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无监督分类：为数据分配类[关闭]

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我有一组来自钻孔的数据，它包含每2米不同地质力学特性的信息 . 我正在尝试创建地质力学域，并将每个点分配给不同的域 .

我试图使用随机森林分类，并且不确定如何将proximty矩阵（或randomForest函数的任何结果）与标签相关联 .

到目前为止我的简陋代码如下：

dh <- read.csv("gt_1_classification.csv", header = T)

#replace all N/A with 0
dh[is.na(dh)] <- 0
library(randomForest)
dh_rf <- randomForest(dh, importance=TRUE, proximity=FALSE, ntree=500, type=unsupervised, forest=NULL)

我希望分类器自己决定域名 .

任何帮助都会很棒！

r machine-learning random-forest unsupervised-learning multilabel-classification

1 回答

  • 0

    Hack-R是正确的 - 首先有必要使用一些聚类（无监督学习）方法来探索数据 . 我使用R内置mtcars数据作为演示提供了一些示例代码：

    # Info on the data
?mtcars
head(mtcars)
pairs(mtcars)    # Matrix plot

# Calculate the distance between each row (car with it's variables)
# by default, Euclidean distance = sqrt(sum((x_i - y_i)^2)
?dist
d <- dist(mtcars)
d # Potentially huge matrix

# Use the distance matrix for clustering
# First we'll try hierarchical clustering
?hclust
c <- hclust(d)
c

# Plot dendrogram of clusters
plot(c)

# We might want to try 3 clusters
# need to specify either k = # of groups
groups3 <- cutree(c, k = 3) # "g3" = "groups 3"
# cutree(hcmt, h = 230) will give same result
groups3
# Or we could do several groups at once
groupsmultiple <- cutree(c, k = 2:5)
head(groupsmultiple)

# Draw boxes around clusters
rect.hclust(c, k = 2, border = "gray")
rect.hclust(c, k = 3, border = "blue")
rect.hclust(c, k = 4, border = "green4")
rect.hclust(c, k = 5, border = "darkred")

# Alternatively we can try K-means clustering
# k-means clustering
?kmeans
km <- kmeans(mtcars, 5)
km

# Graph based on k-means
install.packages("cluster")
require(cluster)
clusplot(mtcars, # data frame
     km$cluster, # cluster data
     color = TRUE, # color
     lines = 3, # Lines connecting centroids
     labels = 2) # Labels clusters and cases

    运行您自己的数据后，请考虑群集的哪个定义捕获您感兴趣的相似程度 . 然后，您可以为每个群集创建一个具有“级别”的新变量，然后为其创建一个受监督的模型 .

    这是使用相同mtcars数据的决策树示例 . 请注意，这里我使用mpg作为响应 - 您可能希望使用基于群集的新变量 .

    install.packages("rpart")
library(rpart)
?rpart
# grow tree 
tree_mtcars <- rpart(mpg ~ ., method = "anova", data = mtcars)
tree_mtcars <- rpart(mpg ~ ., data = mtcars)

tree_mtcars

summary(tree_mtcars) # detailed summary of splits

# Get R-squared
rsq.rpart(tree_mtcars)
?rsq.rpart

# plot tree 
plot(tree_mtcars, uniform = TRUE, main = "Regression Tree for mpg ")
text(tree_mtcars, use.n = TRUE, all = TRUE, cex = .8)
text(tree_mtcars, use.n = TRUE, all = TRUE, cex = .8)

    请注意，虽然信息非常丰富，但基本决策树通常不适合预测 . 如果需要预测，还应该探索其他模型 .

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