Thanks David That's great As a matter of interest, to get a data frame by studies why do you have to do
fitsdf <- as.data.frame(t(as.data.frame(fits))) Why doesn't fitsdf <- as.data.frame(t(fits)) work? Sandy Small ________________________________________ From: David Winsemius [dwinsem...@comcast.net] Sent: 26 October 2010 16:37 To: Small Sandy (NHS Greater Glasgow & Clyde) Cc: r-help@r-project.org Subject: Re: [R] Forcing results from lm into datframe On Oct 26, 2010, at 8:08 AM, Small Sandy (NHS Greater Glasgow & Clyde) wrote: > Hi > > I need some help getting results from multiple linear models into a > dataframe. > Let me explain the problem. > > I have a dataframe with ejection fraction results measured over a > number of quartiles and grouped by base_study. > My dataframe (800 different base_studies) looks like > >> afvtprelvefs > basestudy quartile ef ef_std entropy > CBP0908020 1 21.6 0.53 3.27 > CBP0908020 2 32.5 0.61 3.27 > CBP0908020 3 30.8 0.63 3.27 > CBP0908020 4 33.6 0.37 3.27 > CBP0908022 1 42.4 0.52 1.80 > CBP0908021 1 29.4 0.70 2.63 > CBP0908021 2 29.2 0.42 2.63 > CBP0908021 3 29.7 0.89 2.63 > CBP0908021 4 29.3 0.50 2.63 > CBP0908022 2 45.7 1.30 1.80 > ... > > What I want to do is apply a weighted linear fit to the results from > each base study and get the gradient out of it. I then want to plot > the gradient against the entropy (which is constant for each base > study). > > I can get apply a linear fit with > >> fits <- by(afvtprelvefs, afvtprelvefs$basestudy, function (x) lm >> (ef ~ quartile, data=x, weights=1/ef_std)) > > but how do I get the results from that into a dataframe which I can > use? > > I thought I might get somewhere with >> sapply(fits, "[[", "coefficients") > > But that doesn't give me the basestudy separately so that I can > match up the results with the entropy results. The by objects don't play nicely with as.data.frame so I went to a more "classical" way of runnning the lm call and I added a coef() wrapper to just get the coefficients: > splits <-split(afvtprelvefs, afvtprelvefs$basestudy) > lapply(splits, function (x) coef(lm (ef ~ quartile, data=x, weights=1/ef_std))) $CBP0908020 (Intercept) quartile 20.921397 3.385469 $CBP0908021 (Intercept) quartile 29.31632071 0.01372604 $CBP0908022 (Intercept) quartile 39.1 3.3 > fits <- lapply(splits, function (x) coef(lm (ef ~ quartile, data=x, weights=1/ef_std))) > as.data.frame(fits) CBP0908020 CBP0908021 CBP0908022 (Intercept) 20.921397 29.31632071 39.1 quartile 3.385469 0.01372604 3.3 The split-lapply strategy is reasonably general. You may need to use t() if you were hoping for stufy to be by rows. In this case sapply would have obviated the need for the as.data.frame step at the cost of returning a matrix rather than a data.frame. -- David > > I am sure this must have been answered somewhere before but I have > been unable to find a solution. > Many thanks for your help > > Sandy Small > NHS Greater Glasgow and Clyde > > > ******************************************************************************************************************** > > This message may contain confidential information. If yo...{{dropped: > 24}} > > ______________________________________________ > 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. ******************************************************************************************************************** This message may contain confidential information. If yo...{{dropped:21}} ______________________________________________ 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.