How to filter numpy array by list of indices?

Question

I have a numpy array, filtered__rows , comprised of LAS data [x, y, z, intensity, classification] . I have created a cKDTree of points and have found nearest neighbors, query_ball_point , which is a list of indices for the point and its neighbors.

Is there a way to filter filtered__rows to create an array of only points whose index is in the list returned by query_ball_point ?

Answer 1

It looks like you just need a basic integer array indexing :

filter_indices = [1,3,5]
np.array([11,13,155,22,0xff,32,56,88])[filter_indices] 

Answer 2

numpy.take can be useful and works well for multimensional arrays.

import numpy as np

filter_indices = [1, 2]
array = np.array([[1, 2, 3, 4, 5], 
                  [10, 20, 30, 40, 50], 
                  [100, 200, 300, 400, 500]])

axis = 0
print(np.take(array, filter_indices, axis))
# [[ 10  20  30  40  50]
#  [100 200 300 400 500]]

axis = 1
print(np.take(array, filter_indices, axis))
# [[  2   3]
#  [ 20  30]
# [200 300]]

Answer 3

Using Docs: https://docs.scipy.org/doc/numpy-1.13.0/user/basics.indexing.html The following implementation should work for arbitrary number of dimensions/shapes for some numpy ndarray.

First we need a multi-dimensional set of indexes and some example data:

import numpy as np
y = np.arange(35).reshape(5,7)
print(y) 
indexlist = [[0,1], [0,2], [3,3]]
print ('indexlist:', indexlist)

To actually extract the intuitive result the trick is to use a Transpose:

indexlisttranspose = np.array(indexlist).T.tolist()
print ('indexlist.T:', indexlisttranspose)
print ('y[indexlist.T]:', y[ tuple(indexlisttranspose) ])

Makes the following terminal output:

 y: [[ 0 1 2 3 4 5 6] [ 7 8 9 10 11 12 13] [14 15 16 17 18 19 20] [21 22 23 24 25 26 27] [28 29 30 31 32 33 34]] indexlist: [[0, 1], [0, 2], [3, 3]] indexlist.T: [[0, 0, 3], [1, 2, 3]] y[indexlist.T]: [ 1 2 24]

The tuple... fixes a future warning which we can cause like so:

print ('y[indexlist.T]:', y[ indexlisttranspose ])

 FutureWarning: Using a non-tuple sequence for multidimensional indexing is deprecated; use `arr[tuple(seq)]` instead of `arr[seq]`. In the future this will be interpreted as an array index, `arr[np.array(seq)]`, which will result either in an error or a different result. print ('y[indexlist.T]:', y[ indexlisttranspose ]) y[indexlist.T]: [ 1 2 24]

Answer 4

Do you know how that translates for multi-dimensional arrays?

It can be expanded to multi dimensional arrays by giving a 1d array for every index so for a 2d array

filter_indices=np.array([[1,0],[0,1]])
array=np.array([[0,1],[1,2]])
print(array[filter_indices[:,0],filter_indices[:,1])

will give you : [1,1]

Scipy has an explanation on what will happen if you call: print(array[filter_indices])

Docs - https://docs.scipy.org/doc/numpy-1.13.0/user/basics.indexing.html

Answer 5

最快的方法是X[tuple(index.T)] ，其中X是包含元素的 ndarray， index是希望检索的索引的 ndarray。