[R] Forcing results from lm into datframe

[David Winsemius]

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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