Steven D'Aprano: > since g is not an arbitrary iterator, one can easily do this: > print [(h,len(list(g))) for h,g in groupby(s)] > No need for a special function.
If you look at my first post you can see that I have shown that solution too, but it creates a list that may be long, that may use a lot of of memory, and then throws it away each time. I think that's a bad solution. It goes against the phylosophy of iterators too, they are things created to avoid managing true lists of items too. > If you, the iterator > creator, know enough about the iterator to be sure it has a predictable > length, you know how to calculate it. I don't agree, sometimes I know I have a finite iterator, but I may ignore how many elements it gives (and sometimes they may be a lot). See the simple example with the groupby. Bye, bearophile -- http://mail.python.org/mailman/listinfo/python-list
