[R] Breakdown a number [Broadcast]

[Liaw, Andy]

I meant to send this to R-help, but it went to Gabor instead...

This is my shot at it using diff(), assuming breaks is in non-decreasing
order:

chop <- function(x, breaks) {
    s <- sign(x - breaks)
    if (x <= breaks[1]) {
        ans <- x
    } else {
        numPieces <- sum(s == 1)
        ans <- diff(c(0, breaks[1:numPieces], x))
    }
    ans
}

Some tests:

> chop(1200, c(250, 800))
[1] 250 550 400
> chop(800, c(250, 800))
[1] 250 550
> chop(100, c(250, 800))
[1] 100
> chop(600, c(250, 800))
[1] 250 350

Andy

From: Paul Roebuck
> 
> On Wed, 19 Apr 2006, Gabor Grothendieck wrote:
> 
> > On 4/19/06, Paul Roebuck <roebuck at mdanderson.org> wrote:
> >
> > > Isn't there a builtin method for doing this and, if so, 
> what is it 
> > > called?
> > >
> > >        breakdown <- function(whole) {
> > >            breaks <- c(250, 800)
> > >            pieces <- integer(length(breaks) + 1)
> > >            if (whole > breaks[2]) {
> > >                pieces[3] <- whole - breaks[2]
> > >                whole <- breaks[2]
> > >            }
> > >            if (whole > breaks[1]) {
> > >                pieces[2] <- whole - breaks[1]
> > >                whole <- breaks[1]
> > >            }
> > >            pieces[1] <- whole
> > >
> > >            return(pieces)
> > >        }
> > >
> > > breakdown(1200) # 250 550 400
> >
> > Maybe you could discuss in words what you want but perhaps you are 
> > looking for diff:
> 
> That was rather my problem searching for my answer as I
> was unsure what this was called. I was searching for
> 'bins' or 'breaks' (like in hist method but not for
> plotting) and leading nowhere fast. Alas, I thought
> sample code would go further than my description.
> 
> 
> > > bp <- c(0, 250, 800, 1200)
> > > diff(bp)
> > [1] 250 550 400
> 
> Don't think diff method is going to work either, at
> least in cases where the 'whole' is less than greatest
> 'break'.
> 
> > breakdown2 <- function(x, breaks = c(250, 800)) {
>       diff(c(0, breaks, x))
>   }
> > breakdown(10)		# 10 0 0
> > breakdown2(10)	# 250 550 -790
> >
> > breakdown(400)	# 250 150 0
> > breakdown2(400)	# 250 550 -400
> 
> 
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