similar to: boosting with decision tree

HI, Dear Greg, I AM A NEW to GBM package. Can boosting decision tree be implemented in 'gbm' package? Or 'gbm' can only be used for regression? IF can, DO I need to combine the rpart and gbm command? Thanks so much! -- Sincerely, Changbin -- [[alternative HTML version deleted]]

HI, Guys, I am trying to use the AdaBoosting. M.1 algorithm to integrate three models. I found the sum of weights for each model is not equal to one. How to deal with this? Thanks, any response or suggestions are appreciated! -- Sincerely, Changbin -- [[alternative HTML version deleted]]

HI, Dear Greg and R community, I have one question about the output of gbm package. the output of Boosting should be f(x), from it , how to calculate the probability for each observations in data set? SInce it is stochastic, how can guarantee that each observation in training data are selected at least once? IF SOME obs are not selected, how to calculate the training error? Thanks? --

HI, Dear Andy, I run the RandomFOrest in R, and get the following resutls in variable importance: What is the meaning of MeanDecreaseAccuracy and MeanDecreaseGini? I found they are raw values, they are not scaled to 1, right? Which column if most similar to the variable rel.influence in Boosting? Thanks so much! > fit$importance 0 1

> x <- train[,c( 2:18, 20:21, 24, 27:31)] > y <- train$out > > svm.pr <- svm(x, y, probability = TRUE, method="C-classification", kernel="radial", cost=bestc, gamma=bestg, cross=10) > > pred <- predict(svm.pr, valid[,c( 2:18, 20:21, 24, 27:31)], decision.values = TRUE, probability = TRUE) > attr(pred, "decision.values")[1:4,]

somfunc<- function (file) { aa_som<-scale(file) final.som<-som(data=aa_som, rlen=10000, grid=somgrid(5,4, "hexagonal")) pdf(file="/home/cdu/changbin/file.pdf") #output graphic file. plot(final.som, main="Unsupervised SOM") dev.off() } I have many different files, if I want output pdf file with the same name as for each dataset I feed to the function

HI, Dear R community, Last friday, I used the codes, it works, but today, it does not run? > fit.dimer <- rpart(outcome ~., method="class", data=p.df) Error in `[.data.frame`(frame, predictors) : undefined columns selected DOEs anyone have comments or suggestions? Thanks in advance! -- Sincerely, Changbin -- [[alternative HTML version deleted]]

> head(en.id.pr) valid.gene_id b.pred rf.pred svm.pred 1521 2500151211 0 0 0 366 639679745 0 0 0 1965 2502081603 1 1 1 1420 644148030 1 1 1 1565 2500626489 1 1 1 1816 2501711016 1 1 1 > p.pred <- data.frame(en.id.pr, sum=apply(en.id.pr[,2:4], 1, sum)) #

HI, Dear R community, I want to split a data frame by using two variables: let and g > x = data.frame(num = c(10,11,12,43,23,14,52,52,12,23,21,23,32,31,24,45,56,56,76,45), let = letters[1:5], g = 1:2) > x num let g 1 10 a 1 2 11 b 2 3 12 c 1 4 43 d 2 5 23 e 1 6 14 a 2 7 52 b 1 8 52 c 2 9 12 d 1 10 23 e 2 11 21 a 1 12 23 b 2 13 32 c 1 14

Dear Dr. Sarkar, When I try to run the codes, I found the following problem: > h<- sample(1:14, 319, rep=T) > c<- sample(1:14, 608, rep=T) > n<- sample(1:14, 1140, rep=T) > vt<-c(h, c, n) > ta<-rep(c("h", "c", "n"), c(319, 608, 1140)) > > to<-data.frame(vt,ta) > library(lattice) Attaching package: 'lattice'

svm.fit<-svm(as.factor(out) ~ ., data=all_h, method="C-classification", kernel="radial", cost=bestc, gamma=bestg, cross=10) # model fitting svm.pred<-predict(svm.fit, hh, decision.values = TRUE, probability = TRUE) # find the probability, but can not find. attr(svm.pred, "probabilities") > attr(svm.pred, "probabilities") 1 0 1 0 0 2 0

Dear R community, I have one function, it works for small data set, but does not work on large data set, can anyone help me with this? > #creat new variable by dividing each aa dimer by total_length. > imper<-function(x, file) { + round(x/file$length, 5) + } > dim(test) [1] 999 2402 > test[varname[2:2401]]<-

HI, Dear R community, I am using the following codes to plot, however, the lines code works. But the line was not drawn on the previous plot and did not shown up. How comes? # specify the data for missense simulation x <- seq(0,10, by=1) y <- c(0.952, 0.947, 0.943, 0.941, 0.933, 0.932, 0.939, 0.932, 0.924, 0.918, 0.920) # missense z <- c(0.068, 0.082, 0.080, 0.099, 0.108, 0.107,

HI, Dear R community, My original file has 1932 lines, but when I read into R, it changed to 1068 lines, how comes? cdu@nuuk:~/operon$ wc -l id_name_gh5.txt 1932 id_name_gh5.txt > gene_name<-read.table("/home/cdu/operon/id_name_gh5.txt", sep="\t", skip=0, header=F, fill=T) > dim(gene_name) [1] 1068 3 -- Sincerely, Changbin -- Changbin Du DOE Joint Genome

HI, Dear R community, I am trying to created new variables and put into a data frame through a loop. My original data set: head(first) probe_name chr_id position array1 1 C-7SARK 1 849467 10 2 C-4WYLN 1 854278 10 3 C-3BFNY 1 854471 10 4 C-7ONNE 1 874460 10 5 C-6HYCN 1 874571 10 6 C-7SCGC 1 874609 10 I have

HI, Dear R community, I have one data set like this, What I want to do is to calculate the cumulative coverage. The following codes works for small data set (#rows = 100), but when feed the whole data set, it still running after 24 hours. Can someone give some suggestions for long vector? id reads Contig79:1 4 Contig79:2 8 Contig79:3 13 Contig79:4 14 Contig79:5 17

HI, Dear R community, I have the following codes to calculate the commulative coverage. I want to save the output in a vector, How to do this? test<-seq(10, 342, by=2) #cover is a vector cover_per<-function (cover) { for (i in min(cover):max(cover)) {print(100*sum(ifelse(cover >= i, 1, 0))/length(cover))} } result<-cover_per(test) > result NULL Can anyone help me this this?

> #sort the data by predicted probability > b.order<-bo.id.pred[(order(-predict)),] > b.order[1:20,] gene_id predict 43 637882902 0.07823997 53 638101634 0.66256490 61 639084581 0.08587504 41 637832824 0.02461066 25 637261662 0.11613879 22 637240022 0.06350477 62 639084582 0.02238538 63 639097718 0.06792841 44 637943079 0.04532625 80 640158389 0.06582658 3 637006517 0.57648451

HI, Dear R-community, I have one vector of the following string, > test<-c("H.1.Y", "N.0.E", "G.1.N", "E.0.P", "W.2.G", "W.4.G", "W.3.W", "W.0.N", "D.1.H") > test [1] "H.1.Y" "N.0.E" "G.1.N" "E.0.P" "W.2.G" "W.4.G" "W.3.W"

Dear R community, I have a data set like the following: probe_name chr_id position array1 array2 array3 array4 array5 array6 array7 1 C-3AAAA 10 16566949 10 10 10 10 10 10 10 2 C-3AAAB 17 33478940 10 10 10 10 10 10 10 3 C-3AAAC 3 187369224 10 10 2 10 10 1 10 4 C-3AAAD 8 28375041 10