我想看看 SOM 算法是否可以用于分类预测。 我曾经在下面编写代码,但我发现分类结果远非正确。例如,在测试数据集中,我得到的不仅仅是训练目标变量中的 3 个值。如何创建与训练目标变量对齐的预测模型?
library(kohonen)
library(HDclassif)
data(wine)
set.seed(7)
training <- sample(nrow(wine), 120)
Xtraining <- scale(wine[training, ])
Xtest <- scale(wine[-training, ],
center = attr(Xtraining, "scaled:center"),
scale = attr(Xtraining, "scaled:scale"))
som.wine <- som(Xtraining, grid = somgrid(5, 5, "hexagonal"))
som.prediction$pred <- predict(som.wine, newdata = Xtest,
trainX = Xtraining,
trainY = factor(Xtraining$class))
结果:
$unit.classif
[1] 7 7 1 7 1 11 6 2 2 7 7 12 11 11 12 2 7 7 7 1 2 7 2 16 20 24 25 16 13 17 23 22
[33] 24 18 8 22 17 16 22 18 22 22 18 23 22 18 18 13 10 14 15 4 4 14 14 15 15 4
最佳答案
这可能有帮助:
- SOM 是一种无监督分类算法,因此您不应期望它在包含分类器标签的数据集上进行训练(如果您这样做,它将需要此信息才能工作,并且对未标记的数据集毫无用处)
- 想法是它将输入数字向量“转换”为网络单元号(尝试使用每 3 个网格 1 个再次运行您的代码,您将获得预期的输出)
- 然后您需要将这些网络单位编号转换回您要查找的类别(这是您的代码中缺少的关键部分)
下面的可重现示例将输出经典分类错误。它包括一个用于原始帖子中缺少的“转换回来”部分的实现选项。
不过,对于这个特定的数据集,模型很快就会过度拟合:3 个单元可提供最佳结果。
#Set and scale a training set (-1 to drop the classes)
data(wine)
set.seed(7)
training <- sample(nrow(wine), 120)
Xtraining <- scale(wine[training, -1])
#Scale a test set (-1 to drop the classes)
Xtest <- scale(wine[-training, -1],
center = attr(Xtraining, "scaled:center"),
scale = attr(Xtraining, "scaled:scale"))
#Set 2D grid resolution
#WARNING: it overfits pretty quickly
#Errors are 36% for 1 unit, 63% for 2, 93% for 3, 89% for 4
som_grid <- somgrid(xdim = 1, ydim=3, topo="hexagonal")
#Create a trained model
som_model <- som(Xtraining, som_grid)
#Make a prediction on test data
som.prediction <- predict(som_model, newdata = Xtest)
#Put together original classes and SOM classifications
error.df <- data.frame(real = wine[-training, 1],
predicted = som.prediction$unit.classif)
#Return the category number that has the strongest association with the unit
#number (0 stands for ambiguous)
switch <- sapply(unique(som_model$unit.classif), function(x, df){
cat <- as.numeric(names(which.max(table(
error.df[error.df$predicted==x,1]))))
if(length(cat)<1){
cat <- 0
}
return(c(x, cat))
}, df = data.frame(real = wine[training, 1], predicted = som_model$unit.classif))
#Translate units numbers into classes
error.df$corrected <- apply(error.df, MARGIN = 1, function(x, switch){
cat <- switch[2, which(switch[1,] == x["predicted"])]
if(length(cat)<1){
cat <- 0
}
return(cat)
}, switch = switch)
#Compute a classification error
sum(error.df$corrected == error.df$real)/length(error.df$real)
关于r - 如何使用 SOM 算法进行分类预测,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/45098820/