Quanteda的朴素贝叶斯与插入符号:结果截然不同

JBG*_*ber 4 r supervised-learning text-classification r-caret quanteda

我正在尝试使用这些包quanteda,caret并根据经过训练的样本对文本进行分类.作为试运行,我想比较的内置的朴素贝叶斯分类器quanteda与的那些caret.但是,我似乎caret无法正常工作.

这是一些复制代码.首先是quanteda侧面:

library(quanteda)
library(quanteda.corpora)
library(caret)
corp <- data_corpus_movies
set.seed(300)
id_train <- sample(docnames(corp), size = 1500, replace = FALSE)

# get training set
training_dfm <- corpus_subset(corp, docnames(corp) %in% id_train) %>%
  dfm(stem = TRUE)

# get test set (documents not in id_train, make features equal)
test_dfm <- corpus_subset(corp, !docnames(corp) %in% id_train) %>%
  dfm(stem = TRUE) %>% 
  dfm_select(pattern = training_dfm, 
             selection = "keep")

# train model on sentiment
nb_quanteda <- textmodel_nb(training_dfm, docvars(training_dfm, "Sentiment"))

# predict and evaluate
actual_class <- docvars(test_dfm, "Sentiment")
predicted_class <- predict(nb_quanteda, newdata = test_dfm)
class_table_quanteda <- table(actual_class, predicted_class)
class_table_quanteda
#>             predicted_class
#> actual_class neg pos
#>          neg 202  47
#>          pos  49 202
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不错.没有调整,准确率为80.8%.现在一样(据我所知)caret

training_m <- convert(training_dfm, to = "matrix")
test_m <- convert(test_dfm, to = "matrix")
nb_caret <- train(x = training_m,
                  y = as.factor(docvars(training_dfm, "Sentiment")),
                  method = "naive_bayes",
                  trControl = trainControl(method = "none"),
                  tuneGrid = data.frame(laplace = 1,
                                        usekernel = FALSE,
                                        adjust = FALSE),
                  verbose = TRUE)

predicted_class_caret <- predict(nb_caret, newdata = test_m)
class_table_caret <- table(actual_class, predicted_class_caret)
class_table_caret
#>             predicted_class_caret
#> actual_class neg pos
#>          neg 246   3
#>          pos 249   2
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这里的准确度不仅低得惊人(49.6% - 几乎是机会),因此几乎没有预测过pos级!所以我很确定我在这里缺少一些关键的东西,因为我认为实现应该非常相似,但不确定是什么.

我已经查看了quanteda函数的源代码(希望它可能是构建在caret底层或底层的包),并看到有一些加权和平滑正在进行.如果我在训练之前将其应用于我的dfm(laplace = 0稍后设置),准确性会更好一些.然而也只有53%.

Ken*_*oit 5

答案是插入符号(它使用naive_bayesnaivebayes包)假设的高斯分布,而quanteda::textmodel_nb()是基于更文本适当多项式分布(具有伯努利分布的选项以及).

还引用了textmodel_nb()复制IIR书籍(Manning,Raghavan和Schütze2008)以及Jurafsky和Martin(2018)的另一个例子的文档.看到:

另一个软件包e1071产生与您发现的相同的结果,因为它也基于高斯分布.

library("e1071")
nb_e1071 <- naiveBayes(x = training_m,
                       y = as.factor(docvars(training_dfm, "Sentiment")))
nb_e1071_pred <- predict(nb_e1071, newdata = test_m)
table(actual_class, nb_e1071_pred)
##             nb_e1071_pred
## actual_class neg pos
##          neg 246   3
##          pos 249   2
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然而,插入符号e1071都在密集矩阵上运行,这是他们与稀疏dfm操作的quanteda方法相比如此令人头脑麻木的原因之一.因此,从适当性,效率和(根据您的结果)分类器的性能的角度来看,应该非常清楚哪一个是首选的!

library("rbenchmark")
benchmark(
    quanteda = { 
        nb_quanteda <- textmodel_nb(training_dfm, docvars(training_dfm, "Sentiment"))
        predicted_class <- predict(nb_quanteda, newdata = test_dfm)
    },
    caret = {
        nb_caret <- train(x = training_m,
                          y = as.factor(docvars(training_dfm, "Sentiment")),
                          method = "naive_bayes",
                          trControl = trainControl(method = "none"),
                          tuneGrid = data.frame(laplace = 1,
                                                usekernel = FALSE,
                                                adjust = FALSE),
                          verbose = FALSE)
        predicted_class_caret <- predict(nb_caret, newdata = test_m)
    },
    e1071 = {
        nb_e1071 <- naiveBayes(x = training_m,
                       y = as.factor(docvars(training_dfm, "Sentiment")))
        nb_e1071_pred <- predict(nb_e1071, newdata = test_m)
    },
    replications = 1
)
##       test replications elapsed relative user.self sys.self user.child sys.child
## 2    caret            1  29.042  123.583    25.896    3.095          0         0
## 3    e1071            1 217.177  924.157   215.587    1.169          0         0
## 1 quanteda            1   0.235    1.000     0.213    0.023          0         0
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