有关更多背景信息,请参阅question listed here。
我尝试使用text2vec
构建的文档字词矩阵,使用nb
包训练一个朴素的贝叶斯(caret
)模型。但是,我收到此警告消息:
警告讯息: 在eval(xpr,envir = envir)中: 模型适合Fold01.Rep1失败:usekernel = FALSE,fL = 0,adjust = 1 NaiveBayes.default中的错误(x,y,usekernel = FALSE,fL = param $ fL,...): 变量中至少有一个类的零差异:
请帮助我理解此消息以及我需要采取哪些步骤来避免模型拟合失败。我觉得我需要从DTM中删除更多稀疏术语,但我不确定。
构建模型的代码:
control <- trainControl(method="repeatedcv", number=10, repeats=3, savePredictions=TRUE, classProbs=TRUE)
Train_PRDHA_String.df$Result <- ifelse(Train_PRDHA_String.df$Result == 1, "X", "Y")
(warn=1)
(warnings=2)
t4 = Sys.time()
svm_nb <- train(x = as.matrix(dtm_train), y = as.factor(Train_PRDHA_String.df$Result),
method = "nb",
trControl=control,
tuneLength = 5,
metric ="Accuracy")
print(difftime(Sys.time(), t4, units = 'sec'))
构建文档术语矩阵的代码(Text2Vec):
library(text2vec)
library(data.table)
#Define preprocessing function and tokenization fucntion
preproc_func = tolower
token_func = word_tokenizer
#Union both of the Text fields - learn vocab from both fields
union_txt = c(Train_PRDHA_String.df$MAKTX_Keyword, Train_PRDHA_String.df$PH_Level_04_Description_Keyword)
#Create an iterator over tokens with the itoken() function
it_train = itoken(union_txt,
preprocessor = preproc_func,
tokenizer = token_func,
ids = Train_PRDHA_String.df$ID,
progressbar = TRUE)
#Build Vocabulary
vocab = create_vocabulary(it_train)
vocab
#Dimensional Reduction
pruned_vocab = prune_vocabulary(vocab,
term_count_min = 10,
doc_proportion_max = 0.5,
doc_proportion_min = 0.001)
vectorizer = vocab_vectorizer(pruned_vocab)
#Start building a document-term matrix
#vectorizer = vocab_vectorizer(vocab)
#learn vocabulary from Train_PRDHA_String.df$MAKTX_Keyword
it1 = itoken(Train_PRDHA_String.df$MAKTX_Keyword, preproc_func,
token_func, ids = Train_PRDHA_String.df$ID)
dtm_train_1 = create_dtm(it1, vectorizer)
#learn vocabulary from Train_PRDHA_String.df$PH_Level_04_Description_Keyword
it2 = itoken(Train_PRDHA_String.df$PH_Level_04_Description_Keyword, preproc_func,
token_func, ids = Train_PRDHA_String.df$ID)
dtm_train_2 = create_dtm(it2, vectorizer)
#Combine dtm1 & dtm2 into a single matrix
dtm_train = cbind(dtm_train_1, dtm_train_2)
#Normalise
dtm_train = normalize(dtm_train, "l1")
dim(dtm_train)
答案 0 :(得分:0)
这意味着,当重新采样这些变量时,它们只有一个唯一值。您可以使用preProc = "zv"
来消除警告。对于这些问题,这将有助于获得一个小的,可重现的例子。