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我试图在 text2vec 中使用 hash_vectorizer 函数创建 ngram,当我注意到它不会改变我的 dtm 的尺寸而改变值时。

h_vectorizer = hash_vectorizer(hash_size = 2 ^ 14, ngram = c(2L, 10L))
dtm_train = create_dtm(it_train, h_vectorizer)
dim(dtm_train)

在上面的代码中,无论是 2-10 还是 9-10,尺寸都不会改变。

vocab = create_vocabulary(it_train, ngram = c(1L, 4L))
ngram_vectorizer = vocab_vectorizer(vocab)
dtm_train = create_dtm(it_train, ngram_vectorizer)

在上面的代码中,尺寸发生了变化,但我也想使用 hash_vectorizo​​r 因为它节省了空间。我该如何使用它?

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使用散列时,您可以提前设置输出矩阵的大小。您通过设置hash_size = 2 ^ 14. 这与模型中指定的 ngram 窗口保持相同。但是,输出矩阵中的计数会发生变化。

(作为对以下评论的回应:)在下面,您可以找到一个带有两个非常简单字符串的最小示例,以演示 a 中使用的两个不同 ngram 窗口的不同输出hash_vectorizer。对于二元组,我添加了 a 的输出矩阵以vocab_vectorizer进行比较。你意识到你必须设置一个足够大的散列大小来解释所有项。如果它太小,单个术语的哈希值可能会发生冲突。

您关于始终必须比较一种方法的输出的评论vocab_vectorizer和一种hash_vectorizer方法会导致错误的方向,因为这样您就会失去哈希方法可能产生的效率/内存优势,从而避免生成词汇表。根据您的数据和所需的输出散列可能会将准确性(以及 dtm 中术语的可解释性)与效率相提并论。因此,散列是否合理取决于您的用例(尤其适用于大型集合的文档级别的分类任务)。

我希望这能让您大致了解散列以及您可以或不能从中得到什么。你也可以在quora维基百科(或者这里)查看一些关于散列的帖子。或者也可以参考text2vec.org上列出的详细原始资源。

library(text2vec)
txt <- c("a string string", "and another string")

it = itoken(txt, progressbar = F)


#the following four example demonstrate the effect of the size of the hash
#and the use of signed hashes (i.e. the use of a secondary hash function to reduce risk of collisions)
vectorizer_small = hash_vectorizer(2 ^ 2, c(1L, 1L)) #unigrams only
hash_dtm_small = create_dtm(it, vectorizer_small)
as.matrix(hash_dtm_small)
#    [,1] [,2] [,3] [,4]
# 1    2    0    0    1
# 2    1    2    0    0  #collision of the hash values of and / another

vectorizer_small_signed = hash_vectorizer(2 ^ 2, c(1L, 1L), signed_hash = TRUE) #unigrams only
hash_dtm_small = create_dtm(it, vectorizer_small_signed)
as.matrix(hash_dtm_small)
#     [,1] [,2] [,3] [,4]
# 1    2    0    0    1
# 2    1    0    0    0 #no collision but some terms (and / another) not represented as hash value

vectorizer_medium = hash_vectorizer(2 ^ 3, c(1L, 1L)) #unigrams only
hash_dtm_medium = create_dtm(it, vectorizer_medium)
as.matrix(hash_dtm_medium)
#    [,1] [,2] [,3] [,4] [,5] [,6] [,7] [,8]
# 1    0    0    0    1    2    0    0    0
# 2    0    1    0    0    1    1    0    0 #no collision, all terms represented by hash values


vectorizer_medium = hash_vectorizer(2 ^ 3, c(1L, 1L), signed_hash = TRUE) #unigrams only
hash_dtm_medium = create_dtm(it, vectorizer_medium)
as.matrix(hash_dtm_medium)
#     [,1] [,2] [,3] [,4] [,5] [,6] [,7] [,8]
# 1    0    0    0    1    2    0    0    0
# 2    0   -1    0    0    1    1    0    0 #no collision, all terms represented as hash values
                                            #in addition second hash function generated a negative hash value


#the following two examples deomstrate the difference between 
#two hash vectorizers one with unigrams, one allowing for bigrams
#and one vocab vectorizer with bigrams
vectorizer = hash_vectorizer(2 ^ 4, c(1L, 1L)) #unigrams only
hash_dtm = create_dtm(it, vectorizer)
as.matrix(hash_dtm)
#    [,1] [,2] [,3] [,4] [,5] [,6] [,7] [,8] [,9] [,10] [,11] [,12] [,13] [,14] [,15] [,16]
# 1    0    0    0    0    0    0    0    0    0     0     0     1     2     0     0     0
# 2    0    0    0    0    0    0    0    0    0     1     0     0     1     1     0     0

vectorizer2 = hash_vectorizer(2 ^ 4, c(1L, 2L)) #unigrams + bigrams
hash_dtm2 = create_dtm(it, vectorizer2)
as.matrix(hash_dtm2)
#     [,1] [,2] [,3] [,4] [,5] [,6] [,7] [,8] [,9] [,10] [,11] [,12] [,13] [,14] [,15] [,16]
# 1    1    0    0    1    0    0    0    0    0     0     0     1     2     0     0     0
# 2    0    0    0    0    0    1    1    0    0     1     0     0     1     1     0     0

v <- create_vocabulary(it, c(1L, 2L))
vectorizer_v = vocab_vectorizer(v) #unigrams + bigrams
v_dtm = create_dtm(it, vectorizer_v)
as.matrix(v_dtm)
#   a_string and_another a another and string_string another_string string
# 1        1           0 1       0   0             1              0      2
# 2        0           1 0       1   1             0              1      1


sum(Matrix::colSums(as.matrix(hash_dtm)) > 0)
#[1] 4   - these are the four unigrams a, string, and, another
sum(Matrix::colSums(hash_dtm2) > 0)
#[1] 8   - these are the four unigrams as above plus the 4 bigrams string_string, a_string, and_another, another_string 
sum(Matrix::colSums(v_dtm) > 0)
#[1] 8 - same as hash_dtm2
于 2017-12-14T15:40:47.107 回答