Python NumPy：在矢量化赋值期间计算索引数组

我想将此NumPy操作矢量化：

for j in range(yt):
    for i in range(xt):
        y[j, i] = x[idx[j, i], j, i]

其中

idx

包含指向

x

切片的轴0索引。有什么简单的方法可以做到这一点吗？

您可以使用：

J, I = np.ogrid[:yt, :xt]
x[idx, J, I]

以下是测试：

import numpy as np

yt, xt = 3, 5
x = np.random.rand(10, 6, 7)
y = np.zeros((yt, xt))
idx = np.random.randint(0, 10, (yt, xt))

for j in range(yt):
    for i in range(xt):
        y[j, i] = x[idx[j, i], j, i]

J, I = np.ogrid[:yt, :xt]
np.all(x[idx, J, I] == y)

这里有一种方法是使用-

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运行时测试和验证输出

本节比较了本篇文章中提出的方法和绩效评估，并验证了结果

函数定义-

def original_app(x,idx):
    _,yt,xt = x.shape
    y = np.zeros((yt,xt))
    for j in range(yt):
        for i in range(xt):
            y[j, i] = x[idx[j, i], j, i]
    return y

def ogrid_based(x,idx):
    _,yt,xt = x.shape
    J, I = np.ogrid[:yt, :xt]
    return x[idx, J, I]

def reshape_based(x,idx):                               
    zt,yt,xt = x.shape
    return x.reshape(zt,-1)[idx.ravel(),np.arange(yt*xt)].reshape(-1,xt)

设置输入-

In [56]: # Inputs
    ...: zt,yt,xt = 100,100,100
    ...: x = np.random.rand(zt,yt,xt)
    ...: idx = np.random.randint(0,zt,(yt,xt))
...: 

验证输出-

In [57]: np.allclose(original_app(x,idx),ogrid_based(x,idx))
Out[57]: True

In [58]: np.allclose(original_app(x,idx),reshape_based(x,idx))
Out[58]: True

时间安排-

In [68]: %timeit original_app(x,idx)
100 loops, best of 3: 6.97 ms per loop

In [69]: %timeit ogrid_based(x,idx)
1000 loops, best of 3: 391 µs per loop

In [70]: %timeit reshape_based(x,idx)
1000 loops, best of 3: 230 µs per loop

In [68]: %timeit original_app(x,idx)
100 loops, best of 3: 6.97 ms per loop

In [69]: %timeit ogrid_based(x,idx)
1000 loops, best of 3: 391 µs per loop

In [70]: %timeit reshape_based(x,idx)
1000 loops, best of 3: 230 µs per loop