Sorting dictionary by values without losing information of keys

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How do I sort a dictionary by value? 38 answers

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If you want to keep the sorted result with a key/value structure I recommend collections.OrderedDict :

from collections import OrderedDict
from operator import itemgetter

dct = {'Rick Porcello, Bos SP': 579.0, 'Chris Sale, CWS SP': 575.0, 'Justin Verlander, Det SP': 601.0, 'Madison Bumgarner, SF SP': 617.0, 'Max Scherzer, Wsh SP': 668.0, 'Johnny Cueto, SF SP': 584.0}

OrderedDict(sorted(dct.items(), key=itemgetter(1), reverse=True))

The key=itemgetter(1) defines that you sort by "values" and reverse=True tells sorted to sort in descending order.

This gives:

OrderedDict([('Max Scherzer, Wsh SP', 668.0),
             ('Madison Bumgarner, SF SP', 617.0),
             ('Justin Verlander, Det SP', 601.0),
             ('Johnny Cueto, SF SP', 584.0),
             ('Rick Porcello, Bos SP', 579.0),
             ('Chris Sale, CWS SP', 575.0)])

and can still be accessed like a normal dictionary:

>>> odict['Chris Sale, CWS SP']
575.0

or iterate over it:

>>> for name, value in odict.items():
...     print('{name}: {value}'.format(name=name, value=value))
Max Scherzer, Wsh SP: 668.0
Madison Bumgarner, SF SP: 617.0
Justin Verlander, Det SP: 601.0
Johnny Cueto, SF SP: 584.0
Rick Porcello, Bos SP: 579.0
Chris Sale, CWS SP: 575.0

Given that it sorts the input it will scale with O(n logn) .

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