Hi:
This would be a lot easier to check with a reproducible example, but
here's a simplified version of your problem:
testd <- data.frame(gps = rep(c("ADHALP","ADLCON","ADLARC","BDALAT","BDPARC"),
each = 15),
trt = rep(LETTERS[1:3], each = 5),
y = rnorm(75))
testd$gps <- as.character(testd$gps)
str(testd)
library('plyr')
# The subset is handled by the grouping variable gps:
mlist <- dlply(testd, .(gps), function(d) aov(y ~ trt, data = d))
# Apply the summary function to each groupwise ANOVA:
llply(mlist, summary)
# Equivalent approach:
# splitting the data by gps is equivalent to use of the grouping
variable in dlply()
mlist2 <- lapply(split(testd, testd$gps), function(d) aov(y ~ trt, data = d))
llply(mlist2, summary)
Is that what you had in mind?
Dennis
On Thu, Sep 29, 2011 at 4:37 PM, kelseyann <kreme...@mail.gvsu.edu> wrote:
> Hello, I am using the following script to run an anova for numerous species
> in a table that I have:
>
>
> SiteSpp <-
> c("ADHALP","ADLCON","ADLARC","BDALAT","BDPARC","BDLCON","BDLARC","AWCAQU","AWERUS","AWEANG","AWDPSI","BWCSTA","BWHPAU","BWETRI","BWERUS","BWDFIS","BWPARC","BWLCON","BWLARC","BWJBIG")
>
> n.SiteSpp <- length(SiteSpp)
>
> for (i in (1:n.SiteSpp)) {
> QInfl.aov <- aov(AvgOfnumResponse ~ strTrea*strYear,
> data=QInflAvgbyPlot, subset=(SiteSpp == SiteSpp[i]))
> print(summary(QInfl.aov))
> }
>
>
> The resulting summary just continuously prints the same anova results (For
> ADHALP) over and over again.... Any suggestions?
>
> Thanks,
> Kelsey
>
> --
> View this message in context:
> http://r.789695.n4.nabble.com/For-loop-for-subset-repeating-same-over-and-over-tp3858115p3858115.html
> Sent from the R help mailing list archive at Nabble.com.
>
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>
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