similar to: How to calculate the robust standard error of the dependent variable

I would like to prepare the data for barplot. But I only have the data frame now. x1=rnorm(10,mean=2) x2=rnorm(20,mean=-1) x3=rnorm(15,mean=3) data=data.frame(x1,x2,x3) If there a way to put data within a specific range? The expected result is as follows: range x1 x2 x3 -10-0 2 5 1 (# points in this

Hi, folks, As seen in the following codes: x1=rlnorm(10) x2=rlnorm(10,mean=2) y=rlnorm(10,mean=10)### Fake dataset linmod=lm(log(y)~log(x1)+log(x2)) After the regression, I would like to know the mean of y. Since log(y) is normal and y is lognormal, I need to know the mean and variance of log(y) first. I tried mean (y) and mean(linmod), but either one is what I want. Any tips? Thanks in

Hi, folks, As I understand, Least-squares Estimate (second-moment assumption) and the Method of Maximum Likelihood (full distribtuion assumption) are used for linear regression. I do >?lm, but the help file does not tell me the model employed in R. But in the book 'Introductory Statistics with R', it indicates R estimate the parameters using the method of Least-squares. However it

Apologize for not being clearer earlier. I would like to ask again. Thank Joris and Markleeds for response. Two examples: 1. Function 'var'. In R, it is the sum of square divided by (n-1) but not by n. (I know this in R class) 2. Function 'lm'. In R, it is the residual sum of square divied by (n-2) not by n, the same as in the least squares estimate. But the assumption following

Hi, folks, Please first look at the codes: plan_a=c('apple','orange','apple','apple','pear','bread') plan_b=c('bread','bread','orange','bread','bread','yogurt') value=1:6 data=data.frame(plan_a,plan_b,value) library(plyr) library(reshape) mm=melt(data, id=c('plan_a','plan_b'))

On 08/30/2018 01:56 PM, Joris Meys wrote: > I have to agree with Emil here. && and || are short circuited like in C and > C++. That means that > > TRUE || c(TRUE, FALSE) > FALSE && c(TRUE, FALSE) > > cannot give an error because the second part is never evaluated. Throwing a > warning or error for > > c(TRUE, FALSE) || TRUE > > would mean

I've noticed that I get a warning message every time a package masks some functions from Hmisc. The warning message says : Warning message: In identical(get(., i), get(., lib.pos)) : ignoring non-pairlist attributes This happens with eg: library(plyr) library(xtable) I think I've seen this passing by before, but I'm not sure any more. Just thought I'd mention it. Cheers Joris

Dear all, Suppose my data frame is as follows: id price distance 1 2 4 1 3 5 ... 2 4 8 2 5 9 ... n 3 7 n 8 9 I would like to calculate the rank-order correlation between price and distance for each id. cor(price,distance,method = "spearman") calculate a correlation for all. Then I tried to use apply(data,list='id',cor(price , distance , method =

Hi, I would like to have an index for a column in a matrix encoded in a cell of the same matrix. For example: x = matrix(c(11,12,13,1, 21,22,23,3, 31,32,33,2),byrow=T,ncol=4) In this case, column 4 is the index. I then access the column specified in the index by: > for (i in 1:3) print(x[i,x[i,4]]) [1] 11 [1] 23 [1] 32 > > for (i in 1:3) {x[i,x[i,4]] <- x[i,x[i,4]] + 5} > x

?typeof? is your friend here: > typeof(`[`) [1] "special" > typeof(mc[[1]]) [1] "symbol" > typeof(mc2[[1]]) [1] "special" so mc[[1]] is a symbol, and thus not a primitive. - Lukas > On 28 Mar 2017, at 14:46, Michael Lawrence <lawrence.michael at gene.com> wrote: > > There is a difference between the symbol and the function (primitive >

On 31/01/2018 6:33 AM, Joris Meys wrote: > 3. given your criticism, I'd like your opinion on where I can improve > the documentation of https://github.com/CenterForStatistics-UGent/pim. > I'm currently busy updating the help files for a next release on CRAN, > so your input is more than welcome. After this invitation I sent some private comments to Joris. I would say his

Dear gurus, I was utterly surprised to learn that one of my examples illustrating the need of match.fun() doesn't give me the expected result. center <- function(x,FUN) FUN(x) center(1:10, mean) mean <- 4 center(1:10, mean) Used to give me the error message "could not find function FUN". Now it just works, even though I didn't expect it to. I believe this is at least

Hi all, Sorry to bother you. I'm estimating a discrete choice model in R using the maxBFGS command. Since I wrote the log-likelihood myself, in order to double check, I run the same model in Limdep. It turns out that the coefficient estimates are quite close; however, the standard errors are very different. I also computed the hessian and outer product of the gradients in R using the

Dear Mr. or Ms.,   I used the R-software to run the zero-inflatoin negative binomial model (zeroinfl()) .   Firstly, I introduced one dummy variable to the model as an independent variable, and I got the estimators of parameters. But the results are not satisfied to me. So I introduced three dummy variables to the model. but I could not get the results. And the error message is

Dear all, I'm running a set of nonparametric MDS analyses, using a wrapper for isoMDS, on a 800x800 distance matrix. I noticed that setting the parameter k to larger numbers seriously increases the calculation time. Actually, with k=10 it calculates already longer than for k=2 and k=5 together. It's now calculating for 6 hours, and counting... There is quite a difference between the

On 01/04/2015 1:35 PM, Gabriel Becker wrote: > Joris, > > > The second argument to evalq is envir, so that line says, roughly, "call > environment() to generate me a new environment within the environment > defined by data". I think that's not quite right. environment() returns the current environment, it doesn't create a new one. It is evalq() that created

Hi, I want to give a summary or anova for "arima" model in R, as "summary", and "anova" for "lm". As including various intervention factors in arima(xreg = ) part, I want to assess the significancy of thse factors. I can do it using interrupted analysis of time series by linear regression, but want to see whether arima model works for the data first.

Seriously, if a method gives a wrong result, it's wrong. line() does NOT implement the algorithm of Tukey, even not after the patch. We're not discussing Excel here, are we? The method of Tukey is rather clear, and it is NOT using the default quantile definition from the quantile function. Actually, it doesn't even use quantiles to define the groups. It just says that the groups

Apparently, as.POSIXlt takes one o'clock as the start of the day : > as.POSIXlt(0,origin="1970-01-01") [1] "1970-01-01 01:00:00 CET" > as.POSIXlt(0,origin="1970-01-01 00:00:00") [1] "1970-01-01 01:00:00 CET" > as.POSIXlt(0,origin="1970-01-01 23:59:59") [1] "1970-01-02 00:59:59 CET" Cheers -- Joris Meys Statistical

can anyone know where i can find information on compile hmisc on windows, especially 64 windows? thanks, _________________________________________________________________ The New Busy is not the too busy. Combine all your e-mail accounts with Hotmail. ID28326::T:WLMTAGL:ON:WL:en-US:WM_HMP:042010_4 [[alternative HTML version deleted]]