In that case, using the example data from the prior response all you need is:
coef(lm(t(mat) ~ x)) On Sun, Oct 5, 2008 at 1:18 PM, Mark Kimpel <[EMAIL PROTECTED]> wrote: > Sorry for the vagueness of my question, your interpretation, however, was > spot on. Correct me if I am wrong, but my impression is that apply is a more > compact way of a for loop, but that the way R handles them computationally > are the same. In the article I seem to remember, there was a significant > increase in speed with your second approach, presumably because function > calls are avoided in R and the heavy lifting is done in C. I will use your > second approach anyway, but can I expect increased computational efficiency > with it and, if so, is my reasoning in the prior sentence correct? > > BTW, it appears as though my own attempt was almost correct, but I did not > transpose the matrix. In genomics, our response variables (genes) are the > rows and the predictor values are the column names. The BioConductor > packages I routinely use are very good at hiding this and I just didn't come > to mind. > > Mark > ------------------------------------------------------------ > Mark W. Kimpel MD ** Neuroinformatics ** Dept. of Psychiatry > Indiana University School of Medicine > > 15032 Hunter Court, Westfield, IN 46074 > > (317) 490-5129 Work, & Mobile & VoiceMail > (317) 399-1219 Home > Skype: mkimpel > > ****************************************************************** > > > On Sun, Oct 5, 2008 at 10:28 AM, Duncan Murdoch <[EMAIL PROTECTED]>wrote: > >> On 05/10/2008 10:08 AM, Mark Kimpel wrote: >> >>> I have a large matrix, each row of which needs lm applied. I am certain >>> than >>> I read an article in R-news about this within the last year or two that >>> discussed the application of lm to matrices but I'll be darned if I can >>> find >>> it with Google. Probably using the wrong search terms. >>> >>> Can someone steer me to this article of just tell me if this is possible >>> and, if so, how to do it? My simplistic attempts have failed. >>> >> >> You don't give a lot of detail on what you mean by applying lm to a row of >> a matrix, but I'll assume you have fixed predictor variables, and each row >> is a different response vector. Then you can use apply() like this: >> >> x <- 1:10 >> mat <- matrix(rnorm(200), nrow=20, ncol=10) >> >> resultlist <- apply(mat, 1, function(y) lm(y ~ x)) >> resultcoeffs <- apply(mat, 1, function(y) lm(y ~ x)$coefficients) >> >> >> "resultlist" will contain a list of 20 different lm() results, >> "resultcoeffs" will be a matrix holding just the coefficients. >> >> lm() also allows the response to be a matrix, where the columns are >> considered different components of a multivariate response. So if you >> transpose your matrix you can do it all in one call: >> >> resultmulti <- lm(t(mat) ~ x) >> >> The coefficients of resultmulti will match resultcoeffs. >> >> Duncan Murdoch >> >> Duncan Murdoch >> > > [[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.
