GBM实例1：

#1、构造数据集
# A least squares regression example # create some data  
N<-1000
X1<-runif(N)
X2<-2*runif(N)
X3<-ordered(sample(letters[1:4],N,replace=TRUE),levels=letters[4:1])
X4<-factor(sample(letters[1:6],N,replace=TRUE))
X5<-factor(sample(letters[1:3],N,replace=TRUE))
X6<-3*runif(N)
mu<-c(-1,0,1,2)[as.numeric(X3)]
SNR<-10                         # signal-to-noise ratio
Y<-X1**1.5+2*(X2**.5)+mu
sigma<-sqrt(var(Y)/SNR)
Y<-Y+rnorm(N,0,sigma)

# introduce some missing values
X1[sample(1:N,size=500)]<-NA
X4[sample(1:N,size=300)]<-NA
data<-data.frame(Y=Y,X1=X1,X2=X2,X3=X3,X4=X4,X5=X5,X6=X6)

#2、使用GBM函数建模
library(gbm)

## Loading required package: survival

## Loading required package: lattice

## Loading required package: splines

## Loading required package: parallel

## Loaded gbm 2.1.1

gbm1 <- gbm(Y~X1+X2+X3+X4+X5+X6,     #formula
            data = data,             #dataset
            var.monotone = c(0, 0, 0, 0, 0, 0), #-1:monotone decrease,+1:monotone increase,0:no monotone restrictions
            distribution="gaussian", #see the help for other choices
            n.trees=1000,            #number of trees
            shrinkage=0.05,          #shrinkageor learning rate, 0.001 to 0.1 usually work
            interaction.depth=3,     #1:additive model,2: two-way interactions, etc.
            bag.fraction=0.5,        #subsampling fraction,0.5 is probably best
            train.fraction=0.5,      #fraction of data for training, first train.fraction*N used for training
            n.minobsinnode=10,       # minimum total weight needed in each node
            cv.folds=3,              #do 3-fold cross-validation
            keep.data=TRUE,          #keep a copy of the dataset with the object
            verbose=FALSE,           #don't print out progress
            n.cores=1)               #use only a single core (detecting #cores is error-prone,so avoided here)

#3、使用交叉验证确定最佳迭代次数
best.iter <- gbm.perf(gbm1, method = "cv")

print(best.iter)

## [1] 109

#4、各解释变量的重要程度
summary(gbm1, n.trees=best.iter)

##    var    rel.inf
## X3  X3 67.2320878
## X2  X2 28.7445034
## X1  X1  2.7022822
## X6  X6  0.6273469
## X4  X4  0.5817726
## X5  X5  0.1120072

实例2：请参考《logit,GBM,knn,xgboost对一个信用卡数据的实现》之GBM 
http://chiffon.gitcafe.io/2015/05/20/newtry.html#topofpage
##GBM算法
#1、使用组合算法建模
library(caret)
ctrl <-trainControl(method = "repeatedcv", number=5, repeats=5)
set.seed(300)
model_gbm <-train(V16~., data=train_data, method="gbm", metric="Kappa",trControl=ctrl)
 
#2、建模
pred_gbm <-predict(model_gbm, test_data)
 
#3、模型评估
> table(pred_gbm, test_data$V16)       
pred_gbm  -  +
       - 91 11
       + 16 77
> mean(pred_gbm==test_data$V16)
[1] 0.8615385