You came close.
Here is how it might be done:
individual <- rep(c(1,1,6,8,8,9,9,9,12,12),2)
day <- rep(c(4,17,12,12,17,3,9,22,13,20),2)
condition <- rep(c(0.72, 0.72, 0.67, 0.73, 0.76, 0.65, 0.68, 0.78, 0.73,
0.71),2)
test <- data.frame(individual, day, condition)
#ind.id <- unique(test$individual)
#ind.list <- lapply(1:length(ind.id), function(i){ subset(test,
test$individual==ind.id[i])})
#lms <- lapply(ind.list, lm, formula=condition~day)
require(plyr)
func1 <- function(...)
{
lm(condition~day, data = test)
}
lms <- lapply(1:10, func1)
lms
func2 <- function(fit)
{
F <- (summary(fit))$fstatistic
P <- 1-pf(F[1], F[2],F[3]) # notice how we must calculate the p-value
data.frame(r.squared = summary(fit)$r.squared, p.value = P)
}
ldply(lms, func2)
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On Sun, Apr 29, 2012 at 9:04 AM, Kristi Glover <kristi.glo...@hotmail.com>wrote:
>
> Hello R User,
> I was trying to display r.squared and p value in table from regression,
> but I could not display these parameters in the table (matrix)
>
> for example
> individual <- c(1,1,6,8,8,9,9,9,12,12)
> day <- c(4,17,12,12,17,3,9,22,13,20)
> condition <- c(0.72, 0.72, 0.67, 0.73, 0.76, 0.65, 0.68, 0.78, 0.73, 0.71)
> test <- data.frame(individual, day, condition)
> ind.id <- unique(test$individual)
> ind.list <- lapply(1:length(ind.id), function(i){ subset(test,
> test$individual==ind.id[i])})
> lms <- lapply(ind.list, lm, formula=condition~day)
> ldply(lms, function(x) x$coefficients)
> here I can display coefficients, here I need to write code for r2.squared
>
> I tried with following script
>
> summary(ldply(lms, function(x) )$r.squared, $p value)
> but it did not work.
>
> can any one help me?
>
> thanks
> Kristi
>
>
>
> [[alternative HTML version deleted]]
>
> ______________________________________________
> R-help@r-project.org mailing list
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> PLEASE do read the posting guide
> http://www.R-project.org/posting-guide.html
> and provide commented, minimal, self-contained, reproducible code.
>
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