Re: [R] Using weighted.mean() in aggregate()

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From: Spencer Graves (spencer.graves@pdf.com)
Date: Sun 22 Jun 2003 - 11:02:01 EST


Message-id: <3EF50009.4030704@pdf.com>


> tstdf <- data.frame(Sub =rep(1:2, 2),
+ Length=1:4, Slope=11:14)
> by(tstdf, tstdf$Sub,
+ function(x)weighted.mean(x$Slope, x$Length))
tstdf$Sub: 1
[1] 12.5
------------------------------------------------------------
tstdf$Sub: 2
[1] 13.33333
>
Does this answer your question?

hth. spencer graves

Aleksey Naumov wrote:
> Dear R users, I have a question on using weighted.mean() while aggregating a
> data frame. I have a data frame with columns Sub, Length and Slope:
>
>
>>x[1:5,]
>
> Sub Length Slope
> 1 2 351.547 0.0025284969
> 2 2 343.738 0.0025859390
> 3 1 696.659 0.0015948968
> 4 2 5442.338 0.0026132544
> 5 1 209.483 0.0005304225
>
> and I would like to calculate the weighted.mean of Slope, using Length as
> weights, for each value of Sub. The obvious way:
>
>
>>aggregate(list(Mean.Slope=x$Slope), by=list(Sub=x$Sub), FUN=weighted.mean,
>
> w=x$Length)
>
> does not work. weighted.mean() generates warnings that "longer object length
> is not a multiple of shorter object length in: x * w", from which I conclude
> that weights are not supplied as I intend, instead each subset of Sub, when
> passed to weighted.mean(), receives the whole x$Length as weights, which is
> not correct.
>
> Is there an elegant way to do this, or do I have to have a loop here?
>
> Thank you,
> Aleksey
>

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