Displaying 1 result from an estimated 1 matches for "locuniq1".
2008 Mar 13
4
a more elegant way to get percentages?
...t 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...