计算三个单词的频率答案

【问题标题】：Counting the Frequency of three words计算三个单词的频率
【发布时间】：2019-01-27 16:49:41
【问题描述】：

我有下面的代码来查找两个单词短语的频率。我需要对三个单词短语做同样的事情。

但是，下面的代码似乎不适用于 3 个单词的短语。

from collections import Counter
import re

sentence = "I love TV show makes me happy, I love also comedy show makes me feel like flying"
words = re.findall(r'\w+', sentence)
two_words = [' '.join(ws) for ws in zip(words, words[1:])]
wordscount = {w:f for w, f in Counter(two_words).most_common() if f > 1}
wordscount
{'show makes': 2, 'makes me': 2, 'I love': 2}

【问题讨论】：

标签： python string python-3.x counter

【解决方案1】：

您可以在可迭代的 3 字分组上使用 collections.Counter。后者是通过生成器理解和列表切片构造的。

from collections import Counter

three_words = (words[i:i+3] for i in range(len(words)-2))
counts = Counter(map(tuple, three_words))
wordscount = {' '.join(word): freq for word, freq in counts.items() if freq > 1}

print(wordscount)

{'show makes me': 2}

请注意，我们直到最后才使用str.join，以避免不必要的重复字符串操作。此外，Counter 需要 tuple 转换，因为 dict 键必须是可散列的。

【讨论】：

【解决方案2】：

我建议将功能分解为a seperate function：

def nwise(iterable, n):
    """
    Iterate over n-grams of an iterable.
    Has a bit of an overhead compared to pairwise (although only during
    initialization), so the two functions are implemented independently.
    """
    iterables = [iter(iterable) for _ in range(n)]
    for index, it in enumerate(iterables):
        for _ in range(index):
            next(it)
    yield from zip(*iterables)

那你就可以了

two_words = [" ".join(bigram) for bigram in nwise(words, 2))]

和

three_words = [" ".join(trigram) for trigram in nwise(words, 3))]

等等。然后你可以在上面使用collections.Counter：

three_word_counts = Counter(" ".join(trigram) for trigram in nwise(words, 3))

【讨论】：

这很好。我只是觉得昂贵str.join 应该延迟到最小计数步骤的最终过滤。
@jpp 我怀疑这将是一个问题，但您也可以直接将nwise(words, 3) 输入计数器并按需提供str.join。

【解决方案3】：

试试zip(words, words[1:], words[2:])

例如：

from collections import Counter
import re

sentence = "I love TV show makes me happy, I love also comedy show makes me feel like flying"
words = re.findall(r'\w+', sentence)

three_words = [' '.join(ws) for ws in zip(words, words[1:], words[2:])]
wordscount = {w:f for w, f in Counter(three_words).most_common() if f > 1}
print( wordscount )

输出：

{'show makes me': 2}

【讨论】：

【解决方案4】：

怎么样：

from collections import Counter

sentence = "I love TV show makes me happy, I love also comedy show makes me feel like flying"
words = sentence.split()
r = Counter([' '.join(words[i:i+3]) for i in range(len(words)-3)])

>>> r.most_common()[0] #get the most common 3-words
('show makes me', 2)

【讨论】：