Indexing 2d array with 2d array in Numpy

Question

I have a question that bothers me for a few days. Let's assume we define a 2d array in Numpy:

x = np.array([[0, 1, 2],
              [3, 4, 5],
              [6, 7, 8]])

We also define a 1d array for indexing, let's say:

ind = np.array([2,1])

If we will try x[ind] we will get:

array([[6, 7, 8],
       [3, 4, 5]])

which makes a lot of sense: Row number 2 and row numer 1 of x.

If we will run: x[:,ind] we will get:

array([[2, 1],
       [5, 4],
       [8, 7]])

Again, it makes a lot of sense - we receive column number 2 followed by column number 1

Now we will define the index array as 2d:

ind = np.array([[2,1],
                [2,2]])

If we run x[ind] we get:

array([[[6, 7, 8],
        [3, 4, 5]],

       [[6, 7, 8],
        [6, 7, 8]]])

Again, it makes sense - for each row in the indexing 2d array we receive a 2d array that represent the corresponding rows from the original 2d array x.

However, if we run x[:,ind] we receive the next array:

array([[[2, 1],
        [2, 2]],

       [[5, 4],
        [5, 5]],

       [[8, 7],
        [8, 8]]])

I don't understand this output since it returns specific item in the indexed rows, but not the full rows. I would assume, that just like the case of x[:,ind] when it was 1d array, we will receive 2d arrays that include the original columns from the original x array.

yatu · Accepted Answer · 2020-04-16 20:15:13Z

1

In the last case with the indexing array:

print(ind)
array([[2, 1],
       [2, 2]])

Since ind is a 2D array of shape (2,2), and your taking a full slice along the first axis, with ind you'll be indexing along the columns of A on each of its rows. So for instance by indexing the second row [3, 4, 5] with ind, you'll get the elements at indices 2->5, 1->4, 2->5 and 2->5 again, with the resulting shape being the same as ind, so [[5,4][5,5]].

The same for each of its rows resulting in a (3,2,2) shaped array.

answered Apr 16, 2020 at 20:15

yatu

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5 Comments

lirand Over a year ago

Why the slicing is only on the columns of each row separately and not on the full column, like in the first case I presented of 2d array?

yatu Over a year ago

Because in the first case youre slicing along the first axis. Note that x[ind] is the same as x[ind,:]. So unless you specify it @lirand by taking full slices along the preceding axes (x[:,ind]), you'll be indexing along the first: x[first_axis_ixs, secoind_axis_ixs, ...]

yatu Over a year ago

Makes sense @lirand ?

lirand Over a year ago

Im not quite sure - If I use x[:,ind] I would expect that for each on the indices in the 2d array I will get a full columns i.e for ind[0,0] (2) we will get [2,5,8].T and for ind[0,1] (1)we will get [1,4,7].T. for ind[1,1] and ind[1,2] (the second row) we will get [2,5,8] and [2,5,8] so the out put will be two 3x2 arrays.

yatu Over a year ago

No. It doesn't work like that. It's because you're indexing on the columns for each of the rows, so you only get the indices along the values in each row. You'd get that result with x.T[ind] @lirand

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