задачу на завтра пожалуйста

library (gbm)
df = read.table('c://maxim//2.csv',sep=",", header=TRUE);
df
head(df)
names(df)
 
 
gbm1 <-
  gbm(z~x+y, # formula
      data=df, # dataset
      var.monotone=c(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, # shrinkage or 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,12 gbm
      # first train.fraction*N used for training
      n.minobsinnode = 10, # minimum total weight needed in each node
      cv.folds = 30, # 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)
# check performance using an out-of-bag estimator
# OOB underestimates the optimal number of iterations
best.iter <- gbm.perf(gbm1,method="OOB")
print(best.iter)
# check performance using a 50% heldout test set
best.iter <- gbm.perf(gbm1,method="test")
print(best.iter)
# check performance using 5-fold cross-validation
best.iter <- gbm.perf(gbm1,method="cv")
print(best.iter)
# plot the performance # plot variable influence
summary(gbm1,n.trees=1) # based on the first tree
summary(gbm1,n.trees=best.iter) # based on the estimated best number of trees
# compactly print the first and last trees for curiosity
print(pretty.gbm.tree(gbm1,1))
print(pretty.gbm.tree(gbm1,gbm1$n.trees))
 
r = read.table('c://maxim//0.csv',sep=",", header=TRUE);
r
head(r)
names(r)
 
f.predict <- predict(gbm1,r,best.iter)