python - 使用 python 在句子列表中形成单词的二元组并计算二元组

Question

我需要:1. 形成二元对并将它们存储在列表中2. 找到 id 的总和，其中存在频率最高的前 3 个二元组

我有一个句子列表:

[['22574999', 'your message communication sent']
, ['22582857', 'your message be delivered']
, ['22585166', 'message has be delivered']
, ['22585424', 'message originated communication sent']]

这是我所做的:

for row in messages: 
    sstrm = list(row)
    bigrams=[b for l in sstrm for b in zip(l.split(" ")[:1], l.split(" ")[1:])]
    print(sstrm[0],bigrams)

产生:

22574999 [('your', 'message')]
22582857 [('[your', 'message')]
22585166 [('message', 'has')]
22585424 [('message', 'originated')]

我想要的是:

22574999 [('your', 'message'),('communication','sent')]
22582857 [('[your', 'message'),('be','delivered')]
22585166 [('message', 'has'),('be','delivered')]
22585424 [('message', 'originated'),('communication','sent')]

我想要得到以下结果结果:

频率最高的前 3 个二元组:

('your', 'message') :2 
('communication','sent'):2    
('be','delivered'):2

其中出现频率最高的前 3 个二元组的 id 总和:

('your', 'message'):2           Is included (22574999,22582857)     
('communication','sent'):2      Is included(22574999,22585424)
('be','delivered'):2            Is included (22582857,22585166)

感谢您的帮助!

Answer 1

我想指出的第一件事是，二元组是两个相邻元素的序列。

例如，“狐狸跳过了懒狗”的二元词是:

[("the", "fox"),("fox", "jumped"),("jumped", "over"),("over", "the"),("the ", "懒"),("懒", "狗")]

这个问题可以使用inverted index来建模，其中二元组是帖子，ID 集是帖子列表。

def bigrams(line):
    tokens = line.split(" ")
    return [(tokens[i], tokens[i+1]) for i in range(0, len(tokens)-1)]


if __name__ == "__main__":
    messages = [['22574999', 'your message communication sent'], ['22582857', 'your message be delivered'], ['22585166', 'message has be delivered'], ['22585424', 'message originated communication sent']]
    bigrams_set = set()

    for row in messages:
        l_bigrams = bigrams(row[1])
        for bigram in l_bigrams:
            bigrams_set.add(bigram)

    inverted_idx = dict((b,[]) for b in bigrams_set)

    for row in messages:
        l_bigrams = bigrams(row[1])
        for bigram in l_bigrams:
            inverted_idx[bigram].append(row[0])

    freq_bigrams = dict((b,len(ids)) for b,ids in inverted_idx.items())
    import operator
    top3_bigrams = sorted(freq_bigrams.iteritems(), key=operator.itemgetter(1), reverse=True)[:3]

输出

[(('communication', 'sent'), 2), (('your', 'message'), 2), (('be', 'delivered'), 2)]

尽管此代码可以进行大量优化，但它为您提供了想法。