Question

我有一个看起来如下的numpy操作：

 for i in range(i_max):
    for j in range(j_max):
        r[i, j, x[i, j], y[i, j]] = c[i, j]

其中x，y和c具有相同的形状。

是否可以使用numpy的高级索引来加快此操作？

我尝试使用：

i = numpy.arange(i_max)
j = numpy.arange(j_max)
r[i, j, x, y] = c

然而，我没有得到我预期的结果。

Answer 1

使用linear indexing -

d0,d1,d2,d3 = r.shape
np.put(r,np.arange(i_max)[:,None]*d1*d2*d3 + np.arange(j_max)*d2*d3 + x*d3 +y,c)

基准测试和验证

定义函数 -

def linear_indx(r,x,y,c,i_max,j_max):
    d0,d1,d2,d3 = r.shape
    np.put(r,np.arange(i_max)[:,None]*d1*d2*d3 + np.arange(j_max)*d2*d3 + x*d3 +y,c)
    return r

def org_app(r,x,y,c,i_max,j_max):
    for i in range(i_max):
        for j in range(j_max):
            r[i, j, x[i,j], y[i,j]] = c[i,j]
    return r

设置输入数组和基准 -

In [134]: # Setup input arrays
     ...: i_max = 40
     ...: j_max = 50
     ...: D0 = 60
     ...: D1 = 70
     ...: N = 80
     ...: 
     ...: r = np.zeros((D0,D1,N,N))
     ...: c = np.random.rand(i_max,j_max)
     ...: 
     ...: x = np.random.randint(0,N,(i_max,j_max))
     ...: y = np.random.randint(0,N,(i_max,j_max))
     ...: 

In [135]: # Make copies for testing, as both functions make in-situ changes
     ...: r1 = r.copy()
     ...: r2 = r.copy()
     ...: 

In [136]: # Verify results by comparing with original loopy approach
     ...: np.allclose(linear_indx(r1,x,y,c,i_max,j_max),org_app(r2,x,y,c,i_max,j_max))
Out[136]: True

In [137]: # Make copies for testing, as both functions make in-situ changes
     ...: r1 = r.copy()
     ...: r2 = r.copy()
     ...: 

In [138]: %timeit linear_indx(r1,x,y,c,i_max,j_max)
10000 loops, best of 3: 115 µs per loop

In [139]: %timeit org_app(r2,x,y,c,i_max,j_max)
100 loops, best of 3: 2.25 ms per loop

Answer 2

索引数组需要broadcastable才能生效。唯一需要做的更改是将轴添加到第一个索引i以使形状与其余索引匹配。快速完成此任务的方法是使用None（相当于numpy.newaxis）建立索引：

i = numpy.arange(i_max)
j = numpy.arange(j_max)
r[i[:,None], j, x, y] = c

优化numpy ndarray索引操作

2 个答案: