Question

我正在尝试使用以下代码打印前10个常用词。但是，它不起作用。有关如何修复的想法吗？

def reducer_count_words(self, word, counts):
    # send all (num_occurrences, word) pairs to the same reducer.
    # num_occurrences is so we can easily use Python's max() function.
    yield None, (sum(counts), word)




# discard the key; it is just None
def reducer_find_max_10_words(self, _, word_count_pairs):
    # each item of word_count_pairs is (count, word),
    # so yielding one results in key=counts, value=word

        tmp = sorted(word_count_pairs)[0:10]
        yield tmp

Answer 1

使用collections.Counter及其most_common方法：

>>>from collections import Counter
>>>my_words = 'a a foo bar foo'
>>>Counter(my_words.split()).most_common()
[('foo', 2), ('a', 2), ('b', 1)]

Answer 2

使用collections.most_common()

示例：

most_common([n])
Return a list of the n most common elements and their counts from the most common to the least. If n is not specified, most_common() returns all elements in the counter. Elements with equal counts are ordered arbitrarily:

>>> from collections import Counter
>>> Counter('abracadabra').most_common(3)
[('a', 5), ('r', 2), ('b', 2)]

Answer 3

tmp = sorted(word_count_pairs, key=lambda pair: pair[0], reverse=True)[0:10]

<强>解释

key sorted()参数允许您在比较之前对每个元素运行一个函数。
lambda pair: pair[0]是一个从word_count_pairs中提取数字的函数。
reverse按降序排序，而不是按升序排序。

来源：

除了：如果你有很多不同的单词，排序整个列表以找到前十名是低效的。有更高效的算法。另一个答案中提到的most_common()方法可能会使用更有效的算法。

在python中根据它们的值对元组进行排序

3 个答案: