dtm稀疏度取决于tf / tfidf,同一语料库

时间:2016-11-29 12:35:11

标签: r text-processing tm tf-idf

任何人都可以解释一下吗?

我的理解:

tf >= 0 (absolute frequency value)

tfidf >= 0 (for negative idf, tf=0)



sparse entry = 0

nonsparse entry > 0

因此,使用下面的代码创建的两个DTM中的精确稀疏/非稀疏比例应该相同。

library(tm)
data(crude)

dtm <- DocumentTermMatrix(crude, control=list(weighting=weightTf))
dtm2 <- DocumentTermMatrix(crude, control=list(weighting=weightTfIdf))
dtm
dtm2

可是:

> dtm
<<DocumentTermMatrix (documents: 20, terms: 1266)>>
**Non-/sparse entries: 2255/23065**
Sparsity           : 91%
Maximal term length: 17
Weighting          : term frequency (tf)
> dtm2
<<DocumentTermMatrix (documents: 20, terms: 1266)>>
**Non-/sparse entries: 2215/23105**
Sparsity           : 91%
Maximal term length: 17
Weighting          : term frequency - inverse document frequency (normalized) (tf-idf)

1 个答案:

答案 0 :(得分:3)

稀疏性可能不同。如果TF为零或IDF为零,则TF-IDF值为零,如果每个文档中出现一个术语,则IDF为零。请考虑以下示例:

txts <- c("super World", "Hello World", "Hello super top world")
library(tm)
tf <- TermDocumentMatrix(Corpus(VectorSource(txts)), control=list(weighting=weightTf))
tfidf <- TermDocumentMatrix(Corpus(VectorSource(txts)), control=list(weighting=weightTfIdf))

inspect(tf)
# <<TermDocumentMatrix (terms: 4, documents: 3)>>
# Non-/sparse entries: 8/4
# Sparsity           : 33%
# Maximal term length: 5
# Weighting          : term frequency (tf)
# 
#        Docs
# Terms   1 2 3
#   hello 0 1 1
#   super 1 0 1
#   top   0 0 1
#   world 1 1 1

inspect(tfidf)
# <<TermDocumentMatrix (terms: 4, documents: 3)>>
# Non-/sparse entries: 5/7
# Sparsity           : 58%
# Maximal term length: 5
# Weighting          : term frequency - inverse document frequency (normalized) (tf-idf)
# 
#        Docs
# Terms           1         2         3
#   hello 0.0000000 0.2924813 0.1462406
#   super 0.2924813 0.0000000 0.1462406
#   top   0.0000000 0.0000000 0.3962406
#   world 0.0000000 0.0000000 0.0000000

术语 super 在文档1中出现1次,有2个术语,它出现在3个文档中的2个中:

1/2 * log2(3/2)
# [1] 0.2924813

术语 world 在文档3中出现1次,有4个术语,它出现在所有3个文档中:

1/4 * log2(3/3) # 1/4 * 0
# [1] 0