python - 如何为一个dict字典进行多层级赋值？

Answer 1

You can implement this structure yourself.
In the example below, AutoVivification inherits from dict

class AutoVivification(dict):
    """Implementation of perl's autovivification feature."""
    def __getitem__(self, item):
        try:
            return dict.__getitem__(self, item)
        except KeyError:
            value = self[item] = type(self)()
            return value
            

We can use AutoVivification like this:

item = AutoVivification()
item['20161101']["age"] = 20
item['20161102']['num'] = 30
print item

Output:

{'20161101': {'age': 20}, '20161102': {'num': 30}}

In addition, there is another implementation method of AutoVivification, which is to directly overload dict's __missing__magic method. Think of it as an extension.

class AutoVivification(dict):
    """Implementation of perl's autovivification feature."""
    def __missing__(self, key):
        value = self[key] = type(self)()
        return value
        

One more thing, Python 2.5 and later versions added the collections.defaultdict type, which can customize a more scalable dict type. collections.defaultdict 类型，该类型可以自定义扩展性更强大的dict类型。
文档中指出，其实现原理就是重载了 __missing__The documentation points out that the implementation principle is to overload the

method. AutoVivification can also be expressed like this:

item = defaultdict(dict)  # 其实现与AutoVivification的实现完全一样
item['20161101']["age"] = 20
item['20161102']['num'] = 30
print item

__missing__defaultdict constructs a dict type whose first parameter is its default_factory. When
is called, the return value is constructed using default_factory.

More examples of defaultdict🎜

Answer 2

I will attach the usage of defaultdict package:

from collections import defaultdict

item = defaultdict(dict)
item['20161101']['age'] = 20
print item

Output:

defaultdict(<type 'dict'>, {'20161101': {'age': 20}})

This way you can achieve the desired effect,

Added:
defaultdict() receives a default parameter, which can be a type name or any callable function without parameters
This is very easy to use

item = defaultdict(lambda:0)
print item['num']

Output:

0