similar to: ?rpart

HI, Dear R community, I am writing the following function to create one data set(*tree.pred*) and one vector(*valid.out*) from loops. Later, I want to use the data set from this loop to plot curves. I have tried return, list, but I can not use the *tree.pred* data and *valid.out* vector. auc.tree<- function(msplit,mbucket) { * tree.pred<-data.frame()

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]]<-

> fit.dimer <- rpart(as.factor(out) ~ ., method="class", data=p_df) > > fit.dimer$frame[, "var"] [1] NE WC <leaf> TA <leaf> <leaf> WG WD WW WC [11] <leaf> <leaf> <leaf> CT <leaf> FC <leaf> YG QT <leaf> [21] <leaf> <leaf> NW DP DY <leaf> SK

HI, Dear R community, How to extract the variables actually used in tree construction? I want to extract these variables and combine other variable as my features in next step model building. > printcp(fit.dimer) Classification tree: rpart(formula = outcome ~ ., data = p_df, method = "class") Variables actually used in tree construction: [1] CT DP DY FC NE NW QT SK TA WC WD WG WW

cv.fold<-function(i, size=3, rang=0.3){ cat('Fold ', i, '\n') out.fold.c <-((i-1)*c.each.part +1):(i*c.each.part) out.fold.n <-((i-1)*n.each.part +1):(i*n.each.part) train.cv <- n.cc[-out.fold.c, c(2:2401, 2417)] train.nv <- n.nn[-out.fold.n, c(2:2401, 2417)] train.v<-rbind(train.cv, train.nv) #training data for feature

HI, Dear R community, I have a large data set names dd.txt, the columns are: there are 2402 variables. a1, b1, ..z1, a11, b11, ...z11, a111, b111, ..z111.. IF I dont know the relative position of the columns, but I know I need the following variables: var<-c(a1, c1,a11,b11,f111) Can I use read.columns to read the data into R? I have tried the following codes, but it does not work

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]]

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

Hi, r-community, This morning, I MET the following problem several times when I try to attach the data set. When I closed the current console and reopen the R console, the problem disappear. BUt with the time passed on, the problem occurs again. Can anyone help me with this? > attach(total) The following object(s) are masked from total ( position 3 ) : acid base cell_evalue

HI, GUYS, I used the following codes to run SVM and get prediction on new data set hh. dim(all_h) [1] 2034 24 dim(hh) # it contains all the variables besides the variables in all_h data set. [1] 640 415 require(e1071) svm.tune<-tune(svm, as.factor(out) ~ ., data=all_h, ranges=list(gamma=2^(-5:5), cost=2^(-5:5)))# find the best parameters. bestg<-svm.tune$best.parameters[[1]]

HI, Dear R-community, I AM using the following codes to grow tree and plot tree: # Classification Tree with rpart library(rpart) pdf(file="/home/cdu/changbin/dimer_tree.pdf") # grow tree fit.dimer <- rpart(outcome ~ ., method="class", data=p.dimer[,2:402]) plotcp(fit.dimer) # visualize cross-validation results # plot tree plot(fit.dimer, uniform=TRUE,

> 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

> 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'

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 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 using the following codes to do the som. I tried to label the notes by the majority vote. either through mapping or prediction. I attached my output, the left one dont have any labels in the note, the right one has more than one label in each note. I need to have only one label for each note either by majority vote or prediction. Can anyone give some suggestions or