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我正在使用R 中的 Naive bayes帖子中提到的klaR包的方法：predict

nb_testpred <- predict(mynb, newdata=testdata).

nb_testpred是我的朴素贝叶斯模型，开发于traindata; testdata是剩余的数据。

但是，我收到此错误：

Error in FUN(1:10[[4L]], ...) : subscript out of bounds

我不确定发生了什么 -testdata行数少于traindata，列数相同。

作为参考，我的代码如下所示：

ind       <- sample(2, nrow(mydata), replace=TRUE, prob=c(0.9,0.1))
traindata <- mydata[ind==1,]
testdata  <- mydata[ind==2,]
myformula <- as.factor(dep) ~ X1 + as.factor(X2) + as.factor(X3) + as.factor(X4) + X5 + as.factor(X6) + as.factor(date) + as.factor(hour)
mynb        <- NaiveBayes(myformula, data=traindata)
nb_testpred <- predict(mynb, newdata=testdata) #where I'm getting an error...

数据样本在这里（原始文件有 100,000+ 行）：

sampledata <- structure(list(dep = c(1L, 1L, 0L, 0L, 0L, 0L, 0L, 0L, 0L, 1L, 0L, 0L, 0L, 0L, 0L, 0L, 0L, 1L), X1 = structure(c(2L, 2L, 2L, 1L, 1L, 1L, 1L, 1L, 1L, 2L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 2L), .Label = c("A", "B"), class = "factor"), X2 = c(200L, 200L, 200L, 200L, 200L, 200L, 200L, 200L, 200L, 200L, 200L, 200L, 200L, 200L, 200L, 200L, 
200L, 200L), X3 = structure(c(4L, 2L, 3L, 1L, 1L, 1L, 1L, 1L, 1L, 3L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 3L), .Label = c(".", "1400000", "2400000", "900000"), class = "factor"), X4 = c(0L, 0L, 0L, 3L, 4L, 5L, 5L, 5L, 5L, 0L, 5L, 5L, 5L, 5L, 5L, 5L, 5L, 0L), X5 = c(TRUE, TRUE, FALSE, TRUE, TRUE, TRUE, TRUE, TRUE, FALSE, FALSE, TRUE, TRUE, TRUE, TRUE, TRUE, TRUE, TRUE, TRUE), X6 = c(1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 2L, 2L, 2L, 2L, 2L, 2L, 2L),     date = structure(c(1L, 1L, 2L, 2L, 2L, 2L, 2L, 1L, 2L, 2L, 2L, 2L, 1L, 2L, 2L, 2L, 2L, 1L), .Label = c("9/23/2012", 
"9/24/2012"), class = "factor"), hour = c(18L, 17L, 23L, 8L, 1L, 19L, 19L, 16L, 22L, 2L, 12L, 16L, 15L, 9L, 1L, 9L, 
13L, 19L)), .Names = c("dep", "X1", "X2", "X3", "X4", "X5", "X6", "date", "hour"), class = "data.frame", row.names = c(NA, -18L))

任何帮助将不胜感激！

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1 回答 1

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您可以执行以下操作：

traindata$dep=factor(traindata$dep)
mynb <- NaiveBayes(dep~.,traindata)

然后它可以工作，但是您应该优化您的数据以避免恒定列。

于 2013-03-29T16:59:01.250 回答