如何将Counter对象转换为dict? [英] How to convert Counter object to dict?

问题描述

数据框:

pair = collections.defaultdict(collections.Counter)

例如

pair = {'doc1':  {'word1':4, 'word2':3}, 
        'doc2':  {'word1':2, 'word3':4},
        'doc3':  {'word2':2, 'word4':1},
         ...}

我想保留数据框，但更改此部分的类型{'word1':4, 'word2':3} {'word1':2, 'word3':4}``...现在是Counter，我需要一个dict.

I want to keep the data frame but alter the type of this part {'word1':4, 'word2':3} {'word1':2, 'word3':4}``... It is now a Counter and I need a dict.

我尝试这样做是为了从pair获取数据，但是我不知道如何为每个文档创建dict:

I tried this to get the data from pair, but I do not know how to create a dict for each doc:

new_pair = collections.defaultdict(collections.Counter)
for doc, tab in testing.form.items():
    for word, freq in tab.items():
        new_pair[doc][word] = freq 

我不想更改输出.我只需要每个文档中的，数据类型为dict，而不是Counter.

I do not want to change the output. I just need that in each doc, the data type is dict, not Counter.

推荐答案

Counter已经是dict-或其子类.但是，如果由于某种原因确实确实需要dict，那么它是单线的:

A Counter is already a dict - or, a subclass of it. But, if you really need exactly a dict for some reason, then its a one-liner:

>>> c = Counter(word1=4, word2=3)
>>> c
Counter({'word1': 4, 'word2': 3})
>>> dict(c)
{'word1': 4, 'word2': 3}

任何映射(行为类似于字典的任何内容)都可以传递到dict中，您将获得具有相同内容的dict.无需迭代就可以自己构造它.

Any Mapping (anything that behaves like a dictionary) can be passed into dict, and you will get a dict with the same contents. There is no need to iterate over it to construct it yourself.

这将为您提供一个循环，主体中只有一行，而不是嵌套循环.但是任何形式的代码:

This gives you one loop, with one line in the body instead of a nested loop. But any code of the form:

 thing = a new empty collection
 for elem in old_thing:
    Add something to do with elem to thing

通常可以使用生成器表达式或列表，集合或字典理解在一行中完成.我们正在构建一个dict，因此，一个 dict理解(示例似乎是您最感兴趣的部分).我将把它作为读者的练习. ;-)

Can usually be done in one line using a generator expression or a list, set or dict comprehension. We're building a dict, so a dict comprehension (the Examples section is what you're most interested in) seems likely. I'll leave coming up with it as an exercise for the reader. ;-)

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