From: John Posner on 8 Mar 2010 21:43 On 3/8/2010 9:39 PM, John Posner wrote: <snip> > # gather data > tally_dict = defaultdict(Tally) > for i in range(len(x)): > obj = tally_dict[y[i]] > obj.id = y[i] < statement redundant, remove it > obj.total += x[i] > obj.count += 1 John
From: John Posner on 8 Mar 2010 21:53 On 3/8/2010 9:43 PM, John Posner wrote: > On 3/8/2010 9:39 PM, John Posner wrote: > > <snip> >> obj.id = y[i] < statement redundant, remove it Sorry for the thrashing! It's more correct to say that the Tally class doesn't require an "id" attribute at all. So the code becomes: # from collections import defaultdict class Tally: def __init__(self): self.total = 0 self.count = 0 x = [1 ,2, 8, 5, 0, 7] y = ['a', 'a', 'b', 'c', 'c', 'c'] # gather data tally_dict = defaultdict(Tally) for i in range(len(x)): obj = tally_dict[y[i]] obj.total += x[i] obj.count += 1 # process data result_list = [] for key in sorted(tally_dict): obj = tally_dict[key] mean = 1.0 * obj.total / obj.count result_list.extend([mean] * obj.count) print result_list # John
From: Michael Rudolf on 9 Mar 2010 05:30 Am 08.03.2010 23:34, schrieb dimitri pater  serpia: > Hi, > > I have two related lists: > x = [1 ,2, 8, 5, 0, 7] > y = ['a', 'a', 'b', 'c', 'c', 'c' ] > > what I need is a list representing the mean value of 'a', 'b' and 'c' > while maintaining the number of items (len): > w = [1.5, 1.5, 8, 4, 4, 4] This kinda looks like you used the wrong data structure. Maybe you should have used a dict, like: {'a': [1, 2], 'c': [5, 0, 7], 'b': [8]} ? > I have looked at iter(tools) and next(), but that did not help me. I'm > a bit stuck here, so your help is appreciated! As said, I'd have used a dict in the first place, so lets transform this straight forward into one: x = [1 ,2, 8, 5, 0, 7] y = ['a', 'a', 'b', 'c', 'c', 'c' ] # initialize dict d={} for idx in set(y): d[idx]=[] #collect values for i, idx in enumerate(y): d[idx].append(x[i]) print("d is now a dict of lists: %s" % d) #calculate average for key, values in d.items(): d[key]=sum(values)/len(values) print("d is now a dict of averages: %s" % d) # build the final list w = [ d[key] for key in y ] print("w is now the list of averages, corresponding with y:\n \ \n x: %s \n y: %s \n w: %s \n" % (x, y, w)) Output is: d is now a dict of lists: {'a': [1, 2], 'c': [5, 0, 7], 'b': [8]} d is now a dict of averages: {'a': 1.5, 'c': 4.0, 'b': 8.0} w is now the list of averages, corresponding with y: x: [1, 2, 8, 5, 0, 7] y: ['a', 'a', 'b', 'c', 'c', 'c'] w: [1.5, 1.5, 8.0, 4.0, 4.0, 4.0] Could have used a defaultdict to avoid dict initialisation, though. Or write a custom class: x = [1 ,2, 8, 5, 0, 7] y = ['a', 'a', 'b', 'c', 'c', 'c' ] class A: def __init__(self): self.store={} def add(self, key, number): if key in self.store: self.store[key].append(number) else: self.store[key] = [number] a=A() # collect data for idx, val in zip(y,x): a.add(idx, val) # build the final list: w = [ sum(a.store[key])/len(a.store[key]) for key in y ] print("w is now the list of averages, corresponding with y:\n \ \n x: %s \n y: %s \n w: %s \n" % (x, y, w)) Produces same output, of course. Note that those solutions are both not very efficient, but who cares ;) > thanks! No Problem, Michael
From: Steve Howell on 9 Mar 2010 10:21 On Mar 8, 6:39 pm, John Posner <jjpos...(a)optimum.net> wrote: > On 3/8/2010 5:34 PM, dimitri pater  serpia wrote: > > > Hi, > > > I have two related lists: > > x = [1 ,2, 8, 5, 0, 7] > > y = ['a', 'a', 'b', 'c', 'c', 'c' ] > > > what I need is a list representing the mean value of 'a', 'b' and 'c' > > while maintaining the number of items (len): > > w = [1.5, 1.5, 8, 4, 4, 4] > > > I have looked at iter(tools) and next(), but that did not help me. I'm > > a bit stuck here, so your help is appreciated! > > Nobody expects objectorientation (or the Spanish Inquisition): > Heh. Yep, I avoided OO for this. Seems like a functional problem. My solution is functional on the outside, imperative on the inside. You could add recursion here, but I don't think it would be as straightforward. def num_dups_at_head(lst): assert len(lst) > 0 val = lst[0] i = 1 while i < len(lst) and lst[i] == val: i += 1 return i def smooth(x, y): result = [] while x: cnt = num_dups_at_head(y) avg = sum(x[:cnt]) * 1.0 / cnt result += [avg] * cnt x = x[cnt:] y = y[cnt:] return result > # > from collections import defaultdict > > class Tally: > def __init__(self, id=None): > self.id = id > self.total = 0 > self.count = 0 > > x = [1 ,2, 8, 5, 0, 7] > y = ['a', 'a', 'b', 'c', 'c', 'c'] > > # gather data > tally_dict = defaultdict(Tally) > for i in range(len(x)): > obj = tally_dict[y[i]] > obj.id = y[i] > obj.total += x[i] > obj.count += 1 > > # process data > result_list = [] > for key in sorted(tally_dict): > obj = tally_dict[key] > mean = 1.0 * obj.total / obj.count > result_list.extend([mean] * obj.count) > print result_list > #
From: Steve Howell on 9 Mar 2010 11:29 On Mar 8, 2:34 pm, dimitri pater  serpia <dimitri.pa...(a)gmail.com> wrote: > Hi, > > I have two related lists: > x = [1 ,2, 8, 5, 0, 7] > y = ['a', 'a', 'b', 'c', 'c', 'c' ] > > what I need is a list representing the mean value of 'a', 'b' and 'c' > while maintaining the number of items (len): > w = [1.5, 1.5, 8, 4, 4, 4] > What results are you expecting if you have multiple runs of 'a' in a longer list?
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