# Re: [R] a more elegant way to get percentages?

From: Gabor Grothendieck <ggrothendieck_at_gmail.com>
Date: Thu, 13 Mar 2008 09:45:05 -0400

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 <pisicandru_at_hotmail.com> 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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