我有像下面这样的字典
dict1 = {'a':{'a':20, 'b':30}, 'b':{'a':30, 'b':40}, 'c':{'a':20, 'b':30}, 'd':{'a':30, 'b':40}}
然后在下面的字典中,两个字典是相同的,所以预期的结果如下
result = [['a','c'],['b','d']]
我有像下面这样的字典
dict1 = {'a':{'a':20, 'b':30}, 'b':{'a':30, 'b':40}, 'c':{'a':20, 'b':30}, 'd':{'a':30, 'b':40}}
然后在下面的字典中,两个字典是相同的,所以预期的结果如下
result = [['a','c'],['b','d']]
>>> seen = {}
>>> dict1 = {'a':{'a':20, 'b':30}, 'b':{'a':30, 'b':40}, 'c':{'a':20, 'b':30}, 'd':{'a':30, 'b':40}}
>>> for k in dict1:
fs = frozenset(dict1[k].items())
seen.setdefault(fs, []).append(k)
>>> seen.values() # note: unordered
[['a', 'c'], ['b', 'd']]
如果需要订购:
>>> from collections import OrderedDict
>>> dict1 = {'a':{'a':20, 'b':30}, 'b':{'a':30, 'b':40}, 'c':{'a':20, 'b':30}, 'd':{'a':30, 'b':40}}
>>> seen = OrderedDict()
>>> for k in sorted(dict1):
fs = frozenset(dict1[k].items())
seen.setdefault(fs, []).append(k)
>>> seen.values()
[['a', 'c'], ['b', 'd']]
注意:此代码目前在 Python 2/3 上是交叉兼容的。在 Python 2 上,您可以通过使用来提高效率,.iteritems()
而不是.items()
一个快速的:首先获得不同的值,然后列出理解。
>>> values = []
>>> for k in dict1:
if dict1[k] not in values:
values.append(dict1[k])
>>> values
[{'a': 20, 'b': 30}, {'a': 30, 'b': 40}]
>>> [[k for k in dict1 if dict1[k] == v] for v in values]
[['a', 'c'], ['b', 'd']]