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
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