What is the best way to perform an anti-transpose in python?

Question

Lets say I have an array

a = np.arange(16).reshape((4,4))

0   1  2   3
4   5  6   7
8   9  10  11
12  13 14  15

But I want

15  11  7  3
14  10  6  2
13  9   5  1
12  8   4  0

which is a flip across the secondary diagonal, or a kind of anti-transpose.

How can I do this in numpy?

Evanevander

Have you attempted anything or are you simply getting your homework done by us... – Fortissimo

@VaibhavBajaj self-answered question – Cummings

@AnttiHaapala :O Oops. Didn't see that. – Fortissimo

Answer 1

One could do one of the following:

rot90(a,2).T

rot90(flipud(a),1)

rot90(fliplr(a), -1)

or as hpaulj suggested in the comments (thanks hpaulj)

a[::-1,::-1].T

Here are the speed rankings as ratios of the slowest method after anti-transposing 1000 random 10000x10000 arrays.

1. 63.5% - a[::-1,::-1].T
2. 85.6% - rot90(a,2).T
3. 97.8% - rot90(flipud(a),1)
4. 100% -rot90(fliplr(a),-1)

Evanevander

I'm hoping someone knows a secret numpy way of doing this. – Evanevander

a[::-1,::-1].T; this is a view, sharing the same data buffer, so it will be as fast as possible. To test for view, try changing one element and see if it changes the original. The double reverse slice is the same as np.rot90(a,2). Look at it's code. – Excel

@Excel so you should add that as an answer – Cummings

Answer 2

np.flip(a).T

From the np.flip documentation & @hpaulj 's comment:

flip(m) corresponds to m[::-1,::-1,...,::-1] with ::-1 at all positions.

Tome

Answer 3

Here's another to throw into the mix.

a.ravel('F')[::-1].reshape(a.shape)

Aphasic

Although I like this because of ravel, its also about 200x slower than rot90(fliplr(a),-1). :/ – Evanevander

No worries, I was just exploring other ways of doing it (-: – Aphasic

Answer 4

Try it in this manner,

np=np[::-1] #reverse the array
a = np.arange(16).reshape((4,4))

Horsefly