[R] hairy indexing problem
Ian.Saunders@csiro.au
Ian.Saunders at csiro.au
Wed Jun 5 06:31:57 CEST 2002
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.
You could add this as a column of the data frame and use gsummary to get the
summary statistics.
Ian.
> -----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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