what about using "mapply"?
splitted.value<-with(x.1, split(VALUE, GROUP))
splitted.freq<-with(x.1, split(FREQUENCY, GROUP))
mapply(weighted.mean, splitted.value, w=splitted.freq)
Stefano
>-----Messaggio originale-----
>Da: r-help-bounces at stat.math.ethz.ch
>[mailto:r-help-bounces at stat.math.ethz.ch]Per conto di
>james.holtman at convergys.com
>Inviato: mercoled?? 25 maggio 2005 17.57
>A: Dan Bolser
>Cc: R mailing list; r-help-bounces at stat.math.ethz.ch
>Oggetto: Re: [R] weighted.mean and tapply (again)
>
>
>
>
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>
>> x.1 <- read.table('clipboard',header=T)
>> x.1
> GROUP VALUE FREQUENCY
>1 2 2 78
>2 2 3 40
>3 2 4 16
>4 2 5 3
>5 2 6 1
>6 2 8 1
>7 3 3 19
>8 3 4 10
>9 3 5 19
>10 3 6 4
>> by(x.1, x.1$GROUP, function(x) weighted.mean(x$VALUE,
>x$FREQUENCY))
>x.1$GROUP: 2
>[1] 2.654676
>------------------------------------------------------------
>---------------
>x.1$GROUP: 3
>[1] 4.153846
>>
>
>Jim
>__________________________________________________________
>James Holtman "What is the problem you are trying to
solve?"
>Executive Technical Consultant -- Office of Technology, Convergys
>james.holtman at convergys.com
>+1 (513) 723-2929
>
>
>
>
>
> Dan Bolser
>
>
> <dmb at mrc-dunn.cam.ac. To:
> R mailing list <r-help at stat.math.ethz.ch>
>
> uk> cc:
>
>
> Sent by: Subject:
> [R] weighted.mean and tapply (again)
>
> r-help-bounces at stat.m
>
>
> ath.ethz.ch
>
>
>
>
>
>
>
>
> 05/25/2005 11:33
>
>
>
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>
>
>
>
>I read answers to questions including the words "tapply" and
>"weighted.mean", but I didn't understand either the problem
>(data) or the
>solution provided.
>
>Here is my question ...
>
>> dat[1:10,]
> GROUP VALUE FREQUENCY
>1 2 2 78
>2 2 3 40
>3 2 4 16
>4 2 5 3
>5 2 6 1
>6 2 8 1
>7 3 3 19
>8 3 4 10
>9 3 5 19
>1 3 6 4
>
>
>For each GROUP, I would like to calculate the weighted.mean
>of VALUE using
>the FREQUENCY as the weight, so for the snippet of data
>shown that would
>be...
>
>group.2 <- weighted.mean(c(2,3,4,5,6,8),c(78,40,16,3,1,1))
>group.3 <- weighted.mean(c(3,4,5,6), c(19,10,19,4))
>
>> cbind(rbind(2,3),rbind(group.2,group.3))
> [,1] [,2]
>group.2 2 2.654676
>group.3 3 4.153846
>
>I would like to use tapply to automatically do this across the whole
>dataset (dat) - which includes lots of other distinct
>grouping factors,
>however, like I said, I couldn't understand (and therefore
>apply to my
>data) any of the other solutions I found, so any help here would be
>greatly appreciated!
>
>All the best,
>Dan.
>
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