similar to: help with predict.lda

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, 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()

Hi, I use the lda function from the MASS package to classify some samples according to some chemical properties. If I run lda without cross validation all is ok but, if I run lda with cross validation, the R consol say: resLDA <- ldaRedOx <- lda(Activity ~ TRedOx[,1:6], CV=TRUE, data=dfDataRedOx, subset=train) predLDA <- predict(resLDA, newdata=dfDataRedOx[-train,])$class >

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

> 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,]

Dear R-helpers, if I am right a discriminant analysis can be done with "lda". My questions are: 1. What method to discriminate the groups is used by "lda" (Fisher's linar discriminant function, diagonal linear discriminant analysis, likelihood ratio discriminant rule, ...)? 2. How can I see, which method is used? (Typing just lda does not give me any code). Thank you in

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 all, I obtained a strage result with LDA (MASS) function in R with NIR data. I tried both CV (leave one out cross validation) and splitting my data in odd (training) and even (prediction) sets. In all the cases the minimum error was near to 0. Due to the strange result, I tried with SPSS IBM software and it give me around 11% of minimum error with and without leave one out cross validation.

Hello Using R 2.7.0 on Windows. I am running a linear discriminant analysis as follows <<<< discrim1 <- lda(normvar~ mafmahal+ mrfmahal+ mffmahal+ bafmahal+ brfmahal+ cofmahal+ bmfmahal+ cfmahal+ fractmahal+ antmahal+ absmifmahal+ absifmahal, subset = train) prediction <- predict(discrim1, traintest1[-train,])$class >>> When I do this, I get a warning

> 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 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

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, 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 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? --

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]]<-

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'

I'm trying to classify some observations using lda and I'm getting a strange error. I loaded the MASS package and created a model like so: >train <- mod1[mod1$rand < 1.7,] >classify <- mod1[mod1$rand >= 1.7,] >lda_res <- lda(over_win ~ t1_scrd_a + t1_alwd_a, data=train, CV=TRUE) That works, and all is well until I try to do a prediction for the holdouts:

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]]

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