[R] Problems with computing an aggregated score
Hi all, I have a problem with computing a new variable as an aggregated score of other variables. Say, I have 10 variables for 1,000 observations (people). Every variable may have values between 0 and 8. What I would like to do is computing the mean of the individual top 3 values for every person. Exampe: The values for the 10 variables (v1 to v10) for person A, B and C are as follows: A 0 1 0 2 5 8 3 0 4 0 B 6 4 3 0 0 0 0 5 0 0 C 0 0 8 0 0 8 0 0 0 0 So, I would like to compute a new variable for the mean of the individual top 3 values, which would result for person A, B and C in the following: A (8+5+4)/3 = 5.67 B (6+5+4)/3 = 5 C (8+8+0)/3 = 5.33 Is there any way to do this? Any clues, hints and suggestions are highly appreciated, many thanks in advance Johannes -- View this message in context: http://www.nabble.com/Problems-with-computing-an-aggregated-score-tp24495390p24495390.html Sent from the R help mailing list archive at Nabble.com. __ 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.
Re: [R] Problems with computing an aggregated score
On Wed, Jul 15, 2009 at 3:37 AM, Hatschjohannes.h...@googlemail.com wrote: Hi all, I have a problem with computing a new variable as an aggregated score of other variables. Say, I have 10 variables for 1,000 observations (people). Every variable may have values between 0 and 8. What I would like to do is computing the mean of the individual top 3 values for every person. Exampe: The values for the 10 variables (v1 to v10) for person A, B and C are as follows: A 0 1 0 2 5 8 3 0 4 0 B 6 4 3 0 0 0 0 5 0 0 C 0 0 8 0 0 8 0 0 0 0 So, I would like to compute a new variable for the mean of the individual top 3 values, which would result for person A, B and C in the following: A (8+5+4)/3 = 5.67 B (6+5+4)/3 = 5 C (8+8+0)/3 = 5.33 Is there any way to do this? Any clues, hints and suggestions are highly appreciated, many thanks in advance Johannes Unquestionably the experienced guys are going to give you a better answer, but as a newbie I worked out a step by step answer in a few minutes. This allows you to see the steps: A - c(0, 1, 0, 2, 5, 8, 3, 0, 4, 0) A[order(as.numeric(A), decreasing=TRUE)] A[order(as.numeric(A), decreasing=TRUE)][1:3] sum(A[order(as.numeric(A), decreasing=TRUE)][1:3]) sum(A[order(as.numeric(A), decreasing=TRUE)][1:3])/3 Hope this helps, Mark __ 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.
Re: [R] Problems with computing an aggregated score
Dear Johannes,
If 'x' is your data set, here is one way using apply:
res - apply(x[,-1], 1, function(x){
xo - sort(x, decreasing = TRUE)
mean(xo[1:3])
}
)
names(res) - x[,1]
res
# ABC
# 5.67 5.00 5.33
See ?apply, ?sort and ?mean for more information.
HTH,
Jorge
On Wed, Jul 15, 2009 at 6:37 AM, Hatsch johannes.h...@googlemail.comwrote:
Hi all,
I have a problem with computing a new variable as an aggregated score of
other variables.
Say, I have 10 variables for 1,000 observations (people). Every variable
may
have values between 0 and 8. What I would like to do is computing the mean
of the individual top 3 values for every person.
Exampe: The values for the 10 variables (v1 to v10) for person A, B and C
are as follows:
A 0 1 0 2 5 8 3 0 4 0
B 6 4 3 0 0 0 0 5 0 0
C 0 0 8 0 0 8 0 0 0 0
So, I would like to compute a new variable for the mean of the individual
top 3 values, which would result for person A, B and C in the following:
A (8+5+4)/3 = 5.67
B (6+5+4)/3 = 5
C (8+8+0)/3 = 5.33
Is there any way to do this?
Any clues, hints and suggestions are highly appreciated,
many thanks in advance
Johannes
--
View this message in context:
http://www.nabble.com/Problems-with-computing-an-aggregated-score-tp24495390p24495390.html
Sent from the R help mailing list archive at Nabble.com.
__
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http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
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__
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