On Sat, Feb 06, 2010 at 13:24 -0500, Chris Colbert wrote: > [(match_val_291, identifier_b), (match_val_23, identifier_b), (match_val_22, > identifer_k) .... ] > > Now, what I would like to do is step through this list and find the identifier > which appears first a K number of times.
I think you can use the itertools.groupby(L, lambda el: el[1]) to group elements in your *sorted* list L by the value el[1] (i.e. the identifier) and then iterate through these groups until you find the desired number of instances grouped by the same identifier. Let me exemplify this: >>> from itertools import groupby >>> instances = [(1, 'b'), (2, 'b'), (3, 'a'), (4, 'c'), (5, 'c'), (6, 'c'), >>> (7, 'd')] >>> k = 3 >>> grouped_by_identifier = groupby(instances, lambda el: el[1]) >>> grouped_by_identifier = ((identifier, list(group)) for identifier, group in >>> grouped_by_identifier) >>> k_instances = (group for identifier, group in grouped_by_identifier if >>> len(group) == k) >>> next(k_instances) [(4, 'c'), (5, 'c'), (6, 'c')] >>> next(k_instances) Traceback (most recent call last): File "<input>", line 1, in <module> StopIteration There are certainly millions of ways to do this and most of them will be better than my proposal here, but you might like this approach. Another approach would use itertools.takewhile() or itertools.ifilter() ... Just have a look :-) yours sincerely -- .''`. Wolodja Wentland <wentl...@cl.uni-heidelberg.de> : :' : `. `'` 4096R/CAF14EFC `- 081C B7CD FF04 2BA9 94EA 36B2 8B7F 7D30 CAF1 4EFC
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