Python TF-IDF:更新TF-IDF矩阵的快速方法 [英] Python tf-idf: fast way to update the tf-idf matrix

问题描述

我有一个包含数千行文本的数据集，我的目标是计算tfidf得分，然后计算文档之间的余弦相似度，这就是我在python中使用gensim进行的操作，遵循该教程:

I have a dataset of several thousand rows of text, my target is to calculate the tfidf score and then cosine similarity between documents, this is what I did using gensim in Python followed the tutorial:

dictionary = corpora.Dictionary(dat)
corpus = [dictionary.doc2bow(text) for text in dat]

tfidf = models.TfidfModel(corpus)
corpus_tfidf = tfidf[corpus]
index = similarities.MatrixSimilarity(corpus_tfidf)

假设我们已经建立了tfidf矩阵和相似性，当我们收到一个新文档时，我想在现有数据集中查询其最相似的文档.

Let's say we have the tfidf matrix and similarity built, when we have a new document come in, I want to query for its most similar document in our existing dataset.

问题:有什么方法可以更新tf-idf矩阵，这样我就不必将新的文本文档附加到原始数据集中，而再次重新计算整个事情了?

Question: is there any way we can update the tf-idf matrix so that I don't have to append the new text doc to the original dataset and recalculate the whole thing again?

推荐答案

由于没有其他答案，我将发布解决方案.假设我们处于以下情况:

I'll post my solution since there are no other answers. Let's say we are in the following scenario:

import gensim
from gensim import models
from gensim import corpora
from gensim import similarities
from nltk.tokenize import word_tokenize
import pandas as pd

# routines:
text = "I work on natural language processing and I want to figure out how does gensim work"
text2 = "I love computer science and I code in Python"
dat = pd.Series([text,text2])
dat = dat.apply(lambda x: str(x).lower()) 
dat = dat.apply(lambda x: word_tokenize(x))


dictionary = corpora.Dictionary(dat)
corpus = [dictionary.doc2bow(doc) for doc in dat]
tfidf = models.TfidfModel(corpus)
corpus_tfidf = tfidf[corpus]


#Query:
query_text = "I love icecream and gensim"
query_text = query_text.lower()
query_text = word_tokenize(query_text)
vec_bow = dictionary.doc2bow(query_text)
vec_tfidf = tfidf[vec_bow]

如果我们看:

print(vec_bow)
[(0, 1), (7, 1), (12, 1), (15, 1)]

和:

print(tfidf[vec_bow])
[(12, 0.7071067811865475), (15, 0.7071067811865475)]

FYI ID和文档:

FYI id and doc:

print(dictionary.items())

[(0, u'and'),
 (1, u'on'),
 (8, u'processing'),
 (3, u'natural'),
 (4, u'figure'),
 (5, u'language'),
 (9, u'how'),
 (7, u'i'),
 (14, u'code'),
 (19, u'in'),
 (2, u'work'),
 (16, u'python'),
 (6, u'to'),
 (10, u'does'),
 (11, u'want'),
 (17, u'science'),
 (15, u'love'),
 (18, u'computer'),
 (12, u'gensim'),
 (13, u'out')]

看起来该查询仅选择了现有术语，并使用预先计算的权重为您提供了tfidf得分.因此，我的解决方法是每周或每天重建模型，因为这样做很快.

Looks like the query only picked up existing terms and using pre-calculated weights to give you the tfidf score. So my workaround is to rebuild the model weekly or daily since it is fast to do so.

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