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Explore Stack InternalFrom the numpythe NumPy mailing list, here's another solution:
>>> a array([[1, 2], [0, 0], [1, 0], [0, 2], [2, 1], [1, 0], [1, 0], [0, 0], [1, 0], [2, 2]]) >>> a[np.lexsort(np.fliplr(a).T)] array([[0, 0], [0, 0], [0, 2], [1, 0], [1, 0], [1, 0], [1, 0], [1, 2], [2, 1], [2, 2]]) From the numpy mailing list, here's another solution:
>>> a array([[1, 2], [0, 0], [1, 0], [0, 2], [2, 1], [1, 0], [1, 0], [0, 0], [1, 0], [2, 2]]) >>> a[np.lexsort(np.fliplr(a).T)] array([[0, 0], [0, 0], [0, 2], [1, 0], [1, 0], [1, 0], [1, 0], [1, 2], [2, 1], [2, 2]]) From the NumPy mailing list, here's another solution:
>>> a array([[1, 2], [0, 0], [1, 0], [0, 2], [2, 1], [1, 0], [1, 0], [0, 0], [1, 0], [2, 2]]) >>> a[np.lexsort(np.fliplr(a).T)] array([[0, 0], [0, 0], [0, 2], [1, 0], [1, 0], [1, 0], [1, 0], [1, 2], [2, 1], [2, 2]]) From the numpy mailing list, here's another solution:
>>> a array([[1, 2], [0, 0], [1, 0], [0, 2], [2, 1], [1, 0], [1, 0], [0, 0], [1, 0], [2, 2]]) >>> a[np.lexsort(np.fliplr(a).T)] array([[0, 0], [0, 0], [0, 2], [1, 0], [1, 0], [1, 0], [1, 0], [1, 2], [2, 1], [2, 2]])