"Paul McGuire" <[EMAIL PROTECTED]> wrote in message news:[EMAIL PROTECTED] > So here's how to save the values from the iterators while iterating over the > groupby: > > >>> m = [(x,list(y)) for x,y in groupby([1, 1, 1, 2, 2, 3])] > >>> m > [(1, [1, 1, 1]), (2, [2, 2]), (3, [3])] > > -- Paul > >
Playing some more with groupby. Here's a one-liner to tally a list of integers into a histogram: # create data set, random selection of numbers from 1-10 dataValueRange = range(1,11) data = [random.choice(dataValueRange) for i in xrange(10)] print data # tally values into histogram: # (from the inside out: # - sort data into ascending order, so groupby will see all like values together # - call groupby, return iterator of (value,valueItemIterator) tuples # - tally groupby results into a dict of (value, valueFrequency) tuples # - expand dict into histogram list, filling in zeroes for any keys that didn't get a value hist = [ (k1,dict((k,len(list(g))) for k,g in itertools.groupby(sorted(data))).get(k1,0)) for k1 in dataValueRange ] print hist Gives: [9, 6, 8, 3, 2, 3, 10, 7, 6, 2] [(1, 0), (2, 2), (3, 2), (4, 0), (5, 0), (6, 2), (7, 1), (8, 1), (9, 1), (10, 1)] Change the generation of the original data list to 10,000 values, and you get something like: [(1, 995), (2, 986), (3, 941), (4, 998), (5, 978), (6, 1007), (7, 997), (8, 1033), (9, 1038), (10, 1027)] If you know there wont be any zero frequency values (or don't care about them), you can skip the fill-in-the-zeros step, with one of these expressions: histAsList = [ (k,len(list(g))) for k,g in itertools.groupby(sorted(data)) ] histAsDict = dict((k,len(list(g))) for k,g in itertools.groupby(sorted(data))) -- Paul -- http://mail.python.org/mailman/listinfo/python-list
