> Hello R-users
> The help I received from Petr helped me created this solution to my problems.
>
> t1<-with(mydata ,aggregate(mydata$Y,
> list(mydata$time,mydata$treatment, mydata$expREP, mydata$techREP) ,
> median, na.rm=T)) ### find median by factors
>
> colnames(t1)<-c("time","treatment","
:00:53 +0100
From: "Petr Pikal" <[EMAIL PROTECTED]>
Subject: Re: [R] how to "apply" functions to unbalanced data in long
format by factors..cant get "by" or "aggregate" to work
To: "ALAN SMITH" <[EMAIL PROTECTED]>, r-help@s
On 7 Mar 2007 at 17:25, ALAN SMITH wrote:
Date sent: Wed, 7 Mar 2007 17:25:54 -0600
From: "ALAN SMITH" <[EMAIL PROTECTED]>
To: r-help@stat.math.ethz.ch
Subject: [R] how to "apply" functions to unbal
Here is one way of doing it:
> # create the rows for each unique combination
> x.split <- split(seq(nrow(mydata)), list(mydata$time, mydata$treatment,
+ mydata$expREP, mydata$techREP), drop=TRUE)
> # now go through the list of indices and add the median
> mydata$Y50 <- 0 # add the dummy medi
Hello R users,
Problem...I do not understand how to use "aggregate","by", or the
appropriate "apply" to perform a function on data with more than one
factor on unbalanced data...
I have a data frame in the long format that does not contain balanced
data. The ID is a unique identifier correspon