On Thu, 8 Jan 2009, Neil Beddoe wrote:
You can also use subscripts to get at things with a bit of playing around.
You can, but it is easier not to fight R to do so. Much more transsparent
is:
cf <- coef(summary(<lm fit>)
cf[2,2] # index as a matrix.
You are
a) indexing a matrix as a vector
b) using [[]] on a numeric vector, which is unneded
c) indexing a list by number where indexing by name is simpler and using
the extractor function coef() is even simpler.
summary(lm(x~seq(1,length(x),1)))
Call:
lm(formula = x ~ seq(1, length(x), 1))
Residuals:
Min 1Q Median 3Q Max
-40.0961 -15.5289 -0.6489 12.7488 41.0107
Coefficients:
Estimate Std. Error t value Pr(>|t|)
(Intercept) 165.259602 1.620906 101.955 <2e-16 ***
seq(1, length(x), 1) -0.048711 0.005551 -8.775 <2e-16 ***
---
Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
Residual standard error: 18.19 on 503 degrees of freedom
Multiple R-squared: 0.1328, Adjusted R-squared: 0.131
F-statistic: 77 on 1 and 503 DF, p-value: < 2.2e-16
summary(lm(x~seq(1,length(x),1)))[[4]][[4]]
[1] 0.005551145
summary(lm(x~seq(1,length(x),1)))[[4]][[2]]
[1] -0.04871091
-----Original Message-----
From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org] On
Behalf Of AllenL
Sent: 06 January 2009 19:48
To: r-help@r-project.org
Subject: Re: [R] Two Noobie questions
Thanks for your help!
I combined the above two to get the following, which seems to work (if somewhat
inelegant):
int.List<-unlist(lapply(lmList, function(x) {coef(x)[1]}),use.names=FALSE)
lmList is my list of lm objects.
-Allen
David Winsemius wrote:
On Jan 6, 2009, at 1:50 PM, AllenL wrote:
1. I have a list of lm (linear model) objects. Is it possible to
select, through subscripts, a particular element (say, the intercept)
from all the models? I've tried something like this:
?coef
if your list of models is ml, then perhaps something like this
partially tested idea:
lapply(ml, function(x) coef(x)[1] )
This is what I get using that formulation an available logistic model:
> coef(lr.TC_HDL_BMI)[1]
Intercept
-6.132448
List[[1:length(list)]][1]
All members of the list are similar. My goal is to have a list of the
intercepts and lists of other estimated parameters. Is it better to
convert
to a matrix? How to do this?
2. Connected to this, how do I convert from a list back to a vector?
This
problem arose from using "split" to split a vector by a factor, then
selecting a subset of this (ie. length>10), leaving me with subset
list of
my original. Unsplit(newList, factor) doesn't work, presumably due
to my
removal of some values. Thoughts?
?unlist
> ll <- list(1,2,3,4)
> ll
[[1]]
[1] 1
[[2]]
[1] 2
[[3]]
[1] 3
[[4]]
[1] 4
> unlist(ll)
[1] 1 2 3 4
> str(unlist(ll))
num [1:4] 1 2 3 4
> is.vector(unlist(ll))
[1] TRUE
--
David Winsemius
Thanks!
-Allen
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