splitting a continuous variable into groups of equal number of elements - return numeric vector from bin values

Question

I have a continuous variable that I want to split into bins, returning a numeric vector (of length equal to my original vector) whose values relate to the values of the bins. E

Answer 1

Maybe not much elegant, but should be efficient. Try this function:

myCut<-function(x,breaks,retValues=c("means","highs","lows")) {
    retValues<-match.arg(retValues)
    if (length(breaks)!=1) stop("breaks must be a single number")
    breaks<-as.integer(breaks)
    if (is.na(breaks)||breaks<2) stop("breaks must greater than or equal to 2") 
    intervals<-seq(min(x),max(x),length.out=breaks+1)
    bins<-findInterval(x,intervals,all.inside=TRUE)
    if (retValues=="means") return(rowMeans(cbind(intervals[-(breaks+1)],intervals[-1]))[bins])
    if (retValues=="highs") return(intervals[-1][bins]) 
    intervals[-(breaks+1)][bins]
}
x = c(1,5,3,12,5,6,7)
myCut(x,3)
#[1]  2.833333  6.500000  2.833333 10.166667  6.500000  6.500000  6.500000
myCut(x,3,"highs")
#[1]  4.666667  8.333333  4.666667 12.000000  8.333333  8.333333  8.333333
myCut(x,3,"lows")
#[1] 1.000000 4.666667 1.000000 8.333333 4.666667 4.666667 4.666667

Answer 2

Use ave like this:

Given:

x = c(1,5,3,12,5,6,7)

Mean:

ave(x,cut2(x,g = 3), FUN = mean)
[1] 3.5 3.5 3.5 9.5 3.5 6.0 9.5

Min:

ave(x,cut2(x,g = 3), FUN = min)
[1] 1 1 1 7 1 6 7

Max:

ave(x,cut2(x,g = 3), FUN = max)
[1]  5  5  5 12  5  6 12

Or standard deviation:

ave(x,cut2(x,g = 3), FUN = sd)
[1] 1.914854 1.914854 1.914854 3.535534 1.914854       NA 3.535534

Note the NA result for only one data point in interval.

Hope this is what you need.

NOTE:
Parameter g in cut2 is number of quantile groups. Groups might not have the same amount of data points, and the intervals might not have the same length.
On the other hand, cut splits the interval into several of equal length.