Hi,
You can also use
library(plyr)
ddply(dat,.(V1,V2),summarise,V3=mean(V3),.drop=FALSE)
# V1 V2 V3
#1 C 0 0.5000000
#2 C 1 NaN
#3 G 0 1.0000000
#4 G 1 NaN
#5 I 0 0.3333333
#6 I 1 0.4285714
#7 O 0 1.0000000
#8 O 1 0.0000000
#9 R 0 0.0000000
#10 R 1 0.6666667
#11 T 0 0.8333333
#12 T 1 0.5000000
A.K.
----- Original Message -----
From: Thomas Stewart <tgs.public.m...@gmail.com>
To:
Cc: r-help <r-help@r-project.org>
Sent: Thursday, December 6, 2012 3:17 PM
Subject: Re: [R] Can somebody help me with following data manipulation?
You can directly use the tapply function.
-tgs
tapply(dat[,3],dat[,-3],mean)
On Thu, Dec 6, 2012 at 3:03 PM, Sarah Goslee <sarah.gos...@gmail.com> wrote:
> If I understand what you want correctly, aggregate() should do it.
>
> > aggregate(V3 ~ V1 + V2, "mean", data=dat)
> V1 V2 V3
> 1 C 0 0.5000000
> 2 G 0 1.0000000
> 3 I 0 0.3333333
> 4 O 0 1.0000000
> 5 R 0 0.0000000
> 6 T 0 0.8333333
> 7 I 1 0.4285714
> 8 O 1 0.0000000
> 9 R 1 0.6666667
> 10 T 1 0.5000000
>
> That returns the combinations that actually exist.
>
> If you convert V1 and V2 to factors, thus setting the possible levels,
> all combinations will be returned:
> > dat$V1 <- factor(dat$V1)
> > dat$V2 <- factor(dat$V2)
> > aggregate(V3 ~ V1 + V2, "mean", data=dat)
> V1 V2 V3
> 1 C 0 0.5000000
> 2 G 0 1.0000000
> 3 I 0 0.3333333
> 4 O 0 1.0000000
> 5 R 0 0.0000000
> 6 T 0 0.8333333
> 7 I 1 0.4285714
> 8 O 1 0.0000000
> 9 R 1 0.6666667
> 10 T 1 0.5000000
>
> Sarah
>
> On Thu, Dec 6, 2012 at 2:35 PM, Christofer Bogaso
> <bogaso.christo...@gmail.com> wrote:
> > Dear all, let say I have following data:
> >
> > dat <- structure(list(V1 = structure(c(1L, 4L, 5L, 3L, 3L, 5L, 6L, 6L,
> > 4L, 3L, 5L, 6L, 5L, 5L, 4L, 4L, 6L, 2L, 3L, 4L, 3L, 3L, 2L, 5L,
> > 3L, 6L, 3L, 3L, 6L, 3L, 6L, 1L, 6L, 5L, 2L, 2L), .Label = c("C",
> > "G", "I", "O", "R", "T"), class = "factor"), V2 = c(0L, 0L, 0L,
> > 1L, 1L, 1L, 1L, 0L, 1L, 0L, 1L, 0L, 0L, 1L, 1L, 1L, 0L, 0L, 1L,
> > 1L, 1L, 0L, 0L, 0L, 1L, 0L, 0L, 1L, 0L, 1L, 0L, 0L, 1L, 0L, 0L,
> > 0L), V3 = c(1L, 1L, 0L, 1L, 0L, 1L, 0L, 1L, 0L, 0L, 1L, 1L, 0L,
> > 0L, 0L, 0L, 1L, 1L, 1L, 0L, 0L, 1L, 1L, 0L, 0L, 0L, 0L, 1L, 1L,
> > 0L, 1L, 0L, 1L, 0L, 1L, 1L)), .Names = c("V1", "V2", "V3"), class =
> > "data.frame", row.names = c(NA,
> > -36L))
> >
> > Now I want to get following kind of data frame out of that:
> >
> > dat1 <- structure(list(V1 = structure(c(3L, 3L, 1L, 1L, 2L, 2L), .Label =
> > c("C",
> > "G", "I"), class = "factor"), V2 = c(0L, 1L, 0L, 1L, 0L, 1L),
> > V3 = c(0.333333333, 0.428571429, 0.5, NA, 1, NA)), .Names = c("V1",
> > "V2", "V3"), class = "data.frame", row.names = c(NA, -6L))
> >
> > Basically in 'dat1', the 3rd column is coming from: for 'V1 = I' & 'V2 =
> 0'
> > what is the percentage of '1' for "V3" and so on.....
> >
> > Is there any R function to achieve that directly?
> >
> > Thanks and regards,
> >
>
> ______________________________________________
> R-help@r-project.org mailing list
> https://stat.ethz.ch/mailman/listinfo/r-help
> PLEASE do read the posting guide
> http://www.R-project.org/posting-guide.html
> and provide commented, minimal, self-contained, reproducible code.
>
>
>
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PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
______________________________________________
R-help@r-project.org mailing list
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PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
