search for: nobs

....lm* deviance.lm* dfbeta.lm* [9] dfbetas.lm* drop1.lm* dummy.coef.lm* effects.lm* [13] extractAIC.lm* family.lm* formula.lm* hatvalues.lm [17] influence.lm* kappa.lm labels.lm* logLik.lm* [21] model.frame.lm model.matrix.lm nobs.lm* plot.lm [25] predict.lm print.lm proj.lm* qr.lm* [29] residuals.lm rstandard.lm rstudent.lm simulate.lm* [33] summary.lm variable.names.lm* vcov.lm* Non-visible functions are asterisked > > library(gdata) gdata: read.xls...

Hi all, The nobs method of (MASS:::polr class) takes into account of weight, but nobs method of glm does not. I wonder what is the rationale of such design behind nobs.glm. Thanks in advance. Best Regards. > library(MASS) > house.plr <- polr(Sat ~ Infl + Type + Cont, weights = Freq, data = housing) >...

Dear all, I am studying a bit the various support functions that exist for extracting information from fitted model objects. From the help files it is not completely clear to me whether the number returned by nobs() should be the same as the "nobs" attribute of the object returned by logLik(). If so, then there is a slight inconsistency in the methods for 'nls' objects with logLik.nls() taking zero weights into account while nobs.nls() does not. Admittedly, the help page of nobs() state...

Hi! The nobs() method for glm objects always returns the number of cases with non-null weights in the data, which does not correspond to the number of observations for Poisson regression/log-linear models, i.e. when family="poisson" or family="quasipoisson". This sounds dangerous since nobs...

Dear friends, I'm probably wrong but is there anything better than bootstrap to get a confidence interval of the median from a population with unspecified distribution ? Best wishes Troels Ring, Aalborg, Denmark

...wing... I've got a list (returned from a function) and I would like to "cbind()" the lists together to create a "cross tab" report or simply bind them together somehow the function returns a list that looks like the following: > all$BM $species [1] "BM" $vbar.nobs [1] 3 $vbar.sum [1] 54.05435 $count.nobs [1] 20 $basal.area [1] 26 $expf [1] 5.339182 > so there are different variable types in the list (meaning I can't use cbind?) to create a table with more than one column for stringing together multiple species. I tried to use rbind and got similar r...

Hi, in R-3.0 I get the following error when calling lsfit with more observations than variables, which seems to come from an error in the formatting of the error message (note that this was not happening in 2.15.3): > nobs <- 5; nvar <- 6; lsfit(matrix(runif(nobs*nvar), ncol=nvar), runif(nobs), intercept=FALSE) Error in sprintf(ngettext(nry, "%d response", "%d responses"), ", ", ngettext(ncx, : invalid format '%d'; use format %s for character objects > traceback() 3:...

It seems that confint.default returns an empty data.frame for objects of class mlm. For example: ``` nobs <- 20 set.seed(1234) # some fake data datf <- data.frame(x1=rnorm(nobs),x2=runif(nobs),y1=rnorm(nobs),y2=rnorm(nobs)) fitm <- lm(cbind(y1,y2) ~ x1 + x2,data=datf) confint(fitm) # returns: 2.5 % 97.5 % ``` I have seen proposed workarounds on stackoverflow and elsewhere, but suspect th...

...ne help me with the following problem? Many thanks in advance. # Problem Description: # I want to write functions which take a (character) vector of dataframe names as input argument. # For example, I want to extract the number of observations from a number of dataframes. # I tried the following: nobs.fun <- function (dframe.vec) { nobs.vec <- array(NA,c(length(dframe.vec),1)) for (i in 1:length(dframe.vec)) { nobs.vec[i] <- dim(dframe.vec[i])[1] } return(nobs.vec) } # To show the problem, I create a fake dataframe and store its name (i.e., dframe.1) # in a ve...

In accordance with Venables and Ripley, SAS documentation and other sources AIC with sigma^2 unknown is calculated as: AIC = -2LL + 2* #parameters = n log(RSS/n) + 2p For the fitness data: (http://support.sas.com/ctx/samples/index.jsp?sid=927), SAS gets an AIC of 64.534 with model oxygen = runtime. (SAS STAT User's Guide. Chapter 61. pp 3956, the REG Procedure). This value of AIC accords

I was wondering if it is possible to do the following in a smarter way. I want get the mean value across the columns of a matrix, but I want to do this on subrows of the matrix, given by some vector(same length as the the number of rows). Something like nObs<- 6 nDim <- 4 m <- matrix(rnorm(nObs*nDim),ncol=nDim) fac<-sample(1:(nObs/2),nObs,rep=T) ##loop trough different 'factor' levels for (i in unique(fac)) print(apply(m[fac==i,],2,mean)) Now, the problem is that if a value in 'fac' only occurs once, the 'a...

