在python中声明多维字典 [英] Declaring a multi dimensional dictionary in python

问题描述

我需要在python中制作一个二维字典.例如new_dic[1][2] = 5

I need to make a two dimensional dictionary in python. e.g. new_dic[1][2] = 5

当我创建new_dic = {}并尝试插入值时，我得到一个KeyError:

When I make new_dic = {}, and try to insert values, I get a KeyError:

new_dic[1][2] = 5
KeyError: 1

该怎么做?

推荐答案

多维字典只是一个字典，其中值本身也是字典，创建了嵌套结构:

A multi-dimensional dictionary is simply a dictionary where the values are themselves also dictionaries, creating a nested structure:

new_dic = {}
new_dic[1] = {}
new_dic[1][2] = 5

不过，您每次都必须检测到您已经创建了new_dic[1]，以免意外擦拭嵌套的对象以获取new_dic[1]下的其他键.

You'd have to detect that you already created new_dic[1] each time, though, to not accidentally wipe that nested object for additional keys under new_dic[1].

您可以使用多种技术简化创建嵌套词典的过程；例如，使用 dict.setdefault()

You can simplify creating nested dictionaries using various techniques; using dict.setdefault() for example:

new_dic.setdefault(1, {})[2] = 5

dict.setdefault()仅在密钥仍然丢失的情况下才将密钥设置为默认值，从而使您不必每次都进行测试.

dict.setdefault() will only set a key to a default value if the key is still missing, saving you from having to test this each time.

Simpler仍在使用 collections.defaultdict()类型自动创建嵌套字典:

Simpler still is using the collections.defaultdict() type to create nested dictionaries automatically:

from collections import defaultdict

new_dic = defaultdict(dict)
new_dic[1][2] = 5

这里的

defaultdict只是标准dict类型的子类；每次尝试访问映射中尚不存在的键时，都会调用工厂函数来创建新值.这就是 dict()可调用，它在被调用时会生成一个空字典.

defaultdict is just a subclass of the standard dict type here; every time you try and access a key that doesn't yet exist in the mapping, a factory function is called to create a new value. Here that's the dict() callable, which produces an empty dictionary when called.

演示:

>>> new_dic_plain = {}
>>> new_dic_plain[1] = {}
>>> new_dic_plain[1][2] = 5
>>> new_dic_plain
{1: {2: 5}}
>>> new_dic_setdefault = {}
>>> new_dic_setdefault.setdefault(1, {})[2] = 5
>>> new_dic_setdefault
{1: {2: 5}}
>>> from collections import defaultdict
>>> new_dic_defaultdict = defaultdict(dict)
>>> new_dic_defaultdict[1][2] = 5
>>> new_dic_defaultdict
defaultdict(<type 'dict'>, {1: {2: 5}})

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