按最大值排序字典：类方法 [英] Sorting dictionaries by largest values: class method

问题描述

我想排序一个特定的字典，并返回一个top_n数的列表。该字典是txt文档中的单词集合，key是txt文件中的单个单词，value是文档中出现的次数。

I want to sort a specific dictionary and return a list of the top_n number of occurences. The dictionary is a collection of words from a txt document, with the 'key' being a single word from the txt file and the 'value' being its number of occurrences in the document.

我有 init 方法如下：

def __init__(self:'Collection_of_words', file_name: str) -> None:
    '''  this initializer will read in the words from the file,
    and store them in self.counts'''
    l_words = open(file_name).read().split()
    s_words = set(l_words)
    self.counts = dict([ [word, l_words.count(word)] 
                        for word 
                        in s_words])

现在，我的一个实例方法之一将返回一个列表，显示顶点n出现次数的字符串int参数。我给它一个镜头：

Now, one of my instance methods will return a list of strings of the 'top n' number of occurrences givin some int argument. I gave it a shot:

def top_n_words(self, i):
    '''takes one additional parameter, an int,
    <i> which is the top number of occurences. Returns a list of the top <i> words.'''


    return [ pair[0] 
             for pair 
             in sorted(associations, key=lambda pair: pair[1], reverse=True)[:5]]

然而，每当我运行这个代码，我得到错误，无法弄清楚为什么。我不知道如何排序字典对象（例如，self.counts）

However, whenever i run this code i get errors and cannot figure out why. I'm not sure how to sort dictionary objects(eg. self.counts)

推荐答案

sorted(self.counts, key=lambda pair: pair[1], reverse=True)

迭代 self.counts 给出键，而不是键值对。这意味着 pair [1] 将无法正常工作。您需要 key = self.counts.get 。

Iterating over self.counts gives the keys, not key-value pairs. That means pair[1] won't work. You want key=self.counts.get.

如果您的列表需要包括计数以及密钥，您需要按值排序键值对：

If your list needs to include counts as well as keys, you'll need to instead sort the key-value pairs by values:

sorted(self.counts.items(), key=operator.itemgetter(1), reverse=True)

另外，请注意 collections.Counter 已经您需要什么，并使用线性时间而不是二次方的计数算法。

Also, note that collections.Counter already does what you need, and with a counting algorithm in linear time instead of quadratic.

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