Hello, how is, for example, the Schwarz criterion is defined for kmeans? It should be something like: k <- 2 vars <- 4 nobs <- 100 dat <- rbind(matrix(rnorm(nobs, sd = 0.3), ncol = vars), matrix(rnorm(nobs, mean = 1, sd = 0.3), ncol = vars)) colnames(dat) <- paste("var",1:4) (cl <- kmeans(dat, k)) schwarz <- sum(cl$withinss)+ vars*k*log(nobs) Thanks for your help, Serguei _______...

...oting has occured. So I deleted that line. That's all for now. Thanks for the great package and I hope that this bug report will be useful. Cheers, Berwin ========================= original glm.fit ========================= 1 glm.fit <- 2 function (x, y, weights = rep(1, nobs), start = NULL, etastart = NULL, 3 mustart = NULL, offset = rep(0, nobs), family = gaussian(), 4 control = glm.control(), intercept = TRUE) 5 { 6 x <- as.matrix(x) 7 xnames <- dimnames(x)[[2]] 8 ynames <- names(y) 9 conv <- FALSE 10...

...s a YoYo = unstable = changes from positive to negative. So, if I codify the visits with codes 1,2,4,8 if present at year 1,2,3,4 and similarly the tests positive I get the last2 list codifying the test code corresponding to the visits patterns possible, similarly the last1 list 20 here means NULL nobs <- 400 # visits 0 1 2 3 4 5 6 7 8 9 last1 <- list((20),(1),(2),c(3,2),(4),c(5,4),c(6,4),c(7,6,4),(8),c(9,8), # visits 10 11 12 13 14 15 c(10,8),c(11,10,8),c(12,8),c(13,12,8),c(14,12,8),c(15,14,12,8))...

...34 4 16.2311097 35 4 -2.5781897 36 4 -3.0167290 37 4 -0.1119353 38 4 1.1983126 39 4 -8.8212143 40 4 3.8895263 my code: library(nlme) # Define essential constants. # Number of subject studied. NSubs <- 4 # Number of observations per subject. NObs <- 10 # Between study SD tau <- 4 # Within study SD. sigma <- 8 # END Define essential constants. # Define between subject variation between <- matrix(nrow=10,ncol=1) between <- rnorm(NSubs,0,tau) between # END Define between subject variation. # Define within subject varation....

Full_Name: Ju-Sung Lee Version: 2.2.0 OS: Windows XP Submission from: (NULL) (66.93.61.221) BIC() requires the attribute $nobs from the logLik object but the logLik of a glm(formula,family=binomial()) object does not include $nobs. Adding attr(obj,'nobs') = value, seems to allow BIC() to work. Reproducing the problem: library(nmle); BIC(logLik(glm(1~1,family=binomial())));

Hi, I have a data set like this: Mutant Rep Time OD 02H02 1 0 0.029 02H02 2 0 0.029 02H02 3 0 0.023 02H02 1 8 0.655 02H02 2 8 0.615 02H02 3 8 0.557 02H02 1 12 1.776 02H02 2 12 1.859 02H02 3 12 1.668 02H02 1 16 3.379 02H02 2 16 3.726 02H02 3 16 3.367 306 1 0 0.033 306 2

I have a factor (with "n" observations and "k" levels), with only "nobs" < n of the observations not missing. I would like to produce a (n x k) model matrix with treatment contrasts for this factor, with rows of NAs placeholding the missing observations. If I use model.matrix() I get back a (nobs x k) matrix. Is there an easy way to get the (n x k) without c...

...containing variable names. I am trying to get this into R as a dataset for analysis. z<-"Data/media1y.txt" f=file(zz,'r') # open the file rl = readLines(f,1) # Read the first line colnames<-strsplit(rl, '\t') p = length(colnames[[1]]) # counte the number of columns nobs<-450932 close(f) Using: d1<-matrix(scan(zz,skip=1,sep="\t",fill=TRUE,what=rep("character",p), nlines=nobs),ncol=p,nrow=nobs, byrow=TRUE, dimnames=list(NULL,colnames[[1]])) produces the error Read 5761719 items Warning message: In matrix(scan(zz, skip = 1, sep = "\t&...

...ovariance matrix of the estimated parameters from glm. but perhaps there is a direct solution in one of the packages. i know how to solve this particular problem (i wrote it below) but i'm curious about the covariance matrix of coefficient as it seems to be important. the R code example : ### nObs <- 10 cl <- as.factor( sample(c(1,2,3),nObs,replace=TRUE) ) y <- rnorm(nObs) model <- glm(y ~ cl) b <- model$coefficients H <- c(1,1,-1) # we want to test H0: Hb = 0 ### the following code will NOT run unless we can compute covModelCoeffs #the mean of Hb is mu = H %*% model$co...