I realize there are quite a number of 'how to sort numpy array'-questions on here already. But I could not find how to do it in this specific way.
I have an array similar to this:
array([[1,0,1,],
[0,0,1],
[1,1,1],
[1,1,0]])
I want to sort the rows, keeping the order within the rows the same. So I expect the following output:
array([[0,0,1,],
[1,0,1],
[1,1,0],
[1,1,1]])
You can use dot and argsort:
a[a.dot(2**np.arange(a.shape[1])[::-1]).argsort()]
# array([[0, 0, 1],
# [1, 0, 1],
# [1, 1, 0],
# [1, 1, 1]])
The idea is to convert the rows into integers.
a.dot(2**np.arange(a.shape[1])[::-1])
# array([5, 1, 7, 6])
Then, find the sorted indices and use that to reorder a:
a.dot(2**np.arange(a.shape[1])[::-1]).argsort()
# array([1, 0, 3, 2])
My tests show this is slightly faster than lexsort.
a = a.repeat(1000, axis=0)
%timeit a[np.lexsort(a.T[::-1])]
%timeit a[a.dot(2**np.arange(a.shape[1])[::-1]).argsort()]
230 µs ± 18.1 µs per loop (mean ± std. dev. of 7 runs, 1000 loops each)
192 µs ± 4.1 µs per loop (mean ± std. dev. of 7 runs, 1000 loops each)
Verify correctness:
np.array_equal(a[a.dot(2**np.arange(a.shape[1])[::-1]).argsort()],
a[np.lexsort(a.T[::-1])])
# True
np.packbits is probably a bit faster than doing it manually.
a[np.packbits(a, axis=1).ravel().argsort()] however packbits does not return the same result as the dot version so I cannot confirm how correct it is. Any idea why the results differ? Is it the dtype?packbits pads zeros at the end, not the start (weird). So this means that the relative ordering should remain the same. Let me update.raw = np.packbits(a, axis=1); out = np.zeros((A.shape[0], 1), 'u8'); N = raw.shape[1]; idx = np.s_[:, 7:7-N if N<8 else None:-1] if sys.byteorder=='little' else np.s_[:, :N]; out.view('u1')[idx] = raw and then use out with argsort. Not sure this will still be faster than yours. Feel free to use this. I won't post it myself. No point in needlessly annoying a prospective mod, is there ;-)) Speaking of which, how is the campaign going?
a[np.lexsort(a.T[::-1])]