I am not sure why you would want it, but here it is:
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)
test1 <- data.frame(individual, day, condition)
test1
library(plyr)
result=ddply(test1, "individual", function(x) {
model <- lm(condition ~ day, data = x)
F <- as.numeric((summary(model))$fstatistic)
ss <- sum(is.nan(F)) >0 # any(is.nan(F)) # any(is.na(F)) |
P <- ifelse(!ss, 1-pf(F[1], F[2],F[3]) , NA)
data.frame(coef(model), r.squared = summary(model)$r.squared, p.value =
P)
})
result
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On Sun, Apr 29, 2012 at 3:58 PM, Kristi Glover <kristi.glo...@hotmail.com>wrote:
> Hi Tal,
> Thank you so much for the help. I am afraid that I could not make you
> understand of my problem. The code you wrote is for whole data set, but I
> wanted to do regression based on "Individual" and put coefficient, r2 and p
> value. Individual is group variable.
>
> Here I again run the script
>
>
> > 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)
> > test1 <- data.frame(individual, day, condition)
> > test1
> individual day condition
> 1 1 4 0.72
> 2 1 17 0.72
> 3 6 12 0.67
> 4 8 12 0.73
> 5 8 17 0.76
> 6 9 3 0.65
> 7 9 9 0.68
> 8 9 22 0.78
> 9 12 13 0.73
> 10 12 20 0.71
> > library(plyr)
> > result=ddply(test1, "individual", function(x) {
> + model <- lm(condition ~ day, data = x)
> + coef(model)
> + })
> > result
> individual (Intercept) day
> 1 1 0.7200000 -4.039638e-18
> 2 6 0.6700000 NA
> 3 8 0.6580000 6.000000e-03
> 4 9 0.6242403 6.978799e-03
> 5 12 0.7671429 -2.857143e-03
> >
> in the result table: I want to add two columns with the information of r2
> and p value.
> I hope you understand my problem. thaks for allyour help. I am learning R.
> cheers
> kristi
>
>
> ------------------------------
> From: tal.gal...@gmail.com
> Date: Sun, 29 Apr 2012 12:13:08 +0300
> Subject: Re: [R] r2 and p value dispaly in table
> To: kristi.glo...@hotmail.com
> CC: r-help@r-project.org
>
>
> 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)
>
>
>
>
> ----------------Contact
> Details:-------------------------------------------------------
> Contact me: tal.gal...@gmail.com | 972-52-7275845
> Read me: www.talgalili.com (Hebrew) | www.biostatistics.co.il (Hebrew) |
> www.r-statistics.com (English)
>
> ----------------------------------------------------------------------------------------------
>
>
>
>
> 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]]
>
> ______________________________________________
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> PLEASE do read the posting guide
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>
>
>
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