similar to: collapse rows in a matrix

Hello I have data like this x1 x2 x3 x4 x5 I want to create a matrix similar to a correlation matrix, but with the difference between the two values, like this x1 x2 x3 x4 x5 x1 x2-x1 x3-x1 x4-x1 x5-x1 x2 x3-x2 x4-x2 x5-x2 x3 x4-x3 x5-x3 x4 x5-x4 x5 Then I

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 have a sparse contingency table (most cells are 0): > xtabs(~.,data[,idx:(idx+4)]) , , x3 = 1, x4 = 1, x5 = 1 x2 x1 1 2 3 1 0 0 31 2 0 0 112 3 0 0 94 , , x3 = 2, x4 = 1, x5 = 1 x2 x1 1 2 3 1 0 0 0 2 0 0 0 3 0 0 0 , , x3 = 3, x4 = 1, x5 = 1 x2 x1 1 2 3 1 0 0 0 2 0 0 0 3 0 0 0 , , x3 = 1, x4

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 <-

Hi, Suppose I've got a matrix, and the first few elements look like "x1 + x3 + x4 + x5 + x1:x3 + x1:x4" "x1 + x2 + x3 + x5 + x1:x2 + x1:x5" "x1 + x3 + x4 + x5 + x1:x3 + x1:x5" and so on (have got terms from x1 ~ x14). If I want to replace all the x1 with i7, all x2 with i14, all x3 with i13, for example. Is there an easy way? I tried to put what I want

I work on Windows, R version 2.4.1 I need to leave out many rows and columns when reading a matrix. The 'skip' commands in read.table, seem to be just for skipping rows, if I understand well, but I need to skip many columns as well. I tried something very simple and left one row and one column out, but I don't know how to do it for many rows and columns at a time. Example: zzz #

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 <-

Dear R-users I have a probelm running stepAIC in R1.7.1 I wrote a program which used stepAIC as a part of it, and it worked fine while I was using the previous version of R1.7.0. However, I found the program did not work any more. Now, R produces a message which tells "Error in as.data.frame.default(data) : can't coerce function into a data.frame" every time I run the part of

I'm using specify.model for the sem package. I can't figure out how to represent the residual errors for the observed variables for a CFA model. (Once I get this working I need to add some further constraints.) Here is what I've tried: model.sa <- specify.model() F1 -> X1,l11, NA F1 -> X2,l21, NA F1 -> X3,l31, NA F1 -> X4,l41, NA F1 -> X5, NA, 0.20

Hi, I am new to R for solving optimization problems, I have set of communication channels with limited capacity with two types of costs, fixed and variable cost. Each channel has expected gain for a single communication. I want to determine optimal number of communications for each channel maximizing ROI)return on investment) with overall budget as constraint.60000 is the budget allocated.

#### 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

Dear All, I have got the limits for removing extreme values for each variables using following function . f=function(x){quantile(x, c(0.25, 0.75),na.rm = TRUE) - matrix(IQR(x,na.rm = TRUE) * c(1.5), nrow = 1) %*% c(-1, 1)} #Example: n <- 100 x1 <- runif(n) x2 <- runif(n) x3 <- x1 + x2 + runif(n)/10 x4 <- x1 + x2 + x3 + runif(n)/10 x5 <-

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 ?

I'm pasting below a working R file featuring a function I'd like to polish up. I'm teaching regression this semester and every time I come to something that is very difficult to explain in class, I try to simplify it by writing an R function (eventually into my package "rockchalk"). Students have a difficult time with predict and newdata objects, so right now I'm

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

Hello, I would appreciate if someone could help me resolve the following: 1. df1[!is.na( X1 | X2 | X3 | X4 | X5),][,1:5] # This does not work 2. Is these message harmful? The following object(s) are masked from 'df1 (position 3)': X1, X2, X3, X4, X5 Thanks, Pradip Muhuri #Reproducible Example set.seed(5) df1<-data.frame(matrix(sample(c(1:10,NA),100,replace=TRUE),ncol=5))

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)

Dear All, I am new to R, I have one question which might be easy. I have a large data with more than 250 variable, i am reducing number of variables by redun function as in the example below, n <- 100 x1 <- runif(n) x2 <- runif(n) x3 <- x1 + x2 + runif(n)/10 x4 <- x1 + x2 + x3 + runif(n)/10 x5 <- factor(sample(c('a','b','c'),n,replace=TRUE)) x6 <-

Hola, me gustar?a hacer algo como en el siguiente ejemplo A un df a?adirle una columna que es la transformaci?n de otra, en plan a todo lo que sea x1, x2, x3 lo llamo prueba 1 todo lo que sea x4,x5,x6 lo llamo prueba 2 el resto de x las dejo como est?n. Ser?a algo as? col1 <- c('x1', 'x2', 'x11', 'x1','x33', 'x1','x4', 'x5',