我想基本上有一个我的词典列表的排序列表,如下所示:
similarity = [{'Ben': 49}, {'Moose': 18}, {'Reuven': 39}, {'Cust1': 58}, {'Cust2': 10}, {'Francois': 58}, {'Jim C': 39}, {'Iren': 13}, {'Cust3': 13}]
感谢任何帮助,并提前感谢。
答案 0 :(得分:4)
你应该在这里使用元组列表:
>>> 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)]
然后你可以照常排序。
>>> 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)]
如果需要,您还可以使用单个ordered dictionary来保存值:
>>> 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)])
答案 1 :(得分:1)
这是一种做法。
>>> sorted(similarity, key=lambda x: x.values()[0])
[{'Cust2': 10}, {'Iren': 13}, {'Cust3': 13}, {'Moose': 18}, {'Reuven': 39}, {'Jim C': 39}, {'Ben': 49}, {'Cust1': 58}, {'Francois': 58}]
答案 2 :(得分:0)
我在这方面遇到困难。
所以我建议你用这种方法
a = {'Ben': 49}
b = {'Moose': 18}
ab = {**a, **b}
print(ab)
{'Ben': 49, 'Moose': 18}
1: sorted(ab.values())
[('Ben', 49), ('Moose', 18)]
2: sorted(ab.items())
[18, 49]