> Martin v. Löwis wrote: >>> Hmm, perhaps when using sets as work queues? >> >> A number of comments: >> >> - it's somewhat confusing to use a set as a *queue*, given >> that it won't provide FIFO semantics. >> - there are more appropriate and direct container structures >> available, including a dedicated Queue module (even though >> this might be doing to much with its thread-safety). >> - if you absolutely want to use a set as a work queue, >> then the .pop() method should be sufficient, right? >> >> Regards, >> Martin > > We were using sets to track the tips of a graph, and to compare whether > one node was an ancestor of another one. We were caching that answer > into frozensets, since that made them immutable. If > > res = heads(node1, node2) > if len(res) == 1: > # What is the 'obvious' way to get the node out? > > I posit that there *isn't* an obvious way to get the single item out of > a 1-entry frozenset. > > for x in res: break > list(res)[0] > set(res).pop() > iter(res).next() > [x for x in res][0] > x, = res # I didn't think of this one before recently > > Are all answers, but none of them I would consider *obvious*. At the > least, none of them are obviously better than another, so you look into > the performance characteristics to give you a reason to pick one over > the other. > > res.get() would be a fairly obvious way to do it. Enough that I would > probably never have gone searching for any of the other answers. Though > personally, I think I would call it "set.peek()", but the specific name > doesn't really matter to me. > > John >
When I first wrote Graphine (graph library), I did something very similar to the last solution. The code has since been rewritten to avoid the issue, but it would be nice if it didn't have to be the next time it comes up- the comma is easy to miss, and while the results are common-sense once you get what the line is doing, it doesn't lend itself to immediate comprehension. Geremy Condra _______________________________________________ Python-Dev mailing list Python-Dev@python.org http://mail.python.org/mailman/listinfo/python-dev Unsubscribe: http://mail.python.org/mailman/options/python-dev/archive%40mail-archive.com
