Hi Francesco > Dear Petr, > > thank you so much for your quick reply. I was sure that there were some > smart ways to address my issue. I went through it and took some time to > look at the help for lapply and mapply. > However, some doubts still remain. Following your example, I did: > lll <-vector(mode = "list", length = 3) > mmm <-vector(mode = "list", length = 3) > > yyy <- lapply(lll, function(x) runif(100, min =0, max = 1)) > xxx <- lapply(mmm, function(x) runif(100, min =0, max = 1)) > > but then I get stucking again. It's not clear to me how to pass a lm > command to mapply. I tried to give a look at lapply and sapply, but I did > not manage to go much further. > It would be of big help if you could give me some more hints on this or if > you could provide me with some references. I am sorry, but I find the help > files quite cryptic. Is there a manual or some other source that you would > advice me where I could find some more example on how to deal with similar issues?
You can use for cycle for (i in 1:3) lll[[i]] <-lm(yyy[[i]]~xxx[[i]]) put result of lm to list object lll then lapply(lll, summary) gives you summary of each lm function, if you want to use mapply mmm<-mapply(function(x,y) lm(y~x), xxx, yyy, SIMPLIFY=F) you can see structure of any object by str(mmm) Both results shall be similar, for this case I believe that for loop is easier to understand. Regards Petr > > Thank you very much for your precious support, > f. > > On 27 September 2011 10:08, Petr PIKAL <petr.pi...@precheza.cz> wrote: > Hi > > > Dear R listers, > > > > I am trying to be a new R user, but life is not that easy. > > My problem is the following one: let's assume to have 3 outcome > variables > > (y1, y2, y3) and 3 explanatory ones (x1, x2, x3). > > How can I run the following three separate regressions without having to > > repeat the lm command three times? > > > > fit.1 <- lm(y1 ~ x1) > > fit.2 <- lm(y2 ~ x2) > > fit.3 <- lm(y3 ~ x3) > > > > > > Both the y and x variables have been generated extracting random numbers > > from uniform distributions using a command such as: > > > > y1 <- runif(100, min = 0, max = 1) > > > > I went to several introductory manuals, the manual R for stata users, > > econometrics in R, Introductory statistics with R and several blogs and > help > > files, but I didn't find an answer to my question. > > can you please help me? In Stata I wouldn't have any problem in running > > this as a loop, but I really can't figure out how to do that with R. > You can construct loop with naming through paste, numbers and get in R too > but you will find your life much easier to use R powerfull list > operations. > > Insted of > > y1 <- runif(100, min = 0, max = 1) > ... > > lll <- vector(mode="list", length=3) > lll <- lapply(1, function(x) runif(100, min = 0, max = 1)) > > you can use probably mapply for doing your regression. > Or you can easily access part of the list by loop > > for (i in 1:3) lm(lll[[i]]~xx[[i]]) > > (if you have your x's in list xx) > > Regards > Petr > > > Thanks in advance for all your help. > > Best, > > f. > > > > [[alternative HTML version deleted]] > > > > ______________________________________________ > > R-help@r-project.org mailing list > > https://stat.ethz.ch/mailman/listinfo/r-help > > PLEASE do read the posting guide > http://www.R-project.org/posting-guide.html > > and provide commented, minimal, self-contained, reproducible code. ______________________________________________ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
