你应该使用此元组的列表:
>>> lst = [d.items()[0] for d in similarity]
>>> lst
[('Ben', 49), ('Moose', 18), ('Reuven', 39), ('Cust1', 58), ('Cust2', 10), ('Francois', 58), ('Jim C', 39), ('Iren', 13), ('Cust3', 13)]
然后,你可以整理之如常。
>>> from operator import itemgetter
>>> sorted(lst, key=itemgetter(1))
[('Cust2', 10), ('Iren', 13), ('Cust3', 13), ('Moose', 18), ('Reuven', 39), ('Jim C', 39), ('Ben', 49), ('Cust1', 58), ('Francois', 58)]
如果你愿意,你也可以使用一个单一的,ordered dictionary持有的价值观:
>>> from collections import OrderedDict
>>> OrderedDict(sorted(lst, key=itemgetter(1)))
OrderedDict([('Cust2', 10), ('Iren', 13), ('Cust3', 13), ('Moose', 18), ('Reuven', 39), ('Jim C', 39), ('Ben', 49), ('Cust1', 58), ('Francois', 58)])
我能问你为什么使用单次入境词典列表?这些可以合并成一个单一的字典? – senderle 2012-04-14 22:19:45
@senderle:或者也许他可以使用元组列表... – rubik 2012-04-14 22:31:09