similar to: count data with a specific range

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 linmod=y~x+z summary(linmod) The summary of linmod shows the standard error of the coefficients. How can we get the sd of y and the robust standard errors in R? Thanks! [[alternative HTML version deleted]]

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'))

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

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

?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 >

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

Im developing an application with R and Gtk+. It's just a simple GUI which helps new users to interactuate with R. Thing is, when you do a statistical analysis, I also want to provide a HTML report, but HTMLStart doesnt work propperly when executing from TinnR. It does create the file but not empty, I've tried some examples from different websites, and it's always the same.. it works

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]]

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.

HI, I want to import 1.5G CSV file in R. But the following error comes: 'Victor allocation 12.4 size' How to read the large CSV file in R . Any one can help me? -- View this message in context: http://r.789695.n4.nabble.com/Large-Data-tp2254130p2254130.html Sent from the R help mailing list archive at Nabble.com.

All, How can we find a distance matrix for categorical data ie. given a csv below var1 var2 var3 var4 element1-1 yes x a k element1-2 no y b l element1-3 maybe y c m how can i compute the distance matrix between all the elements Actually i need it to create clusters on top

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

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, Before I file this as a bug, I wanted to check if I didn't miss something. The help page of lag() says that the function returns a time series object. It actually does return something that looks like a ts object (the attribute tsp is set). But when using a vector, the class "ts" is not added to the result: > avec <- 1:10 > lag(avec) [1] 1 2 3 4 5 6 7 8

Hi all, I have a distribution, and take a sample of it. Then I compare that sample with the mean of the population like here in "Wilcoxon signed rank test with continuity correction": > wilcox.test(Sample,mu=mean(All), alt="two.sided") Wilcoxon signed rank test with continuity correction data: AlphaNoteOnsetDists V = 63855, p-value = 0.0002093 alternative hypothesis:

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

Dear all, I want to apply a weighted median on a huge dataset, and I remember a function from the package R.basic that could do this using an internal sorting algorithm qsort. This speeded things up quite a bit. Alas, I can't find that package anywhere anymore. There is a weighted.median function in the package limma too, but I didn't use that before. Anybody who knows what happened to

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