hairy indexing problem - R-help

Bill Venables · 2002-06-05T05:03:06Z

> -----Original Message----- > From: Ian.Saunders at csiro.au [mailto:Ian.Saunders at csiro.au] > Sent: Wednesday, June 05, 2002 2:32 PM > To: seniorr at aracnet.com; R-help at stat.math.ethz.ch > Subject: RE: [R] hairy indexing problem > > There's probably a better way, but ... > > apply(outer(subject,subject,FUN="=="),1,sum) > > will give you a vector of the counts for each value of subject, so would > be > 2 2 4 4 4 4 2 2 ... > in your example. [WNV] I think there could be a better

Bill Venables

Tue, Jun 4, 2002 10:03 PM #

[WNV]  I think there could be a better way.  

	Make sure subject is a factor:

		subject <- as.factor(data$subject)

	and then your "replications class" factor is

		reps <- factor(table(subject)[subject])

	The next step could be 

		fooMeans <- tapply(data$foo, reps, mean)

[WNV]  Yep, that too.   gsummary is part of the nlme package which
has to be loaded.

[WNV]  Bill.

-----Original Message-----
From: Russell Senior [mailto:seniorr at aracnet.com]
Sent: Wednesday, 5 June 2002 10:31 AM
To: R-help at stat.math.ethz.ch
Subject: [R] hairy indexing problem

I've got a data frame that looks like this:

   subject   foo   bar
      2      1.7   3.2
      2      2.3   4.1
      3      7.6   2.3
      3      7.1   3.3
      3      7.3   2.3
      3      7.4   1.3
      5      6.2   6.1
      5      3.4   6.9
     ...

That is, I've got multiple rows per subject.  I need to compute
summaries within categories where the subject has the same number of
rows.  For example, subject 2 and 5 both have two rows.  I need to
compute mean for those four values of foo.  This looks like a good
candidate for index vectors, but I need some help.  I've tried
something like:

  table(data) -> tmp

and:

  tmp[tmp == 2]

and even:

  as.numeric(attr(tmp[tmp == 2],"names"))

to get a vector of subject numbers that have two rows in the original
data frame.  But I am getting stuck there.  I want some kind of
"is.member" function to use in a subsequent index vector expression,
like:

  i <- as.numeric(attr(tmp[tmp == 2],"names"))
  data[is.member($subject,i)]$foo

but there isn't an is.member() function.  Can someone please give me a
pointer on the canonical way to do this?

Thanks!

-- 
Russell Senior         ``The two chiefs turned to each other.        
seniorr at aracnet.com      Bellison uncorked a flood of horrible       
                         profanity, which, translated meant, `This is
                         extremely unusual.' ''                      
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