similar to: Different results from random.Forest with test option and using predict function

I read everything I could get my hands on. Windows domain user authenticates into linux. RHEL4.6 looks at the AD for authentication and allows them to see the samba shared folders. Using winbind and samba. Now that it is functioning and I can set a samba shared folder , HOW do I make that folder so that only the members in a certain group can see the folder. Seems everyone can see it,

Hello, I have spent considerable time trying to figure out that which I am about to describe. This included searching Help, consulting my various R books, and trail and (always) error. I have been assuming I would need to use a loop (looping over columns) but perhaps and apply function would do the trick. I have unsuccessfully tried both. A scaled down version of my situation is as follows:

I'm trying to use boot.stepAIC for feature selection; I need to be able to specify the name of the dependent variable programmatically, but this appear to fail: In R-Studio with MS R Open 3.4: library(bootStepAIC) #Fake data n<-200 x1 <- runif(n, -3, 3) x2 <- runif(n, -3, 3) x3 <- runif(n, -3, 3) x4 <- runif(n, -3, 3) x5 <- runif(n, -3, 3) x6 <- runif(n, -3, 3) x7

SImplify your call to lm using the "." argument instead of manipulating formulas. > strt <- lm(y1 ~ ., data = dat) and you do not need to explicitly specify the "1+" on the rhs for lm, so > frm2<-as.formula(paste(trg," ~ ", paste(xvars,collapse = "+"))) works fine, too. Anyway, doing this gives (but see end of output)" bst <-

Failed? What was the error message? Cheers, Bert Bert Gunter "The trouble with having an open mind is that people keep coming along and sticking things into it." -- Opus (aka Berkeley Breathed in his "Bloom County" comic strip ) On Tue, Aug 22, 2017 at 8:17 AM, Stephen O'hagan <SOhagan at manchester.ac.uk> wrote: > I'm trying to use boot.stepAIC for

The error is "the model fit failed in 50 bootstrap samples Error: non-character argument" Cheers, SOH. On 22/08/2017 17:52, Bert Gunter wrote: > Failed? What was the error message? > > Cheers, > > Bert > > > Bert Gunter > > "The trouble with having an open mind is that people keep coming along > and sticking things into it." > -- Opus (aka

Until I get a fix that works, a work-around would be to rename the 'y1' column, used a fixed formula, and rename it back afterwards. Thanks for your help. SGO. -----Original Message----- From: Bert Gunter [mailto:bgunter.4567 at gmail.com] Sent: 22 August 2017 20:38 To: Stephen O'hagan <SOhagan at manchester.ac.uk> Cc: r-help at r-project.org Subject: Re: [R] boot.stepAIC

It seems that if you build the formula as a character string, and postpone the "as.formula" into the lm call, it works. instead of frm1 <- as.formula(paste(trg,"~1")) use frm1a <- paste(trg,"~1") and then strt <- lm(as.formula(frm1a),dat) regards, Heinz Stephen O'hagan wrote/hat geschrieben on/am 23.08.2017 12:07: > Until I get a fix that works, a

OK, here's the problem. Continuing with your example: strt1 <- lm(y1 ~1, dat) strt2 <- lm(frm1,dat) > strt1 Call: lm(formula = y1 ~ 1, data = dat) Coefficients: (Intercept) 41.73 > strt2 Call: lm(formula = frm1, data = dat) Coefficients: (Intercept) 41.73 Note that the formula objects of the lm object are different: strt2 does not evaluate the formula. So

Dear R users, I have a problem with reshaping data. I know such questions have been asked before, but I can't get it right, neither with the reshape function nor with the melt function. My dataset has about 407 variables and about 48000 cases. Each case looks as follows: V1 v2 v3 v4 v5 v6 v7 x1 y1 x2 y2 .... x200 y200 V1 is unique, v2-v7 are

Hello all, I've got a scatter plot, for which I create a regression line using: reg<- lm(yvars~xvars) and can plot the regression through the scatter just fine. I'd like to add two additional lines on the scatter plot: one being regressionline+standard deviation, the other being regressionline-standard deviation, thus creating some kind of 'band' for the scatter. Any help will

Dear R-users, While making a prediction using the randomForest function (package randomForest) I'm getting the following error message: "Error in predict.randomForest(model, newdata = CV) : No forest component in the object" Here's my complete code. For reproducing this task, please find my 2 data sets attached ( http://www.nabble.com/file/p17855119/data.rar data.rar ).

Requires patch 1 and 2 of the python series: https://www.redhat.com/archives/libguestfs/2017-January/msg00126.html This is the perl implementation along the same lines. We still haven't decided if patch 1 of the python series should change OCaml to report errno as reliable or not, but perhaps we can commit that patch as-is now and then touch things up further when we actually get set_error

Dear all; Is there any way to make this to work?: .x<-rnorm(50,10,3) .y<-.x+rnorm(50,0,1) X<-data.frame(.x,.y) colnames(X)<-c("Xvar","Yvar") Ytext<-"Yvar" lm(Ytext~Xvar,data=X) # doesn't run lm(Yvar~Xvar,data=X) # does run The main idea is to use Ytext as input in a function, so you just type "Yvar" and the model should fit....

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I am getting a difference in results when running some analysis using by and tapply compare to using a for loop. I've tried searching the web but had no luck with the keywords I used. I've attached a simple example below to illustrates my problem. I get a difference in the mean of yvar, diff and the p-value using tapply & by compared to a for loop. I cannot see what I am doing wrong.

I have the following code: ``` rm(list=ls()) N = 30000 xvar <- runif(N, -10, 10) e <- rnorm(N, mean=0, sd=1) yvar <- 1 + 2*xvar + e plot(xvar,yvar) lmMod <- lm(yvar~xvar) print(summary(lmMod)) domain <- seq(min(xvar), max(xvar))??? # define a vector of x values to feed into model lines(domain, predict(lmMod, newdata = data.frame(xvar=domain)))??? # add regression line, using

Dear users, I have a double for loop that does exactly what I want, but is quite slow. It is not so much with this simplified example, but IRL it is slow. Can anyone help me improve it? The data and code for foo_reg() are available at the end of the email; I preferred going directly into the problematic part. Here is the code (I tried to simplify it but I cannot do it too much or else it

I am running R version 2.15.1 in Windows XP I am having problems with a function I'm trying to create to: 1. subset a data.frame based on function arguments (colname & parmname) 2. rename the PARMVALUE column in the data.frame based on function argument (xvar) 3. generate charts plotvar <- function(parentdf,colname, parmname,xvar,yvar ){ subdf <-

Hi there, Could someone help me with the following programming problem..? I have written a function that works for my intended purpose, but it is quite closely tied to a particular dataframe and the names of the variables in this dataframe. However, I'd like to use the same function for different dataframes and variables. My problem is that I'm not quite sure how to tell my function in