Percentile for Each Observation w/r/t Grouping Variable

Question

I have some data that looks like the following. It is grouped by variable \"Year\" and I want to extract the percentiles of each observation of Score, with respect to t

Answer 1

I found a method, but it requires a loop.

group.pctiles <- function(group.var, comparable) {
    unique.vals <- unique(group.var)
    pctiles <- vector(length = length(group.var))
    for (i in 1:length(unique.vals)) {
        slice <- which(group.var == unique.vals[i])
        F <- ecdf(comparable[slice])
        group.pctiles <- F(comparable[slice])
        pctiles[slice] <- group.pctiles
    }
    return(pctiles)
}

group.var is the variable that groups the data. In my example in my question, it is Year. comparable contains the values we want to find the percentiles for. In my question, comparable would be Score.

For the following data, I get the result below:

Year,School,Fees
2000,10,1000
2008,1,1050
2008,4,2000
2000,3,1700
2000,1,2000

> group.pctiles(dat, dat$Year, dat$Fees)
[1] 0.3333333 0.5000000 1.0000000 0.6666667 1.0000000

Then, I can cbind these percentiles back into the original data.frame for analysis, reporting, etc.

Anyone have a solution that doesn't require a loop?

Answer 2

Using data.table is pretty straight-forward as well. Just for completeness and also as an easy way to find the data.table solution.

library(data.table)
year <- rep(2001:2005, 2)
score <- round(rnorm(10, 35, 3))

dt <- data.table(score)


dt[, .(Percentile = ecdf(score)(score)), by = list(year)]

Answer 3

Following up on Vince's solution, you can also do this with plyr or by:

ddply(df, .(years), function(x) transform(x, percentile=ecdf(x$scores)(x$scores)))

Answer 4

I may be misunderstanding, but I think it can be done this way:

> years = c(2006, 2006, 2006, 2006, 2001, 2001, 2001, 2001, 2001)
> scores = c(13, 65, 23, 34, 78, 56, 89, 98, 100)
> tapply(scores, years, quantile)
$`2001`
  0%  25%  50%  75% 100% 
  56   78   89   98  100 

$`2006`
   0%   25%   50%   75%  100% 
13.00 20.50 28.50 41.75 65.00 

Is this right?

I mean the actual percentile of each observation. – Ryan Rosario

Edit:

I think this may do it then:

> tapply(scores, years, function(x) { f = ecdf(x); sapply(x, f) })
$`2001`
[1] 0.4 0.2 0.6 0.8 1.0

$`2006`
[1] 0.25 1.00 0.50 0.75

With your data:

> tapply(scores, years, function(x) { f = ecdf(x); sapply(x, f) })
$`2000`
[1] 0.3333333 0.6666667 1.0000000

$`2008`
[1] 0.5 1.0

Edit 2:

This is probably faster:

tapply(scores, years, function(x) { f = ecdf(x); f(x) })

f() is vectorized :-)

Last, modification, I promise :-). If you want names:

> tapply(scores, years, function(x) { f = ecdf(x); r = f(x); names(r) <- x; r })
$`2000`
     1000      1700      2000 
0.3333333 0.6666667 1.0000000 

$`2008`
1500 2000 
 0.5  1.0 

Answer 5

How about something like:

Year <- c(2000,2008,2008,2000,2000)
Fees <- c(1000,1050,2000,1700,2000)
dat <- data.frame(Fees,Year,result=NA)
res <- tapply(Fees,Year,function(x) rank(x,ties.method="max")/length(x))
for(i in 1:length(res))
   dat[Year==as.numeric(names(res)[i]),"result"] <-res[[i]]

which yields:

  Fees Year    result
1 1000 2000 0.3333333
2 1050 2008 0.5000000
3 2000 2008 1.0000000
4 1700 2000 0.6666667
5 2000 2000 1.0000000

Answer 6

You can also do something like this:

# first I'll create two dummy variables (Year, Score)
year <- rep(2001:2005, 2)
score <- round(rnorm(10, 35, 3))

# then coerce variables to data frame
d <- data.frame(year, score)

# then you can use split() function to apply
# function to each stratum of grouping variable
sapply(split(score, year), function(x) quantile(x, probs=seq(.1, .9, .1)))

Output will go something like this:

     2001 2002 2003 2004 2005
10%  34.3 32.1 34.3 29.6 36.1
20%  34.6 32.2 34.6 30.2 36.2
30%  34.9 32.3 34.9 30.8 36.3
40%  35.2 32.4 35.2 31.4 36.4
50%  35.5 32.5 35.5 32.0 36.5
60%  35.8 32.6 35.8 32.6 36.6
70%  36.1 32.7 36.1 33.2 36.7
80%  36.4 32.8 36.4 33.8 36.8
90%  36.7 32.9 36.7 34.4 36.9

You can utilize t() function to transpose rows and columns if you prefer. Writing a function will be a good way to tackle this kind of problems. I strongly recommend plyr package written by Hadley Wickam.

Hope this helps! All the best!