> On Thu, Jun 4, 2009 at 8:23 AM, Alan G Isaac <ais...@american.edu> wrote: >> a[(a==b[:,None]).sum(axis=0,dtype=bool)]
On 6/4/2009 8:35 AM josef.p...@gmail.com apparently wrote: > If b is large this creates a huge intermediate array True enough, but one could then use fromiter: setb = set(b) itr = (ai for ai in a if ai in setb) out = np.fromiter(itr, dtype=a.dtype) I suspect (?) that b would have to be pretty big relative to a for the repeated testing to be more costly than sorting a. 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 a different note, I think a name change is needed for your function. (Compare intersect1d_nu to see the potential confusion. And btw, what is the use case for intersect1d, which gives neither a set intersection nor a multiset intersection?) Cheers, Alan Isaac _______________________________________________ Numpy-discussion mailing list Numpy-discussion@scipy.org http://mail.scipy.org/mailman/listinfo/numpy-discussion
