row in enumerate(ixs) for j in row), dtype=X.dtype, count=ixs.size).reshape(ixs.shape)
1.42 ms ± 9.94 µs per loop (mean ± std. dev. of 7 runs, 1000 loops each)
@mxbi,我添加了一些时间,我的结果与你的不太一致,你应该检查一下
这是一个更大的数组:
In [33]: X = np.arange(10000*100000, dtype=np.int32).reshape(10000,-1)
In [34]: ixs = np.random.randint(0, 100000, (10000, 2))
In [35]: ixs.sort(axis=1)
In [36]: X.shape
Out[36]: (10000, 100000)
In [37]: ixs.shape
Out[37]: (10000, 2)
有一些结果:
In [42]: %timeit np.array([row[ix] for row, ix in zip(X, ixs)])
11.4 ms ± 177 µs per loop (mean ± std. dev. of 7 runs, 100 loops each)
In [43]: %timeit X[np.arange(len(ixs)), ixs.T].T
596 µs ± 17.8 µs per loop (mean ± std. dev. of 7 runs, 1000 loops each)
In [44]: %timeit X.take(ixs+np.arange(0, X.shape[0]*X.shape[1], X.shape[1])[:,None])
540 µs ± 16.6 µs per loop (mean ± std. dev. of 7 runs, 1000 loops each)
现在,我们使用500个索引而不是两个,我们看到列表推导式开始胜出:
In [45]: ixs = np.random.randint(0, 100000, (10000, 500))
In [46]: ixs.sort(axis=1)
In [47]: %timeit np.array([row[ix] for row, ix in zip(X, ixs)])
93 ms ± 1.2 ms per loop (mean ± std. dev. of 7 runs, 1 loop each)
In [48]: %timeit X[np.arange(len(ixs)), ixs.T].T
133 ms ± 638 µs per loop (mean ± std. dev. of 7 runs, 10 loops each)
In [49]: %timeit X.take(ixs+np.arange(0, X.shape[0]*X.shape[1], X.shape[1])[:,None])
87.5 ms ± 1.13 ms per loop (mean ± std. dev. of 7 runs, 10 loops each)
X [:,ixs]
。 有人可以在此基础上继续吗? - Billixs2 = np.array(((0, 1), (0, 3), (1, 0), ...))
?如果可以的话,那么X[ixs2[:,0], ixs2[:,1]]
应该能得到你所需的结果。 - Billnumba
JIT的直接for循环实现吗? - juanpa.arrivillaga