如何计算落在树的每个节点中的观测值 [英] How to count the observations falling in each node of a tree
问题描述
我目前正在处理MMST包中的葡萄酒数据。我已经将整个数据集分为训练和测试，并构建了像以下代码这样的树：
I am currently dealing with wine data in MMST package. I have split the whole dataset into training and test and build a tree like the following codes:
library("rpart")
library("gbm")
library("randomForest")
library("MMST")
data(wine)
aux <- c(1:178)
train_indis <- sample(aux, 142, replace = FALSE)
test_indis <- setdiff(aux, train_indis)
train <- wine[train_indis,]
test <- wine[test_indis,] #### divide the dataset into trainning and testing
model.control <- rpart.control(minsplit = 5, xval = 10, cp = 0)
fit_wine <- rpart(class ~ MalicAcid + Ash + AlcAsh + Mg + Phenols + Proa + Color + Hue + OD + Proline, data = train, method = "class", control = model.control)
windows()
plot(fit_wine,branch = 0.5, uniform = T, compress = T, main = "Full Tree: without pruning")
text(fit_wine, use.n = T, all = T, cex = .6)
我可以得到这样的图像：
And I could get a image like this:
每个节点下的数字是多少（例如下的0/1/48 Grignolino）是什么意思？
如果我想知道每个节点有多少训练和测试样本，我应该在代码中写些什么？
What does the number under each node (for example 0/1/48 under Grignolino) mean? If I want to know how many training and testing sample fall into each node, what should I write in the codes?
推荐答案
数字表示该节点中每个类的成员数。因此，标签 0/1/48告诉我们，类别1（Barabera，我推断）有0种情况，类别2（Barolo）只有一个例子，类别3（Grignolino）有48种。
The numbers indicate the number of members of each class in that node. So, the label "0 / 1 / 48" tells us that there are 0 cases of category 1 (Barabera, I infer), only one example of category 2 (Barolo), and 48 of category 3 (Grignolino).
您可以使用 summary（fit_wine）获取有关树和每个节点的详细信息。
有关更多详细信息，请参见？summary.rpart 。
You can get detailed information about the tree and each node using summary(fit_wine).
See ?summary.rpart for more details.
您还可以使用 predict（）（它将调用 predict.rpart（） ），以查看树如何对数据集进行分类。例如， predict（fit_wine，train，type = class）。或将其包装在表格中以便于查看 table（predict（fit_wine，train，type = class），train [， class]）
You can additionally use predict() (which will call predict.rpart()) to see how the tree categorizes a dataset. For example, predict(fit_wine, train, type="class"). Or wrap it in a table for easy viewing table(predict(fit_wine, train, type = "class"),train[,"class"])
如果您特别想知道观察结果落在哪个叶节点上，则此信息存储在 fit_wine $ where 中。对于数据集中的每种情况， fit_wine $其中包含代表叶子节点的 fit_wine $ frame 行号案子落在哪里。因此，我们可以使用以下每种情况获取叶子信息：
If you specifically want to know which leaf node an observation falls on, this information is stored in fit_wine$where. For each case in the data set,fit_wine$where contains the row number of fit_wine$frame that represents the leaf node where the case falls. So we can get the leaf information for each case with:
trainingnodes <- rownames(fit_wine$frame)[fit_wine$where]
为了获取测试数据的叶子信息，我曾经运行 predict（）与 type = matrix 并进行推断。令人困惑的是，这返回了一个通过连接预测类而产生的矩阵，该类在拟合树中该节点处的计数以及类概率。因此，对于此示例：
In order to get the leaf info for test data, I used to run predict() with type="matrix" and infer it. This returns, confusingly, a matrix produced by concatenating the predicted class, the class counts at that node in the fitted tree, and the class probabilities. So for this example:
testresults <- predict(fit_wine, test, type = "matrix")
testresults <- data.frame(testresults)
names(testresults) <- c("ClassGuess","NofClass1onNode", "NofClass2onNode",
"NofClass3onNode", "PClass1", "PClass2", "PClass2")
据此，我们可以推断出不同的节点，例如从 unique（testresults [，2：4] ）却不雅致。
From this, we can infer the different nodes, e.g., from unique(testresults[,2:4]) but it is inelegant.
但是，Yuji在前面的问题中对此有一个巧妙的技巧。他复制了rpart对象，并用其中的节点代替了类，因此运行预测返回的节点不是类：
However, Yuji has a clever hack for this at a previous question. He copies the rpart object and substitutes the nodes in for the classes, so running predict returns the node not the class:
nodes_wine <- fit_wine
nodes_wine$frame$yval = as.numeric(rownames(nodes_wine$frame))
testnodes <- predict(nodes_wine, test, type="vector")
我在这里包括了解决方案，但是人们去应该投票给他。
I've included the solution here, but people go should upvote him .
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