getting the opposite diagonal of a numpy array

Question

So in numpy arrays there is the built in function for getting the diagonal indices, but I can\'t seem to figure out how to get the diagonal starting from the top right rathe

Answer 1

There is

In [47]: np.diag(np.fliplr(array))
Out[47]: array([ 4,  8, 12, 16, 20])

or

In [48]: np.diag(np.rot90(array))
Out[48]: array([ 4,  8, 12, 16, 20])

Of the two, np.diag(np.fliplr(array)) is faster:

In [50]: %timeit np.diag(np.fliplr(array))
100000 loops, best of 3: 4.29 us per loop

In [51]: %timeit np.diag(np.rot90(array))
100000 loops, best of 3: 6.09 us per loop

Answer 2

Here are two ideas:

step = len(array) - 1

# This will make a copy
array.flat[step:-step:step]

# This will make a veiw
array.ravel()[step:-step:step]

Answer 3

Here is a simple way using numpy slicing. I personally find it not to hard on the eyes (but concede that fliplr is a little more descriptive!).

Just to highlight this example's contribution to existing answers, I ran the same simple benchmark.

In [1]: import numpy as np

In [3]: X = np.random.randint(0, 10, (5, 5))

In [4]: X
Out[4]: 
array([[7, 2, 7, 3, 7],
       [8, 4, 5, 9, 6],
       [0, 2, 9, 0, 4],
       [8, 2, 1, 0, 3],
       [3, 1, 0, 7, 0]])

In [5]: Y = X[:, ::-1]

In [6]: Z1 = np.diag(Y)

In [7]: Z1
Out[7]: array([7, 9, 9, 2, 3])

Now to compare to the current fastest solution given.

In [8]: step = len(X) - 1

In [9]: Z2 = np.take(X, np.arange(step, X.size-1, step))

In [10]: Z2
Out[10]: array([7, 9, 9, 2, 3])

In [11]: np.array_equal(Z1, Z2)
Out[11]: True

Benchmarks

In [12]: %timeit np.diag(X[:, ::-1])
1.92 µs ± 29.5 ns per loop (mean ± std. dev. of 7 runs, 1000000 loops each)

In [13]: %timeit step = len(X) - 1; np.take(X, np.arange(step, X.size-1, step))
2.21 µs ± 246 ns per loop (mean ± std. dev. of 7 runs, 100000 loops each)

Initial comparisons indicate that my solution is additionally linear in complexity, while using the second 'step' solution is not:

In [14]: big_X = np.random.randint(0, 10, (10000, 10000))

In [15]: %timeit np.diag(big_X[:, ::-1])
2.15 µs ± 96.3 ns per loop (mean ± std. dev. of 7 runs, 1000000 loops each)

In [16]: %timeit step = len(big_X) - 1; np.take(big_X, np.arange(step, big_X.size-1, step))
100 µs ± 1.85 µs per loop (mean ± std. dev. of 7 runs, 10000 loops each)

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I generally use this method to either flip images (mirroring them), or to convert between opencv's format of (channels, height, width) to matplotlib's format of (height, width , channels). So for a three dimensional image, it'd simply be flipped = image[:, :, ::-1]. Of course you can generalise it to flip along any dimension, by putting the ::-1 part in the desired dimension.