Question

I’ve got a set of numbers from 0 to 1. Given a value X in the set, I’d like to find the range values (high and low) where Y% of the values in the set are within high and low and where X is the mid point.

So let’s say the numbers are evenly distributed. Given X=0.4 and Y=20%, I need a function that will give me:

high = 0.5
low = 0.3

How can I do that in R?

Answer 1

Update: In light of the extra info from the comments, this will do what the OP wants:

foobar <- function(x, mid, y) {
    ## x, input data on range 0,1
    ## mid, midpoint X in OP's Q
    ## y, % of points around mid
    sx <- sort(x)
    want <- sx >= mid
    ## what do you want to do if y% of x is not integer?
    num <- floor(((y/100) * length(x)) / 2)
    high <- if((len <- length(want[want])) == 0) {
        1
    } else {
        if(len < num) {
            tail(sx, 1)
        } else {
            sx[want][num]
        }
    }
    low <- if((len <- length(want[!want])) == 0) {
        0
    } else {
        if(len < num) {
            head(sx, 1)
        } else {
            rev(sx[!want])[num]
        }
    }
    res <- c(low, high)
    names(res) <- c("low","high")
    res
}

Which gives the following on a sample of random values on interval 0,1:

> set.seed(1)
> x <- runif(20)
> sort(x)
 [1] 0.06178627 0.17655675 0.20168193 0.20597457 0.26550866 0.37212390
 [7] 0.38003518 0.38410372 0.49769924 0.57285336 0.62911404 0.66079779
[13] 0.68702285 0.71761851 0.76984142 0.77744522 0.89838968 0.90820779
[19] 0.94467527 0.99190609
> foobar(x, 0.4, 20)
      low      high 
0.3800352 0.5728534

The OP has answered the Qs below and the version of the function above does as was requested and in light of comments.

There are a couple of issues to deal with:

• What do you want to do if y% of the data is not an integer? At the moment, if y% of the data evaluates to say 4.2 I am rounding down to floor(4.2) but we could round up to ceiling(4.2).
• What do you want to do if there are 0 values above or below the chosen mid point? At the moment the code returns the stated extremes (0,1) in those cases.
• What do you want to do if there are some values above/below the mid point but not enough in a given direction to encompass y/2% in any one direction? At the moment I return the extreme points of the data that lie above/below the mid point. This is a little inconsistent with the previous point though, should we return the extremes 0, 1 in this case too?

---

Original: This will give you what you want, assuming the assumptions you state (evenly distributed on range 0,1)

foo <- function(x, y) {
    ## x is the mid-point
    ## y is the % range about x, i.e. y/2 either side of x
    x + (c(-1,1) * (((y/100) / 2) * 1))
}

> foo(0.4, 20)
[1] 0.3 0.5

We could extend the function to allow an arbitrary range with defaults 0, 1:

bar <- function(x, y, min = 0, max = 1) {
    ## x is the mid-point
    ## y is the % range about x, i.e. y/2 either side of x
    ## min, max, the lower and upper bounds on the data
    stopifnot(x >= min & x <= max)
    x + (c(-1,1) * (((y/100) / 2) * (max - min)))
}

> bar(0.4, 20)
[1] 0.3 0.5
> bar(0.6, 20, 0.5, 1)
[1] 0.55 0.65
> bar(0.4, 20, 0.5, 1)
Error: x >= min & x <= max is not TRUE