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Python NLTK:使用BigramConsolutionFinder从数据帧的文本字段显示常用短语(ngrams)的频率

word frequency nltk pandas python

PineNuts0 · 技术社区 · 7 年前

我有以下标记化数据帧示例:

No  category    problem_definition_stopwords
175 2521       ['coffee', 'maker', 'brewing', 'properly', '2', '420', '420', '420']
211 1438       ['galley', 'work', 'table', 'stuck']
912 2698       ['cloth', 'stuck']
572 2521       ['stuck', 'coffee']

我成功地运行了下面的代码以获取ngram短语。

finder = BigramCollocationFinder.from_documents(df['problem_definition_stopwords'])

# only bigrams that appear 1+ times
finder.apply_freq_filter(1) 

# return the 10 n-grams with the highest PMI
finder.nbest(bigram_measures.pmi, 10)

结果如下所示,pmi排名前10位:

[('brewing', 'properly'), ('galley', 'work'), ('maker', 'brewing'), ('properly', '2'), ('work', 'table'), ('coffee', 'maker'), ('2', '420'), ('cloth', 'stuck'), ('table', 'stuck'), ('420', '420')]

我希望上面的结果出现在一个包含频率计数的数据框中,显示这些二元图发生的频率。

所需输出示例:

ngram                    frequency
'brewing', 'properly'    1
'galley', 'work'         1
'maker', 'brewing'       1
'properly', '2'          1
...                      ...

如何在Python中执行上述操作?

1 回复 | 直到 7 年前

blacksite 7 年前

这应该够了。。。

首先,设置数据集(或类似的数据集):

import pandas as pd
from nltk.collocations import *
import nltk.collocations
from nltk import ngrams
from collections import Counter

s = pd.Series(
    [
        ['coffee', 'maker', 'brewing', 'properly', '2', '420', '420', '420'],
        ['galley', 'work', 'table', 'stuck'],
        ['cloth', 'stuck'],
        ['stuck', 'coffee']
    ]
)

finder = BigramCollocationFinder.from_documents(s.values)
bigram_measures = nltk.collocations.BigramAssocMeasures()

# only bigrams that appear 1+ times
finder.apply_freq_filter(1) 

# return the 10 n-grams with the highest PMI
result = finder.nbest(bigram_measures.pmi, 10)

使用 nltk.ngrams 要重新创建ngrams列表:

ngram_list = [pair for row in s for pair in ngrams(row, 2)]

使用 collections.Counter 要计算每个ngram在整个语料库中出现的次数:

counts = Counter(ngram_list).most_common()

构建一个看起来像你想要的数据框架:

pd.DataFrame.from_records(counts, columns=['gram', 'count'])
                   gram  count
0            (420, 420)      2
1       (coffee, maker)      1
2      (maker, brewing)      1
3   (brewing, properly)      1
4         (properly, 2)      1
5              (2, 420)      1
6        (galley, work)      1
7         (work, table)      1
8        (table, stuck)      1
9        (cloth, stuck)      1
10      (stuck, coffee)      1

然后,您可以进行过滤,只查看由您的 finder.nbest 电话:

df = pd.DataFrame.from_records(counts, columns=['gram', 'count'])
df[df['gram'].isin(result)]