On 08/10/06, Robert Kern <[EMAIL PROTECTED]> wrote: > Bill Baxter wrote: > > Yes, that'd be > > a[b] += c > > No, I'm afraid that fancy indexing does not do the loop that you are thinking > it > would (and for reasons that we've discussed previously on this list, *can't* > do > that loop). That statement reduces to something like the following:
So the question remains, is there a for-loop-free way to do this? (This, specifically, is: for i in range(len(b)): a[b[i]]+=c[i] where b[i] may contain repetitions.) I didn't find one, but came to the conclusion that for loops are not necessarily slower than fancy indexing, so the way to do this one is just to use a for loop. A. M. Archibald ------------------------------------------------------------------------- Take Surveys. Earn Cash. Influence the Future of IT Join SourceForge.net's Techsay panel and you'll get the chance to share your opinions on IT & business topics through brief surveys -- and earn cash http://www.techsay.com/default.php?page=join.php&p=sourceforge&CID=DEVDEV _______________________________________________ Numpy-discussion mailing list Numpy-discussion@lists.sourceforge.net https://lists.sourceforge.net/lists/listinfo/numpy-discussion