[R] Whiskers on the default boxplot {graphics}

[David Winsemius]

I agree. I was convinced by Ehlers' example that type =2 was a better  
match to fivenum's result

-- 
David..


On May 13, 2010, at 1:36 PM, Joshua Wiley wrote:

> On Thu, May 13, 2010 at 7:55 AM, David Winsemius <dwinsemius at comcast.net 
> > wrote:
>> Yes, and experimentation leads me to the conclusion that the only  
>> possible
>> candidate for matching up the results of fivenum[c(2,4]  with  
>> quantile(y,
>> c(1,3)/4, type=i) is for type=5. I'm not able to prove that to  
>> myself from
>> mathematical arguments. since I do not quite understand the  
>> formalism in the
>> quantile page. If the match is not exact, this would be a tenth  
>> definition
>> of IQR.
>
> David,
>
> Here is some sample data, and the most parsimonious code I could come
> up with for how quantile() computes the quartiles when using type=5.
> The code for fivenum() seems simple enough, but I am not quite able to
> make enough sense of the code for type=5 from quantile() to say
> confidently why they are different.
>
> I am open to the possibility that my attempts to extract relevant code
> from quantile were flawed, but my tentative conclusion is that
> quantile(x, type=5) != fivenum(x).
>
> ##########################
> x <- c(0.643796386452606, -0.605277531056206, -0.339239367816402,
> 1.12408365699422, 0.615753476531243, -1.10545696568758,
> 0.666533406841698, 1.42794492209271, 0.624752921945051,
> 2.02317205214712, -0.365586657432646, 0.821742701084307,
> -0.874753498321076, -0.0298783402061118, 1.18037670706428,
> -0.178274986836195, 0.308703365439049, 0.619700844646392,
> 0.54977981430092, -1.82161514610448, -1.28413556650749,
> -0.0443852992196351, 0.704196760556652, -1.88596816676741,
> -0.420811351737096)
> oldx <- x #this is just a backup because x will be transformed
>
> ##Start from quantile()
> probs <- c(0, 0.25, 0.5, 0.75, 1)
> type <- 5
> n <- length(x)
> switch(type - 3, {
>  a <- 0
>  b <- 1
> }, a <- b <- 0.5, a <- b <- 0, a <- b <- 1, a <- b <- 1/3, a <- b <-  
> 3/8)
> fuzz <- 4 * .Machine$double.eps
> nppm <- a + probs * (n + 1 - a - b)
> j <- floor(nppm + fuzz)
> h <- nppm - j
> h <- ifelse(abs(h) < fuzz, 0, h)
> x <- sort(x, partial = unique(c(1, j[j > 0L & j <= n], (j + 1)[j > 0L
> & j < n], n)))
> x <- c(x[1L], x[1L], x, x[n], x[n])
> qs <- x[j + 2L]
> qs[h == 1] <- x[j + 3L][h == 1]
> other <- (h > 0) && (h < 1)
> if (any(other)) qs[other] <- ((1 - h) * x[j + 2L] + h * x[j + 3L]) 
> [other]
> ##End from quantile
>
> qs # from the calculations above
> quantile(oldx, type=5) #this should match qs
> fivenum(oldx) #the 25% does not match
> ############################################
>
> <everything else snipped>
>
> Josh

David Winsemius, MD
West Hartford, CT