在 Kohonen package 中使用 xyf() function 在 R 代碼中做一些預測

Question

我在 R 中使用 xyf() 和 predict.kohonen() 時遇到了一些問題

以下是 package kohonen 2.0.19 版文檔的示例副本。 你可以在https://www.rdocumentation.org/packages/kohonen/versions/2.0.19/topics/xyf看到它

但是我已經嘗試過並遇到了一些問題。

1.When I use xyf(),Error in factor(wine.classes[training]): object'wine.classes' not found 問題

2.然后當嘗試預測（）時，找不到object 'xyf.wines'

3.我將 xyf() 更改為 xyf.wines <- xyf(Xtraining, vintages[training], grid = somgrid(5, 5, "hexagonal")) 它起作用了，我得到了 xyf.wines

4.但是當我試圖預測時，我又失敗了。 Rstudio said Error in FUN(X[[i]], ...): Data type not allowed: should be a matrix or a factor 5.由於第 4 步失敗，我無法使用 table()

library(kohonen)
data(wines)
set.seed(7)
wine.classes<-vintages
training <- sample(nrow(wines), 120)
Xtraining <- scale(wines[training,])
Xtest <- scale(wines[-training,],center = attr(Xtraining, "scaled:center"),
           scale = attr(Xtraining, "scaled:scale"))
xyf.wines <- xyf(Xtraining,factor(wine.classes[training]),
             grid = somgrid(5, 5, "hexagonal"))
xyf.prediction <- predict(xyf.wines, newdata=Xtest)
table(wine.classes[-training], xyf.prediction$prediction)

Answer 1

library(kohonen)
data(wines)
set.seed(7)
#wine.classes<-vintages
training <- sample(nrow(wines), 120)
Xtraining <- scale(wines[training,])
Xtest <- scale(wines[-training,],center = attr(Xtraining, "scaled:center"),
               scale = attr(Xtraining, "scaled:scale"))
Xtraining_data <- list(measurements = Xtraining,
                       vintages = vintages[training])
Xtest_data <- list(measurements = Xtest,
                   vintages = vintages[-training])
xyf.wines <- xyf(Xtraining,vintages[training],
             grid = somgrid(5, 5, "hexagonal"))
xyf.prediction <- predict(xyf.wines, newdata=Xtest_data)
table(vintages[-training], xyf.prediction$predictions[[2]])