Hi Alain,
here is a one-liner for a df without the rowSum column
df<-data.frame(id=c("x01","x02","x03","x04","x05","x06"),a=c(1,2,NA,4,5,6),b=c(2,4,6,8,10,NA),c=c(NA,3,9,12,NA,NA))
(df$wSum<-apply(sweep(df[,-1],1,rowSums(df[,-1],na.rm=T),"/"),1,function(x)sum(x*w,na.rm=T)))
hth.
Am 06.02.2013 14:17, schrieb D. Alain:
> Dear R-List,
>
>
> I am sure there must be a very simple way to do this - I just do not know
> how...
> This is what I want to do:
>
>
> #my dataframe
>
> df<-data.frame(id=c("x01","x02","x03","x04","x05","x06"),a=c(1,2,NA,4,5,6),b=c(2,4,6,8,10,NA),c=c(NA,3,9,12,NA,NA),sum=c(3,9,15,24,15,6))
>
> id a b c sum
> 1 x01 1 2 NA 3
> 2 x02 2 4 3 9
> 3 x03 NA 6 9 15
> 4 x04 4 8 12 24
> 5 x05 5 10 NA 15
> 6 x06 6 NA NA 6
>
>
> #my weights
> w<-c(10.5,9,12.8)
>
> #now I want to calculate the proportion of the rowsum = "sum" of every other
> number in the row, that is df[1,2]/df[1,5], df[1,3]/df[1,5], ...
>
>
> #e.g.
>
>
> df.prop
>
>
> id a b c sum
> 1 x01 0.33 0.66 NA 3
> 2 x02 0.22 0.44 0.33 9
> 3 x03 NA 0.4 0.6 15
> 4 x04 0.16 0.33 0.5 24
> 5 x05 0.33 0.66 NA 15
> 6 x06 1 NA NA 6
>
> #and then calculate a rowsum, were each column is weighed by its associated
> factor in the vector "w", i.e. all entries in column df$a should be weighed
> by the factor 10.5, df$b by 9 and df$c by 12.8 and then summed up to a
> weighed rowsum (wrowsum=10.5*0.33+9*0.66).
>
> Any suggestions how to achieve this in a simple way with dataframes that have
> 9000 rows and 44 columns (so I cannot do this row by row) - sorry, if this is
> to easy a question.
>
> Thank you very much in advance!
>
> Best wishes
>
> Alain
> [[alternative HTML version deleted]]
>
>
>
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