Assuming your x is as follows:
x <- data.frame(locat = c("a", "b", "b", "c", "c", "c", "c", "d", "d", "d"),
val = c(5, 5, 15, 5, 20, 5, 10, 5, 15, 10))
Try this:
x$percent1 <- ave(x$val, x$locat, FUN = function(x) 100*x/sum(x))
On Thu, Mar 13, 2008 at 9:36 AM, Monica Pisica <[EMAIL PROTECTED]> wrote:
>
> Hi,
>
> I am trying to get percentages in a more elegant way. I have a data.frame
> with locations and values (counts) of species at that location. Each location
> is repeated for each species i have values for and i would like to get
> percentages of each species at that location. I am not sure if i am clear in
> my explanations so i will paste my code below:
>
> #####################
>
> > x
> locat val
> 1 a 5
> 2 b 5
> 3 b 15
> 4 c 5
> 5 c 20
> 6 c 5
> 7 c 10
> 8 d 5
> 9 d 15
> 10 d 10
> > loc1 <- x$locat
> > n <- length(loc1)
> > locuniq1 <- unique(loc1)
> > m <- length(locuniq1)
> > counts <- seq(1:m)
> >
> > for (i in 1:m) {
> + count <- 0
> + for (j in 1:n) {
> + if (loc1[j]==locuniq1[i]) count <- count+1
> + counts[i] <- count
> + }
> + }
> >
> > percent1 <- rep(0,n)
> > j <- 0
> > for (i in 1:m) {
> +
> + b <- x[(j+1):(j+counts[i]),]
> + total <- sum(b$val)
> + percent1[(j+1):(j+counts[i])] <- round(apply(as.matrix(b$val), 1,
> function(x) {x*100/total}),2)
> + j = j+counts[i]
> + }
> > x1 <- cbind(x, percent1) # this is the result i want
> > x1
> locat val percent1
> 1 a 5 100.00
> 2 b 5 25.00
> 3 b 15 75.00
> 4 c 5 12.50
> 5 c 20 50.00
> 6 c 5 12.50
> 7 c 10 25.00
> 8 d 5 16.67
> 9 d 15 50.00
> 10 d 10 33.33
> >
> ################
>
> I am wondering if there is any way to do it more efficiently, much more that
> the first loop which gives how many times each location is present in the
> data.frame is slow if you have a larger data.frame and not only 10 rows.
>
> Thanks for any input and sorry if the email is on the long side,
>
> Monica
>
>
> _________________________________________________________________
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>
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