Reverse diagonal on numpy python

Question

let's say I have this: (numpy array)

a=
[0  1  2  3],
[4  5  6  7],
[8  9 10  11]

to get [1,1] which is 5 its diagonal is zero; according to numpy, a.diagonal(0)= [0,5,10]. How do I get the reverse or the right to left diagonal [2,5,8] for [1,1]? Is this possible? My original problem is an 8 by 8 (0:7).. I hope that helps

there are two answers that appear not generalized enough becuase your sample a is small. Do you want the answer for [2,2] to be [7, 10]? related question, can you be rectangular in the other direction (tall not wide)? — Phil Cooper
– Phil Cooper, Commented Nov 23, 2013 at 16:46

falsetru · Accepted Answer · 2013-11-23 16:32:46Z

18

Get a new array each row reversed.

>>> import numpy as np
>>> a = np.array([
...     [0, 1, 2, 3],
...     [4, 5, 6, 7],
...     [8, 9, 10, 11]
... ])
>>> a[:, ::-1]
array([[ 3,  2,  1,  0],
       [ 7,  6,  5,  4],
       [11, 10,  9,  8]])
>>> a[:, ::-1].diagonal(1)
array([2, 5, 8])

or using numpy.fliplr:

>>> np.fliplr(a).diagonal(1)
array([2, 5, 8])

edited Nov 23, 2013 at 16:32

answered Nov 23, 2013 at 16:24

falsetru

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6 Comments

wim Over a year ago

Note for the general case this requires calculating the offset from the other side of the array, e.g. you will need to use a.shape in your calculations

falsetru Over a year ago

@wim, You're right. Your answer is more elegant in regard to this matter. Thank you for comment. +1

MasterWizard Over a year ago

In my program, I dont get to know the "1" inside the diagonal function? How do I get that?

falsetru Over a year ago

@AhmedNassar, You can use max(a.shape[1]-a.shape[0], 0) instead of 1.

MasterWizard Over a year ago

Sorry, I'm confused how can I use this. i might've not been clear, let's say I want to get [3,6,9] what operation should I take to get it straight away, without looking at the table.

|

wim · Accepted Answer · 2013-11-23 16:30:33Z

5

Flip the array upside-down and use the same:

np.flipud(a).diagonal(0)[::-1]

answered Nov 23, 2013 at 16:30

wim

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Comments

Akavall · Accepted Answer · 2013-11-23 17:25:21Z

4

Another way to achieve this is to use np.rot90

import numpy as np

a = np.array([[0,  1,  2,  3],
              [4,  5,  6,  7],
              [8,  9, 10,  11]])            

my_diag = np.rot90(a).diagonal(-1)

Result:

>>> my_diag
array([2, 5, 8])

answered Nov 23, 2013 at 17:25

Akavall

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Comments

Phil Cooper · Accepted Answer · 2014-01-06 15:33:10Z

A number of answers so far. @Akavall is closest as you need to rotate or filip and transpose (equivilant operations). I haven't seen a response from the OP regarding expected behavior on the "long" part of the rectangle.

Generalized solution for a square matrix:

a = array([[ 0,  1,  2,  3,  4],
           [ 5,  6,  7,  8,  9],
           [10, 11, 12, 13, 14],
           [15, 16, 17, 18, 19],
           [20, 21, 22, 23, 24]])
>>> [(i, np.rot90(a).diagonal(2*i-a.shape[0]+1)) for i in range(a.shape[0])]
[(0, array([0])),
 (1, array([ 2,  6, 10])),
 (2, array([ 4,  8, 12, 16, 20])),
 (3, array([14, 18, 22])),
 (4, array([24]))]

As a function:

def reverse_diag(arr, n):
    idx = 2*n - arr.shape[0]+1
    return np.rot90(arr).diagonal(idx)

original matrix can be made square with a[:np.min(a.shape),:np.min(a.shape)]

EDIT: OP indicated the array is square.... Final Answer is the above

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