similar to: cumsum while maintaining NA

Dear R-helpers, I have a dataframe like this: ID X1 X2 X3 X4 X5 X6 49 1 1 1 0 NA NA 50 1 1 1 1 NA 1 I would like to convert a missing value (NA) that follows a 0 (zero) or another missing value (NA) into a 0 (zero). So, the above lines would be converted to: ID X1 X2 X3 X4 X5 X6 49 1 1 1 0 0 0 50 1 1 1 1 NA 1 I have been struggling with

Dette er en melding med flere deler i MIME-format. --=_alternative 004C4E4A00257091_= Content-Type: text/plain; charset="US-ASCII" Yes. so (x1*x2*x3*x4*x5*x6*x7*x8)^2 = (x1+x2+x3+x4+x5+x6+x7+x8)^8 ? and there is a difference in (x1*x2*x3*x4*x5*x6*x7*x8)^2 and (x1*x2*x3*x4*x5*x6*x7*x8) althoug the resulting formulas are the same, or? This fikses my problem, but R still crashes for the

On Wed, 5 Oct 2005 Hallgeir.Grinde at elkem.no wrote: > 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 <- runif(1000) > y <-

On Wed, 5 Oct 2005 Hallgeir.Grinde at elkem.no wrote: > Yes. > so (x1*x2*x3*x4*x5*x6*x7*x8)^2 = (x1+x2+x3+x4+x5+x6+x7+x8)^8 ? Yes in the sense that the simplified formula given by terms() is the same. > and there is a difference in > (x1*x2*x3*x4*x5*x6*x7*x8)^2 > and > (x1*x2*x3*x4*x5*x6*x7*x8) > althoug the resulting formulas are the same, or? The first is reduced to the

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 <-

> From: Peter Dalgaard > > Hallgeir.Grinde at elkem.no writes: > > > 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)

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

Hi, I wonder if anyone can help me with this issue. I am using R2BayesX. It seems that the model can maximally contain 20 interactions. When the number of interaction terms exceed 20, the code stops working. Here is a piece of toy code. rm(list=ls()) library(BayesX) library(R2BayesX) #data generating model f2<-function(x1,x2,x3,x4) { y<-2*sin(pi*x1)*1.5+exp(2*x2)/3+2 * sin(4 * pi * (x3

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

Dear R-list, I'm not sure what I've found about a function in DAAG package is a bug. When I was using cv.lm(DAAG) , I found there might be something wrong with it. The problem is that we can't use it to deal with a linear model with more than one predictor variable. But the usage documentation hasn't informed us about this. The code illustrates my discovery: > library(DAAG)

#### 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 R-List, I tried to fit a partial credit model using the "pcmdat" from eRm-package comparing the results of mirt, eRm, ltm and winsteps. The results where quite different, though. I cannot figure out what went wrong and I do not know which result I can rely on. This is what I did in R library(mirt) #load(file="u3.RData")

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 <-

Is there a way to apply paste to?list(form1 = EQ1, form2 = EQ2, form3 = EQ3, form4 = EQ4)?such that I don't have to write form1=EQ1 for all my models?(I might have a list of 20 or more)? I also need the EQs to read the formulas associated with them. For example, below, I was able to automate the name assignment but I could not figure out how to?to set up the list using?paste or other

Yo copio y pego este código y me sale correctamente. Se me ocurre que pueda deberse a la versión de R ¿cuál usas? El 10/09/2020 a las 17:51, Samura . escribió: > Gracias por las respuestas. > > Probé lo de hacer la función y no me salía. Pensaba que hacía algo mal. > Ahora con el código de Marcelino tampoco me sale. > > col1 <- c('x1', 'x2', 'x11',

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

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)

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',

Dear R-help community, If I have a data.frame df as follows: > df x1 x2 x3 x4 x5 x6 1 5 5 1 1 2 1 2 5 5 5 5 1 5 3 1 5 5 5 5 5 4 5 5 1 4 5 5 5 5 1 5 2 4 1 6 5 1 5 4 5 1 7 5 1 5 4 4 5 8 5 1 1 1 1 5 9 1 5 1 1 2 5 10 5 1 5 4 5 5 11 1 5 5 2 1 1 12 5 5 5 4 4 1 13 1 5 1 4 4 1 14 1 1 5 4 5 5 15 1 5 5 4

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 <-