1.利用key排序 d = {'d1':2, 'd2':4, 'd4':1,'d3':3,} for k in sorted(d): print(k,d[k]) d1 2d2 4d3 3d4 1 2.利用value排序:__getitem__ d = {'d1':2, 'd2':4, 'd4':1,'d3':3,} for k in sorted(d,key=d.__getitem__): print(k,d[k]) d4 1d1 2d3 3d2 4 反序:reverse=True d = {'
#字典值相加 def union_dic(*objs): _keys = set(sum([obj.keys() for obj in objs],[])) _total = {} for _key in _keys: _total[_key] = sum([obj.get(_key,0) for obj in objs]) return _total
2种方式,update()和items()方式 In [14]: a Out[14]: {'a': 1, 'b': 2, 'c': 3} In [15]: c = {'d': 4} In [16]: a.update(c) In [17]: a Out[17]: {'a': 1, 'b': 2, 'c': 3, 'd': 4} In [18]: a = {'a': 1} In [19]: a Out[19]: {'a': 1} In [20]: c Out[20]: {'d': 4} In [2