slice operation for numpy in python2

Question

If I have a numpy array X:

array([[ 0.13263767,  0.23149757,  0.57097612],
       [ 0.49629958,  0.67507182,  0.6758823 ]])

And an index array Y:

array([1, 2, 1])

I could use X[0:,Y] to index the first row of X, and it will output the:

array([ 0.23149757,  0.57097612,  0.23149757])

my question is, if I have an index array Z with 2 dimensions:

array([[1, 2, 1],
       [0, 1, 2]])

I would like to use first row to of Z to index the first row of X, and second row of Z to index the second row of X (Z and X have the same rows). So one way is to use command as follow:

Row_0 = X[0:, Z[0]]
Row_1 = X[1:, Z[1]]

I was wondering if there is a simple way to do this. Thanks

Hi, it seems that X[0,Y] could only get the first row index result of X. tricks proposed by @Jaime work well. — Lei Hua
– Lei Hua, Commented Aug 25, 2016 at 7:56
I meant that you should change it in the question since X[0:,Y] doesn't output what you claim it does. — Ohad Eytan
– Ohad Eytan, Commented Aug 25, 2016 at 8:22

Jaime · Accepted Answer · 2016-08-24 08:50:13Z

3

You can use fancy indexing to achieve that:

>>> X[[[0], [1]], Z]
array([[ 0.23149757,  0.57097612,  0.23149757],
       [ 0.49629958,  0.67507182,  0.6758823 ]])

The trick is that the array indexing the first dimension must broadcast with the one indexing the second one. In this case:

>>> np.array([[0], [1]]).shape
(2, 1)
>>> Z.shape
(2, 3)

So the return will be of the broadcast shape, (2, 3) with indices taken from the first array for the first dimension, and from the second array for the second dimension.

For more general cases, you can get the same result as:

>>> X[np.arange(Z.shape[0])[:, None], Z]
array([[ 0.23149757,  0.57097612,  0.23149757],
       [ 0.49629958,  0.67507182,  0.6758823 ]])

answered Aug 24, 2016 at 8:50

Jaime

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