Question

我有一个名为wordCounts的词典，它将单词映射到它出现的次数，如何在dict中获得最高n个单词，同时允许超过n是一个平局？

Answer 1

正如之前的回答所说，您可以转换为Counter以使此数据集更易于处理。

>>> from collections import Counter
>>> d = {"d":1,"c":2,"a":3,'b':3,'e':0,'f':1}
>>> c = Counter(d)
>>> c
Counter({'b': 3, 'a': 3, 'c': 2, 'f': 1, 'd': 1, 'e': 0})

Counter有most_common(n)方法，该方法将采用n最常见的元素。请注意，它将排除关系。因此：

>>> c.most_common(4)
[('b', 3), ('a', 3), ('c', 2), ('f', 1)]

要包含所有等于第n个元素的值，您可以执行以下操作，而无需转换为Counter。这非常麻烦，但应该可以解决这个问题。

from collections import Counter

def most_common_inclusive(freq_dict, n):
    # find the nth most common value
    nth_most_common = sorted(c.values(), reverse=True)[n-1]
    return { k: v for k, v in c.items() if v >= nth_most_common }

您可以按如下方式使用：

>>> d = {'b': 3, 'a': 3, 'c': 2, 'f': 1, 'd': 1, 'e': 0}
>>> most_common_inclusive(d, 4)
{'d': 1, 'b': 3, 'c': 2, 'f': 1, 'a': 3}

Answer 2

一种解决方案可能是：

from collections import Counter, defaultdict


list_of_words = ['dog', 'cat', 'moo', 'dog', 'pun', 'pun']
def get_n_most_common(n, list_of_words):
    ct = Counter(list_of_words)
    d = defaultdict(list)
    for word, quantity in ct.items():
        d[quantity].append(word)
    most_common = sorted(d.keys(), reverse= True)
    return [(word, val) for val in most_common[:n] for word in d[val]]

测试：

 >> get_n_most_common(2, list_of_words)
 => [('pun', 2), ('dog', 2), ('moo', 1), ('cat', 1)]
 >> get_n_most_common(1, list_of_words)
 => [('pun', 2), ('dog', 2)]

Answer 3

MooingRawr走在正确的轨道上，但现在我们需要获得最高的n结果

l = []
for i, (word, count) in enumerate(sorted(d.items(), reverse=True, key=lambda x: x[1])):  
    if i >= n and count<l[-1][1]:
        break
    l.append((word, count))

Python在字典中获取最高值

3 个答案: