[R] Rank-based p-value on large dataset

[Adaikalavan Ramasamy]

One solution is to cut() 'x' according to the breaks defined by 'y'.
Using cut with labels=FALSE is really fast. See a simulation below.

However the accuracy depends on the number of ties you have in your
"empirical" distribution. I have tried to simulate with the round()
function below.

# simulate data
y <- round(rnorm(80000),  5) # empirical distribution with some ties
x <- round(rnorm(130000), 5) # observed values with some ties

# current way
system.time({
  p1 <- numeric( length(x) )
  for(i in 1:length(x)){ p1[i] <- sum( x[i] < y )/length(y) }
})
[1] 484.67  25.08 512.04   0.00   0.00

# suggested solution
system.time({
  break.points <- c(-Inf, unique(sort(y)), Inf)
  p2 <- cut( x, breaks=break.points, labels=FALSE )
  p2 <- 1 - p2/length(break.points)
})
[1] 0.27 0.01 0.28 0.00 0.00
 

head( cbind( p1, p2 ) )
           p1         p2
[1,] 0.658375 0.65813482
[2,] 0.144000 0.14533705
[3,] 0.815500 0.81436898
[4,] 0.412500 0.41269640
[5,] 0.553625 0.55334516
[6,] 0.044500 0.04510897
...

cor(p1, p2)
[1] 0.9999987


The difference arises mainly because I had to use a unique breakpoints
in cut(). You may want to investigate further if you are interested.
Please let me know if you find anything good or bad about this
suggestion as I am using it as part of my codes too. Thank you.

Regards, Adai


On Thu, 2005-03-03 at 17:22 -0500, Sean Davis wrote:
> I have a fairly simple problem--I have about 80,000 values (call them 
> y) that I am using as an empirical distribution and I want to find the 
> p-value (never mind the multiple testing issues here, for the time 
> being) of 130,000 points (call them x) from the empirical distribution. 
>   I typically do that (for one-sided test) something like
> 
> loop over i in x
> p.val[i] = sum(y>x[i])/length(y)
> 
> and repeat for all i.  However, length(x) is large here as is 
> length(y), so this process takes quite a long time.  Any suggestions?
> 
> Thanks,
> Sean
> 
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