我有以下示例标记化数据框:
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']
我成功运行了以下代码以找出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)
结果显示如下,前10个pmi:
[('brewing', 'properly'), ('galley', 'work'), ('maker', 'brewing'), ('properly', '2'), ('work', 'table'), ('coffee', 'maker'), ('2', '420'), ('cloth', 'stuck'), ('table', 'stuck'), ('420', '420')]
我希望以上结果出现在一个包含频率计数的数据帧中,该频率计数显示了这些二元数据发生的频率。
采样所需的输出:
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?
答案 0 :(得分:0)
这应该做...
首先,设置您的数据集(或类似数据集):
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
重新创建ngram列表:
ngram_list = [pair for row in s for pair in ngrams(row, 2)]
使用collections.Counter
来计算每个ngram在整个语料库中出现的次数:
counts = Counter(ngram_list).most_common()
构建一个看起来像您想要的数据框:
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,您可以过滤以仅查看由finder.nbest
调用产生的ngram:
df = pd.DataFrame.from_records(counts, columns=['gram', 'count'])
df[df['gram'].isin(result)]