使用两个1d数组有效索引2d numpy数组 [英] Efficiently index 2d numpy array using two 1d arrays

问题描述

我有一个大的2d numpy数组和两个1d数组，它们代表2d数组中的x/y索引.我想使用这些1d数组对2d数组执行操作. 我可以使用for循环来做到这一点，但是在大型数组上工作时非常慢.有没有更快的方法?我尝试将1d数组简单地用作索引，但这没有用.参见以下示例:

I have a large 2d numpy array and two 1d arrays that represent x/y indexes within the 2d array. I want to use these 1d arrays to perform an operation on the 2d array. I can do this with a for loop, but it's very slow when working on a large array. Is there a faster way? I tried using the 1d arrays simply as indexes but that didn't work. See this example:

import numpy as np

# Two example 2d arrays
cnt_a   =   np.zeros((4,4))
cnt_b   =   np.zeros((4,4))

# 1d arrays holding x and y indices
xpos    =   [0,0,1,2,1,2,1,0,0,0,0,1,1,1,2,2,3]
ypos    =   [3,2,1,1,3,0,1,0,0,1,2,1,2,3,3,2,0]

# This method works, but is very slow for a large array
for i in range(0,len(xpos)):
    cnt_a[xpos[i],ypos[i]] = cnt_a[xpos[i],ypos[i]] + 1

# This method is fast, but gives incorrect answer
cnt_b[xpos,ypos] = cnt_b[xpos,ypos]+1


# Print the results
print 'Good:'
print cnt_a
print ''
print 'Bad:'
print cnt_b

输出为:

Good:
[[ 2.  1.  2.  1.]
 [ 0.  3.  1.  2.]
 [ 1.  1.  1.  1.]
 [ 1.  0.  0.  0.]]

Bad:
[[ 1.  1.  1.  1.]
 [ 0.  1.  1.  1.]
 [ 1.  1.  1.  1.]
 [ 1.  0.  0.  0.]]

对于cnt_b数组，numpy显然不能正确求和，但是我不确定如何解决此问题而不求助于用于计算cnt_a的(v.低效)for循环.

For the cnt_b array numpy is obviously not summing correctly, but I'm unsure how to fix this without resorting to the (v. inefficient) for loop used to calculate cnt_a.

推荐答案

我们可以计算线性索引，然后累加到使用 np.add.at .因此，以xpos和ypos作为数组，这是一个实现-

We could compute the linear indices, then accumulate into zeros-initialized output array with np.add.at. Thus, with xpos and ypos as arrays, here's one implementation -

m,n = xpos.max()+1, ypos.max()+1
out = np.zeros((m,n),dtype=int)
np.add.at(out.ravel(), xpos*n+ypos, 1)

样品运行-

In [95]: # 1d arrays holding x and y indices
    ...: xpos    =   np.array([0,0,1,2,1,2,1,0,0,0,0,1,1,1,2,2,3])
    ...: ypos    =   np.array([3,2,1,1,3,0,1,0,0,1,2,1,2,3,3,2,0])
    ...: 

In [96]: cnt_a   =   np.zeros((4,4))

In [97]: # This method works, but is very slow for a large array
    ...: for i in range(0,len(xpos)):
    ...:     cnt_a[xpos[i],ypos[i]] = cnt_a[xpos[i],ypos[i]] + 1
    ...:     

In [98]: m,n = xpos.max()+1, ypos.max()+1
    ...: out = np.zeros((m,n),dtype=int)
    ...: np.add.at(out.ravel(), xpos*n+ypos, 1)
    ...: 

In [99]: cnt_a
Out[99]: 
array([[ 2.,  1.,  2.,  1.],
       [ 0.,  3.,  1.,  2.],
       [ 1.,  1.,  1.,  1.],
       [ 1.,  0.,  0.,  0.]])

In [100]: out
Out[100]: 
array([[2, 1, 2, 1],
       [0, 3, 1, 2],
       [1, 1, 1, 1],
       [1, 0, 0, 0]])

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