基于 Python 中印象的 N-gram 分析

Question

这是我的示例数据集的样子：

我的目标是了解有多少印象与一个词、两个词、三个词、四个词、五个词和六个词相关联。我用过运行 N-gram 算法，但它只有 returns 计数。这是我当前的 n-gram 代码。

def find_ngrams(text, n):
    word_vectorizer = CountVectorizer(ngram_range=(n,n), analyzer='word')
    sparse_matrix = word_vectorizer.fit_transform(text)
    frequencies = sum(sparse_matrix).toarray()[0]
    ngram = 

pd.DataFrame(frequencies,index=word_vectorizer.get_feature_names(),columns=
['frequency'])
ngram = ngram.sort_values(by=['frequency'], ascending=[False])
return ngram

one = find_ngrams(df['query'],1)
bi = find_ngrams(df['query'],2)
tri = find_ngrams(df['query'],3)
quad = find_ngrams(df['query'],4)
pent = find_ngrams(df['query'],5)
hexx = find_ngrams(df['query'],6)

我想我需要做的是： 1. 将查询拆分为一个词到六个词。 2. 对拆分词附加印象。 3. 重新组合所有拆分的单词并总结印象。

以第二个查询"dog common diseases and how to treat them"为例。应该拆分为：

(1) 1-gram: dog, common, diseases, and, how, to, treat, them;
(2) 2-gram: dog common, common diseases, diseases and, and how, how to, to treat, treat them;
(3) 3-gram: dog common diseases, common diseases and, diseases and how, and how to, how to treat, to treat them;
(4) 4-gram: dog common diseases and, common diseases and how, diseases and how to, and how to treat, how to treat them;
(5) 5-gram: dog common diseases and how, the queries into one word, diseases and how to treat, and how to treat them;
(6) 6-gram: dog common diseases and how to, common diseases and how to treat, diseases and how to treat them;

Answer 1

方法来了！不是最有效的，但是，我们不要过早优化。这个想法是使用 apply 为所有 ngram 获取一个包含新列的新 pd.DataFrame，将其与旧数据框连接起来，并进行一些堆叠和分组。

import pandas as pd

df = pd.DataFrame({
    "squery": ["how to feed a dog", "dog habits", "to cat or not to cat", "dog owners"],
    "count": [1000, 200, 100, 150]
})

def n_grams(txt):
    grams = list()
    words = txt.split(' ')
    for i in range(len(words)):
        for k in range(1, len(words) - i + 1):
            grams.append(" ".join(words[i:i+k]))
    return pd.Series(grams)

counts = df.squery.apply(n_grams).join(df)

counts.drop("squery", axis=1).set_index("count").unstack()\
    .rename("ngram").dropna().reset_index()\
    .drop("level_0", axis=1).groupby("ngram")["count"].sum()

最后一个表达式将 return 变成 pd.Series，如下所示。

    ngram
a                       1000
a dog                   1000
cat                      200
cat or                   100
cat or not               100
cat or not to            100
cat or not to cat        100
dog                     1350
dog habits               200
dog owners               150
feed                    1000
feed a                  1000
feed a dog              1000
habits                   200
how                     1000
how to                  1000
how to feed             1000
how to feed a           1000
how to feed a dog       1000
not                      100
not to                   100
not to cat               100
or                       100
or not                   100
or not to                100
or not to cat            100
owners                   150
to                      1200
to cat                   200
to cat or                100
to cat or not            100
to cat or not to         100
to cat or not to cat     100
to feed                 1000
to feed a               1000
to feed a dog           1000

漂亮的方法

这个可能更有效一些，但它仍然从 CountVectorizer 中具体化了密集的 n-gram 向量。它将每一列上的那个数乘以印象数，然后将这些列相加以获得每 ngram 的总印象数。它给出了与上面相同的结果。需要注意的一件事是，具有重复 ngram 的查询也算作双倍。

import numpy as np
from sklearn.feature_extraction.text import CountVectorizer

cv = CountVectorizer(ngram_range=(1, 5))
ngrams = cv.fit_transform(df.squery)
mask = np.repeat(df['count'].values.reshape(-1, 1), repeats = len(cv.vocabulary_), axis = 1)
index = list(map(lambda x: x[0], sorted(cv.vocabulary_.items(), key = lambda x: x[1])))
pd.Series(np.multiply(mask, ngrams.toarray()).sum(axis = 0), name = "counts", index = index)

Answer 2

这样的事情怎么样：

def find_ngrams(input, n):
    # from http://locallyoptimal.com/blog/2013/01/20/elegant-n-gram-generation-in-python/
    return zip(*[input[i:] for i in range(n)])

def impressions_by_ngrams(data, ngram_max):
    from collections import defaultdict
    result = [defaultdict(int) for n in range(ngram_max)]
    for query, impressions in data:
        words = query.split()
        for n in range(ngram_max):
           for ngram in find_ngrams(words, n + 1):
                result[n][ngram] += impressions
    return result

示例：

>>> data = [('how to feed a dog', 10000),
...         ('see a dog run',     20000)]
>>> ngrams = impressions_by_ngrams(data, 3)
>>> ngrams[0]   # unigrams
defaultdict(<type 'int'>, {('a',): 30000, ('how',): 10000, ('run',): 20000, ('feed',): 10000, ('to',): 10000, ('see',): 20000, ('dog',): 30000})
>>> ngrams[1][('a', 'dog')]  # impressions for bigram 'a dog'
30000

基于 Python 中印象的 N-gram 分析

N-gram analysis based on impression in Python

python

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