numpys fancy indexing pattern

Question

I am trying to get my head around numpy's fancy indexing. While trying to approach the following I am currently unable to solve the problem:

Given the following np.array t.

t = np.array([[6, 1, 8],
           [4, 3, 7],
           [9, 5, 2]])

I want to achieve the following pattern using fancy indexing

array([[8, 1, 8],
       [1, 8, 1],
       [8, 1, 8]])

With my closest approach getting to

array([[8, 1, 8],
       [8, 1, 8],
       [8, 1, 8]])

Using t[:,[2,1,2]][[0,0,0]]

how to tackle such problems?

What is "fancy indexing"? You mean Python's standard-since-forever list slicing syntax? Because if so, there's tons of explainers on the web about how to use that already. Including on SO, of course. E.g. https://stackoverflow.com/questions/509211/understanding-slice-notation — Mike 'Pomax' Kamermans, Jun 18 '21 at 20:43
Thanks for pointing out. In fancy indexing (in difference to list slicing) one passes arrays as indices. — Pierre, Jun 18 '21 at 20:58
Fancy indexing is definitely a python datascience term (albeit rarely used). Simple google search would've shown that though. No need for sass. There is a good read on it here: https://jakevdp.github.io/PythonDataScienceHandbook/02.07-fancy-indexing.html @Mike'Pomax'Kamermans — Gunner Stone, Jun 18 '21 at 21:12
No sass, just good old genuine confusion about an unusual term I've never seen used despite having worked with numpy on more than several occasions. — Mike 'Pomax' Kamermans, Jun 18 '21 at 21:57
@Mike'Pomax'Kamermans, that link is for `numpy`, not base Python. https://numpy.org/doc/stable/reference/arrays.indexing.html#advanced-indexing uses the name **advanced indexing**. For lists there are only 2 options - slices and individual indices. — hpaulj, Jun 19 '21 at 00:42

score 0 · Answer 1 · answered Jun 18 '21 at 21:57

I found two approaches which solve the problem.

Map the indices with the corresponding points to a 2d array of row and column coordinates.

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

t[row,col]

array([[8, 1, 8],
       [1, 8, 1],
       [8, 1, 8]])

which is the same as using

t[  [[0,0,0],[0,0,0],[0,0,0]]  ,  [[2,1,2],[1,2,1],[2,1,2]]  ]

Start by creating a 1d Array including only required numbers and then cast to a 2d array, allthough is approach triggers a Future Warning: 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.
```
  t[0,[1,2]] [[  [[1,0,1],[0,1,0],[1,0,1]]  ]]

  array([[8, 1, 8],
         [1, 8, 1],
         [8, 1, 8]])
```

Your first solution can also be reduced to `t[0, col]`. – ngwalton Jun 18 '21 at 22:44 — ngwalton, Jun 18 '21 at 22:44

hpaulj · Answer 2 · 2021-06-19T00:55:25.820

In [198]: t = np.array([[6, 1, 8],
     ...:            [4, 3, 7],
     ...:            [9, 5, 2]])

Looks like you want to take 2 elements, and repeat them.

In [199]: t[0,[2,1]]                 # select the elements
Out[199]: array([8, 1])

Then taking advantage of how np.resize "pads" a larger array:

In [200]: np.resize(_,(3,3))
Out[200]: 
array([[8, 1, 8],
       [1, 8, 1],
       [8, 1, 8]])

That's not a very general solution, but then I don't know how you'd imagine generalizing the problem

Another way to expand and reshape

In [217]: np.repeat(t[[[0]],[2,1]],5,0).ravel()[:9].reshape(3,3)
Out[217]: 
array([[8, 1, 8],
       [1, 8, 1],
       [8, 1, 8]])

numpys fancy indexing pattern

2 Answers2