R: outlier cleaning for each column in a dataframe by using quantiles 0.05 and 0.95

Question

I am a R-novice. I want to do some outlier cleaning and over-all-scaling from 0 to 1 before putting the sample into a random forest.

g<-c(1000,60,50,60,50,40,50,60,70,60,40,70,50,60,50,70,10)

If i do a simple scaling from 0 - 1 the result would be:

> round((g - min(g))/abs(max(g) - min(g)),1)

 [1] 1.0 0.1 0.0 0.1 0.0 0.0 0.0 0.1 0.1 0.1 0.0 0.1 0.0 0.1 0.0 0.1 0.0

So my idea is to replace the values of each column that are greater than the 0.95-quantile with the next value smaller than the 0.95-quantile - and the same for the 0.05-quantile.

So the pre-scaled result would be:

g<-c(**70**,60,50,60,50,40,50,60,70,60,40,70,50,60,50,70,**40**)

and scaled:

> round((g - min(g))/abs(max(g) - min(g)),1)

 [1] 1.0 0.7 0.3 0.7 0.3 0.0 0.3 0.7 1.0 0.7 0.0 1.0 0.3 0.7 0.3 1.0 0.0

I need this formula for a whole dataframe, so the functional implementation within R should be something like:

> apply(c, 2, function(x) x[x`<quantile(x, 0.95)]`<-max(x[x, ... max without the quantile(x, 0.95))

Can anyone help?

Spoken beside: if there exists a function that does this job directly, please let me know. I already checked out cut and cut2. cut fails because of not-unique breaks; cut2 would work, but only gives back string values or the mean value, and I need a numeric vector from 0 - 1.

for trial:

a<-c(100,6,5,6,5,4,5,6,7,6,4,7,5,6,5,7,1)

b<-c(1000,60,50,60,50,40,50,60,70,60,40,70,50,60,50,70,10)

c<-cbind(a,b)

c<-as.data.frame(c)

Regards and thanks for help,

Rainer

Answer 1

Please don't do this. This is not a good strategy for dealing with outliers - particularly since it's unlikely that 10% of your data are outliers!

Answer 2

I can't think of a function in R that does this, but you can define a small one yourself:

foo <- function(x)
{
    quant <- quantile(x,c(0.05,0.95))
    x[x < quant[1]] <- min(x[x >= quant[1]])
    x[x > quant[2]] <- max(x[x <= quant[2]])
    return(round((x - min(x))/abs(max(x) - min(x)),1))
}

Then sapply this to each variable in your dataframe:

sapply(c,foo)
       a   b
 [1,] 1.0 1.0
 [2,] 0.7 0.7
 [3,] 0.3 0.3
 [4,] 0.7 0.7
 [5,] 0.3 0.3
 [6,] 0.0 0.0
 [7,] 0.3 0.3
 [8,] 0.7 0.7
 [9,] 1.0 1.0
[10,] 0.7 0.7
[11,] 0.0 0.0
[12,] 1.0 1.0
[13,] 0.3 0.3
[14,] 0.7 0.7
[15,] 0.3 0.3
[16,] 1.0 1.0
[17,] 0.0 0.0

Edit: This answer was meant to solve the programming problem. In regard to actually using it I fully agree with Hadley