> On Thu, Jun 4, 2009 at 10:13 AM, Alan G Isaac <ais...@american.edu> wrote: >> Or if a stable order is not important (I don't >> recall if the OP specified), one could just >> np.intersect1d(a, np.unique(b))
On 6/4/2009 10:50 AM josef.p...@gmail.com apparently wrote: > This requires that also `a` has only unique elements. > intersect1d_nu doesn't require unique elements. >>> a array([1, 1, 2, 3, 3, 4]) >>> b array([1, 4]) >>> np.intersect1d(a, np.unique(b)) array([1, 1, 3, 4]) (And thus my question about intersect1d...) Cheers, Alan _______________________________________________ Numpy-discussion mailing list Numpy-discussion@scipy.org http://mail.scipy.org/mailman/listinfo/numpy-discussion
