使用另一个2d数组的索引提取二维数组的元素

Question

I have a 2d numpy array: data.shape==(n,8), and another ind.shape=(n,4). The ind array is the same length as data, and contains indices like [4,3,0,6]. How can I create another array with shape==(n,4) containing the elements from data specified by the indices from ind? My actual arrays are pretty long (shape[0]), so the loop is slow. There must be a better way than loops?

import numpy as np
# Example data
data = np.array([[ 0.44180102, -0.05941365,  2.1482739 , -0.56875081, -1.45400572,
        -1.44391254, -0.33710766, -0.44214518],
       [ 0.79506417, -2.46156966, -0.09929341, -1.07347179,  1.03986533,
        -0.45745476,  0.58853107, -1.08565425],
       [ 1.40348682, -1.43396403,  0.8267174 , -1.54812358, -1.05854445,
         0.15789466, -0.0666025 ,  0.29058816]])
ind = np.array([[3, 4, 1, 5],
                [4, 7, 0, 1],
                [5, 1, 3, 6]])

# This is the part I want to vectorize:
out = np.zeros(ind.shape)
for i in range(ind.shape[0]):
    out[i,:] = data[i,ind[i,:]]

# This should be good
assert np.all(out == np.array([[-0.56875081, -1.45400572, -0.05941365, -1.44391254],
                        [ 1.03986533, -1.08565425,  0.79506417, -2.46156966],
                        [ 0.15789466, -1.43396403, -1.54812358, -0.0666025 ]]))

Answer 1

This can be easily done if we index into the raveled data array:

out = data.ravel()[ind.ravel() + np.repeat(range(0, 8*ind.shape[0], 8), ind.shape[1])].reshape(ind.shape)

Explanation

It might be easier to understand if it is broken down into three steps:

indices = ind.ravel() + np.repeat(range(0, 8*ind.shape[0], 8), ind.shape[1])
out = data.ravel()[indices]
out = out.reshape(ind.shape)

ind has the information on the elements from data that we want. Unfortunately, it is expressed in 2-D indices. The first line above converts these into indices of the 1-D raveled data . The second line above selects those elements out of the raveled array data . The third line restores the 2-D shape to out . The 2-D indices represented by ind is converted to ind indices has the indices

Answer 2

What you want is something like this:

import numpy as np
data = np.array([[ 0.4, -0.1,  2.1, -0.6, -1.5, -1.4, -0.3, -0.4],
                 [ 0.8, -2.5, -0.1, -1.1,  1. , -0.5,  0.6, -1.1],
                 [ 1.4, -1.4,  0.8, -1.5, -1.1,  0.2, -0.1,  0.3]])
expected = np.array([[-0.6, -1.5, -0.1, -1.4],
                     [ 1. , -1.1,  0.8, -2.5],
                     [ 0.2, -1.4, -1.5, -0.1]])

indI = np.array([[0, 0, 0, 0],
                 [1, 1, 1, 1],
                 [2, 2, 2, 2]])
indJ = np.array([[3, 4, 1, 5],
                 [4, 7, 0, 1],
                 [5, 1, 3, 6]])
out = data[indI, indJ]
assert np.all(out == expected)

Notice that indI and indJ are the same shape and that

out[i, j] == data[indI[i, j], indJ[i, j]]

for all i and j .

You might have noticed that that indI is very repetitive . Because of numpy's broadcasting magic you can simply indI to:

indI = np.array([[0],
                 [1],
                 [2]])

You can build this kind of indI array a few different ways, here is my favorite:

a, b = indJ.shape
indI, _ = np.ogrid[:a, :0]
out = data[indI, indJ]