On 19 Mrz., 09:45, Peter Otten <__pete...@web.de> wrote:
> Christian wrote:
> > as beginner in python , I struggle somewhat to filter out only the
> > maximum in the values for and get hmax.
> > h = {'abvjv': ('asyak', 0.9014230420411024),
> > 'afqes': ('jarbm', 0.9327883839839753),
> > 'aikdj': ('jarbm', 0.9503941616408824),
> > 'ajbhn': ('jarbm', 0.9323583083061541),
> > 'ajrje': ('jbhdj', 0.9825125732711598),
> > 'anbrw': ('jarbm', 0.950801828672098)}
>
> > hmax = {'abvjv': ('asyak', 0.9014230420411024),
> > 'ajrje': ('jbhdj', 0.9825125732711598),
> > 'anbrw': ('jarbm', 0.950801828672098)}
>
> You can create an intermediate dict:
>
> >>> d = {}
> >>> for k, (k2, v) in h.items():
>
> ... d.setdefault(k2, []).append((v, k))
> ...>>> import pprint
> >>> pprint.pprint(d)
>
> {'asyak': [(0.9014230420411024, 'abvjv')],
> 'jarbm': [(0.9323583083061541, 'ajbhn'),
> (0.950801828672098, 'anbrw'),
> (0.9327883839839753, 'afqes'),
> (0.9503941616408824, 'aikdj')],
> 'jbhdj': [(0.9825125732711598, 'ajrje')]}
>
> Now find the maximum values:
>
> >>> for k, pairs in d.items():
>
> ... v, k2 = max(pairs)
> ... assert k2 not in hmax
> ... hmax[k2] = k, v
> ...>>> pprint.pprint(hmax)
>
> {'abvjv': ('asyak', 0.9014230420411024),
> 'ajrje': ('jbhdj', 0.9825125732711598),
> 'anbrw': ('jarbm', 0.950801828672098)}
>
> > Maybe it easier when i change the structure of h?
>
> Maybe. Here's one option:
>
> >>> pprint.pprint(data)
>
> [('jarbm', 0.9323583083061541, 'ajbhn'),
> ('jarbm', 0.950801828672098, 'anbrw'),
> ('jarbm', 0.9327883839839753, 'afqes'),
> ('asyak', 0.9014230420411024, 'abvjv'),
> ('jbhdj', 0.9825125732711598, 'ajrje'),
> ('jarbm', 0.9503941616408824, 'aikdj')]>>> data.sort()
> >>> from itertools import groupby
> >>> def last(items):
>
> ... for item in items: pass
> ... return item
> ...>>> dmax = [last(group) for key, group in groupby(data, key=lambda item:
> item[0])]
> >>> pprint.pprint(dmax)
>
> [('asyak', 0.9014230420411024, 'abvjv'),
> ('jarbm', 0.950801828672098, 'anbrw'),
> ('jbhdj', 0.9825125732711598, 'ajrje')]
Thanks a lot.
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