如果密钥存在，则按键对字典进行排序，如果密钥不存在于列表的末尾

Question

Been scratching my head on how to accomplish this,

sorted_dict = sorted(dict['values'],
                    key=lambda k: k['a']['b'])

Now in this dict some values of a have ab value, while some don't. I want it to sort by b values and if it doesn't exist, just put it at the back of the list. Is there anyway to do without some complex code such as splitting the values of a for those that have ab value and those that don't?

Answer 1

不。

key=lambda k: ('b' not in k['a'], k['a'].get('b', None))