Hi Arun,

Thank you so much for your help.  In the end, I wound up using:

sp.data$spores<-as.numeric(as.character(sp.data$spores))

The stats guy on campus helped me :)

Be well,
Melissa
________________________________________
From: arun kirshna [via R] [ml-node+s789695n4646457h12@n4.nabble.com]
Sent: Wednesday, October 17, 2012 6:51 AM
To: Moore, Melissa
Subject: Re: In mean.default(X[[50L]], ...) :   argument is not numeric or logical: returning NA

HI,
Try this:
set.seed(1)
sp.data<-data.frame(iso=factor(rep(1:4,each=4)),spec=factor(sample(1:5,16,replace=TRUE)),temp=factor(rep(c(28,30,32,34),each=4)),site=factor(sample(1:6,16,replace=TRUE)),spores=factor(sample(1:50,16,replace=FALSE)))
ave.sp <- aggregate(sp.data$spores,
by=list(iso=sp.data$iso, spec=sp.data$spec, temp=sp.data$temp,
site=sp.data$site), FUN=mean, na.rm=T)

#  In mean.default(X[[15L]], ...) :
  #argument is not numeric or logical: returning NA

 sp.data1<-within(sp.data,{spores<-as.numeric(as.character(spores))})
 str(sp.data1)
#'data.frame': 16 obs. of  5 variables:
# $ iso   : Factor w/ 4 levels "1","2","3","4": 1 1 1 1 2 2 2 2 3 3 ...
# $ spec  : Factor w/ 5 levels "1","2","3","4",..: 2 2 3 5 2 5 5 4 4 1 ...
# $ temp  : Factor w/ 4 levels "28","30","32",..: 1 1 1 1 2 2 2 2 3 3 ...
# $ site  : Factor w/ 6 levels "1","2","3","4",..: 5 6 3 5 6 2 4 1 2 3 ...
# $ spores: num  25 10 40 32 37 5 47 18 35 27 ...

ave.sp1 <- aggregate(sp.data1$spores,
by=list(iso=sp.data1$iso, spec=sp.data1$spec, temp=sp.data1$temp,
site=sp.data1$site), FUN=mean, na.rm=T)
 head(ave.sp1)
# iso spec temp site    x
#1   2    4   30    1 18.0
#2   3    2   32    1 44.0
#3   2    5   30    2  5.0
#4   3    4   32    2 35.0
#5   1    3   28    3 40.0
#6   3    1   32    3 24.5
A.K.

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