similar to: lm on matrix data

Hi all, I would like to do cross validation in random forest using rfcv function. As the documentation for this package says: rfcv(trainx, trainy, cv.fold=5, scale="log", step=0.5, mtry=function(p) max(1, floor(sqrt(p))), recursive=FALSE, ...) however I don't know how to build trianx and trainy for my data set, and I could not understand the way trainx is built in the package

Hi all, I would like to do cross validation in random forest using rfcv function. As the documentation for this package says: rfcv(trainx, trainy, cv.fold=5, scale="log", step=0.5, mtry=function(p) max(1, floor(sqrt(p))), recursive=FALSE, ...) however I don't know how to build trianx and trainy for my data set, and I could not understand the way trainx is built in the package

Any responds?! On Wednesday, August 23, 2017 5:50 AM, Elahe chalabi via R-help <r-help at r-project.org> wrote: Hi all, I would like to do cross validation in random forest using rfcv function. As the documentation for this package says: rfcv(trainx, trainy, cv.fold=5, scale="log", step=0.5, mtry=function(p) max(1, floor(sqrt(p))), recursive=FALSE, ...) however I

I have a data set with 10 variables, and about 8000 instances (or objects/rows/samples). In addition I have one more ('class') variable that I have about 10 instances for, but for which I wish to impute values for. I am a little confused how to go about doing this, mostly as I'm not well-versed in it. Do I train the SOM with a data object that contains just the first 10 variables

All, The kohonen predict function is returning NA for SOM predictions regardless of data used... even the package example for a SOM using wine data is returning NA's Does anyone have a working example SOM. Also, what is the purpose of trainY, what would be the dependent data for an unsupervised SOM? As may be apparent to you by my questions, I am very new to kohonen maps and am very grateful

Dear all, I am confused with the behaviour of survfit with newdata option. I am using the latest version R-2-15-0. In the simple example below I am building a coxph model on 90 patients and trying to predict 10 patients. Unfortunately the survival curve at the end is for 90 patients. Could somebody please from the survival package confirm that this behaviour is as expected or not - because I

Hi, I'd like to download some data files from a remote server, the problem here is that some of the files actually don't exist, which I don't know before try. Just wondering if a function in R could tell me if a file exists on a remote server? I searched this mailing list and after read severals mails, still clueless. Any help will be highly appreciated. B.C.

Dear All, I was trying to format a numeric vector (100*1) by using outd <- format(x=m, sci=F, digits=2) > outd[1:10] [1] " 0.01787758" "-0.14760306" "-0.45806041" "-0.67858525" "-0.64591748" [6] "-0.05918100" "-0.25632276" "-0.15980138" "-0.08359873" "-0.37866688" >m[1:10] [1]

Hi all, I used the following code and data to get auc values for two sets of predictions: library(caret) > table(predicted1, trainy) trainy hard soft 1 27 0 2 11 99 > aucRoc(roc(predicted1, trainy)) [1] 0.5 > table(predicted2, trainy) trainy hard soft 1 27 2 2 11 97 > aucRoc(roc(predicted2, trainy)) [1] 0.8451621 predicted1: 1 1 2

Hi, I have the following two measurements stored in mat: > print(mat) [,1] [,2] [1,] -14.80976 -265.786 [2,] -14.92417 -54.724 [3,] -13.92087 -58.912 [4,] -9.11503 -115.580 [5,] -17.05970 -278.749 [6,] -25.23313 -219.513 [7,] -19.62465 -497.873 [8,] -13.92087 -659.486 [9,] -14.24629 -131.680 [10,] -20.81758 -604.961 [11,] -15.32194 -18.735 To calculate the ranking

Hi, I'm trying to plot something in the following way and would like if you could help: I'd like in a same plot window, two plots are shown, the left one is a bird-view plot of the whole data, the right half keep changing, i.e., different plots will be shown up on request, so that when I select/click on some where in the left plot, the right plot will be the corresponding plot. What I

Sir, This query is related to randomForest regression using R. I have a dataset called qsar.arff which I use as my training set and then I run the following function - rf=randomForest(x=train,y=trainy,xtest=train,ytest=trainy,ntree=500) where train is a matrix of predictors without the column to be predicted(the target column), trainy is the target column.I feed the same data

Hi, I am trying to construct a svmpoly model using the "caret" package (please see code below). Using the same data, without changing any setting, I am just changing the seed value. Sometimes it constructs the model successfully, and sometimes I get an ?Error in indexes[[j]] : subscript out of bounds?. For example when I set seed to 357 following code produced result only for 8

I am learning the package "caret", after I do the "rfe" function, I get the error ,as follows: Error in `[.data.frame`(x, , retained, drop = FALSE) : undefined columns selected In addition: Warning message: In predict.lm(object, x) : prediction from a rank-deficient fit may be misleading I try to that manual example, that is good, my data is wrong. I do not know what

Hi, I have a least square fitting problem with linear inequality constraints. pcls seems capable of solving it so I tried it, unfortunately, it is stuck with the following error: > M <- list() > M$y = Dmat[,1] > M$X = Cmat > M$Ain = as.matrix(Amat) > M$bin = rep(0, dim(Amat)[1]) > M$p=qr.solve(as.matrix(Cmat), Dmat[,1]) > M$w = rep(1, length(M$y)) > M$C = matrix(0,0,0)

Hi, I am having problems passing arguments to method="gbm" using the train() function. I would like to train gbm using the laplace distribution or the quantile distribution. here is the code I used and the error: gbm.test <- train(x.enet, y.matrix[,7], method="gbm", distribution=list(name="quantile",alpha=0.5), verbose=FALSE,

Hi, Suppose that I have a list where each component is a list of two matrices. I also have a vector of weights. How can I collapse my list of lists into a single list of two matrices where each matrix in the result is the weighted sum of the corresponding matrices. I could use a loop but this is a nested calculation so I was hoping there is a more efficient way to do this. To help clarify,

Dear All, I have a data, suppose it is an N*M matrix data. All I want is to classify it into, let see, 3 classes. Which method(s) do you think is(are) appropriate for this purpose? Any reference will be welcome! Thanks! Best, Baoqiang Cao

Dear R-helper, How could I count only some variable was exist after running sample (random) function. For example, > testx <- factor(c("Game","Paper","Internet","Time","Money")) > for(i in 1:2) { + x <- sample(testx,replace=TRUE) + print(x) + } [1] Money Money Time Internet Time Levels: Game Internet

Hello Please see the example below > class(testX) [1] "matrix" > class(testX[1,]) [1] "numeric" Why not matrix? What am I missing here? Is there a way to keep the same class? The reason for the question is that I want to implement a k-step ahead prediction for my own routines and R wrecks does not seem to like [1,] as shown below. >