similar to: subset using noncontiguous variables by name (not index)

Dette er en melding med flere deler i MIME-format. --=_alternative 004613C000257091_= Content-Type: text/plain; charset="US-ASCII" And some more informastion I forgot. R does not crash if I write out the formula: set.seed(123) x1 <- runif(1000) x2 <- runif(1000) x3 <- runif(1000) x4 <- runif(1000) x5 <- runif(1000) x6 <- runif(1000) x7 <- runif(1000) x8 <-

Hi, Using S=1000 and simdata <- replicate(S, generate(3000)) #If you want both "m1" and "m0" #here the missing values are 0 res1<-sapply(seq_len(ncol(simdata.psm1)),function(i) {x1<-merge(simdata.psm0[,i],simdata.psm1[,i],all=TRUE); x1[is.na(x1)]<-0; x1}) res1[,997:1000] #????? [,1]???????? [,2]???????? [,3]???????? [,4]??????? #x1??? Numeric,3000 Numeric,3000

#### I checked through every 3 factor * 3 loading case. #### While, 4 factor * 3 loading failed. #### the data is 6 factor * 3 loading require(sem); cor18<-read.moments(); 1 .68 1 .60 .58 1 .01 .10 .07 1 .12 .04 .06 .29 1 .06 .06 .01 .35 .24 1 .09 .13 .10 .05 .03 .07 1 .04 .08 .16 .10 .12 .06 .25 1 .06 .09 .02 .02 .09 .16 .29 .36 1 .23 .26 .19 .05 .04 .04 .08 .09 .09 1 .11 .13 .12 .03 .05 .03

Does anyone know how to pass a list of parameters into a function? for example: somefun=function(x1,x2,x3,x4,x5,x6,x7,x8,x9){ ans=x1+x2+x3+x4+x5+x6+x7+x8+x9 return(ans) } somefun(1,2,3,4,5,6,7,8,9) # I would like this to work: temp=c(x3=3,x4=4,x5=5,x6=6,x7=7,x8=8,x9=9) somefun(x1=1,x2=2,temp) # OR I would like this to work: temp=list(x3=3,x4=4,x5=5,x6=6,x7=7,x8=8,x9=9)

Hi, You might need to check library(boot).? I have never used that before.? So, I can't comment much.? It is better to post on R-help list.? I had seen your postings on Nabble in the past.? Unfortunately those postings were not accepted in R-help.? You have to directly post at ? r-help at r-project.org after registering at: https://stat.ethz.ch/mailman/listinfo/r-help ?

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 <-

R-help, I have a list whose elements are data frames. I want to change the colnames attribute in each element of this list but an error message comes up: > lapply(LD_strataNew,function(x) dimnames(x)[[2]][-1]) <- as.roman(1:9)[-6] Error in lapply(LD_strataNew, function(x) dimnames(x)[[2]][-1]) <- as.roman(1:9)[-6] : could not find function "lapply<-" >

Dear friends, I'm doing a simulation on logistic regression model, but the programs can't work well,please help me to correct it and give some suggestions. My programs: data<-matrix(rnorm(400),ncol=8) #sample size is 50 data<-data.frame(data) names(data)<-c(paste("x",1:8,sep="")) #8 independent variables,x1-x8; #logistic regression model is

Dear friends, After running the lm() model, we can get summary resluts like the following: Coefficients: Estimate Std. Error t value Pr(>|t|) x1 0.11562 0.10994 1.052 0.2957 x2 -0.13879 0.09674 -1.435 0.1548 x3 0.01051 0.09862 0.107 0.9153 x4 0.14183 0.08471 1.674 0.0975 . x5 0.18995 0.10482 1.812 0.0732 . x6 0.24832 0.10059 2.469 0.0154 * x7

Hi, i'm trying to simplify some R code but i got stucked in this: test<-data.frame(cbind(id,x1,x2,x3,x4,x5,x6,x7)) test > test id x1 x2 x3 x4 x5 x6 x7 1 1 36 26 21 32 31 27 31 2 2 45 21 46 50 22 36 29 3 3 49 47 35 44 33 31 46 4 4 42 32 38 28 39 45 32 5 5 29 42 39 48 25 35 34 6 6 39 31 30 37 46 43 44 7 7 41 40 25 23 42 40 24 8 8 27 29 47 34 26 38 28 9 9 25 35 29 36

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

Suppose I have the following: x1<-as.vector(rnorm(10)) x2<-as.vector(rnorm(10)) x3<-as.vector(rnorm(10)) x4<-as.vector(rnorm(10)) x5<-as.vector(rnorm(10)) x6<-as.vector(rnorm(10)) x7<-as.vector(rnorm(10)) x8<-as.vector(rnorm(10)) x9<-as.vector(rnorm(10)) x10<-as.vector(rnorm(10)) I would like the mean of x1 to x6 for each vector position. I would do something else

Hello! I have a dataset like this: X1 X2 X3 X4 X5 X6 X7 X8 1 2 2 1 2 3 2 6 2 3 2 5 7 9 1 3 1 9 12 6 1 1 3 6 The columns X1-X6 contains ordinary numeric values. X7 contains the number of the first column that the rowMeans should be calculated from and X8 contains the last column

Hi R users I have serveral binary variables (e.g., X1, X2, X3, X4, X5, X,6, and X7) and one continuous variable (e.g., Y1). I combined these variables using data.frame() mydata <- data.frame(X1,X2,X3,X4,X5,X6,X7,Y1) after that, I sorted this data.frame rank.by.Y1<-order(mydata[,8]) sorted.mydata<-mydata[rank.by.Y1,] after that, I replaced Y1's values with values ranging from 1

hello r-helpers: there is a .txt file: x1 x2 x3 x4 x5 x6 x7 x8 x9 x10 x11 y1 17 5 77 18 19 24 7 24 24 72 52 100 2 6 72 18 17 15 4 12 18 35 42 97.2 17 2 58 10 5 3 4 3 3 40 28 98 17 2 69 14 13 12 4 6 6 50 37 93 2 3 75 20 38 18 6 12 18 73 67 99 14 4 59 16 18 9 4 3 15 47 40 99.95 17 4 87 18 17 12 4 15 12 69 46 100 14 3 74 15 9 12 1 15 12 44 35 98 17 6 76 15 33 21 15 9 18 46 41 100 17 5 76 17 22 18 1

Hi All, I've encountered an issue where tail merging MIs is causing a problem with the post-RA MI scheduler dependency analysis and I'm not sure of the best way to address the problem. In my case, the branch folding pass (lib/CodeGen/BranchFolding.cpp) is merging common code from BB#14 and BB#15 into BB#16. It's clear that there are 4 common instructions (marked with an *) in BB#14

Dear R-List, (I am not sure whether this list is the right place for my question...) I have a dataframe df.cfa

I've just found the lavaan package, and I really appreciate it, as it seems to succeed with models that were failing in sem::sem. I need some clarification, however, in the output, and I was hoping the list could help me. I'll go with the standard example from the help documentation, as my problem is much larger but no more complicated than that. My question is, why is there one latent

Hi everyone! I have a data frame with 1112 time series and I am going to randomly sampling r samples for z times to compose different portfolio size(r securities portfolio). As for r=2 and z=10000,that's: z=10000 A=seq(1:1112) x1=sample(A,z,replace =TRUE) x2=sample(A,z,replace =TRUE) M=cbind(x1,x2) # combination of 2 series Because in a portfolio with x1[i]=x2[i],(i=1,2,...,10000) means a 1