压缩成字典时需要独立的对象 [英] Independent objects needed when zipping into dict

问题描述

我有一个物品清单:

my_list = ['first', 'second', 'third']

我需要将此项目转换为字典(以便我可以通过其名称访问每个元素)，并将多个计数器与每个元素相关联:counter1，counter2，counter3.

I need to convert this items into a dictionary (so I can access each element by it's name), and associate multiple counters to each element: counter1, counter2, counter3.

所以我执行以下操作:

counter_dict = {'counter1': 0, 'counter2': 0, 'counter3': 0}

my_dict = dict(zip(mylist, [counter_dict]*len(mylist)))

通过这种方式，我可以获得嵌套字典

This way I obtain a nested dictionary

{
  'first': {
    'counter1': 0,
    'counter2': 0,
    'counter3': 0
  },
  'second': {
    'counter1': 0,
    'counter2': 0,
    'counter3': 0
  },
  'third': {
    'counter1': 0,
    'counter2': 0,
    'counter3': 0
  }
}

这里的问题是每个计数器字典不是独立的，所以当我更新一个计数器时，所有计数器都将更新.

The problem here is that each counter dictionary is not independent, so when I update one counter, all of them are updated.

以下行不仅更新 counter2 的时间为 second ，而且还更新 counter2 的 first 和第三

The following line not only updates the counter2 of second but also updates counter2 of first and third

my_dict['second']['counter2'] += 1


{
  'first': {
    'counter1': 0,
    'counter2': 1,
    'counter3': 0
  },
  'second': {
    'counter1': 0,
    'counter2': 1,
    'counter3': 0
  },
  'third': {
    'counter1': 0,
    'counter2': 1,
    'counter3': 0
  }
}

我知道通过执行 [counter_dict] * len(mylist)，我将所有词典都指向一个词典.

I am aware that by doing [counter_dict]*len(mylist) I am pointing all dictionaries to a single one.

问题是，如何才能实现创建独立词典所需要的功能?最后，我需要:

Question is, how can I achieve what I need creating independent dictionaries? At the end I need to:

• 将列表中的每个元素作为键访问
• 对于每个元素，我可以独立更新多个计数器

非常感谢

推荐答案

尝试以下方法:

my_dict = {k: counter_dict.copy() for k in my_list}

代替压缩值-足以遍历键并使用dict压缩将计数器复制到值.

Instead of zipping values - it's enough to iterate over keys and copy your counters to values using dict compression.

但是请注意，这仅适用于一级counter_dict.如果需要嵌套字典，请使用 copy 包中的 deepcopy :

But note, that this will work only with one level counter_dict. If needed nested dictionary - use deepcopy from copy package:

from copy import deepcopy
my_dict = {k: deepcopy(counter_dict) for k in my_list}

另一种方法-使用来自 collections 的 defaultdict ，因此您甚至不需要创建带有预定义计数器的字典:

Another aproach - using defaultdict from collections, so you dont even need to create dictionary with predefined counters:

from collections import defaultdict
my_dict = {k: defaultdict(int) for k in my_list}

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