Re: [R] intervals from cut() as numerics?

From: Dimitrios Rizopoulos <Dimitris.Rizopoulos_at_med.kuleuven.be>
Date: Sat 20 May 2006 - 23:44:54 EST

as an alternative, you can have a look inside cut.default and use the part that produces the breaks, i.e.,

breaks <- 10
groups <- cut(x, breaks = breaks)
max.bias <- as.vector(tapply(error, groups, mean))

# from cut.default()
nb <- as.integer(breaks + 1)
dx <- diff(rx <- range(x, na.rm = TRUE)) breaks <- round(seq(rx[1] - dx/1000, rx[2] + dx/1000, len = nb), 2) mat <- cbind(breaks[1:(nb - 1)], breaks[2:nb])

plot(x, error, type = "n")
abline(h = 0, col = "grey")
panel.smooth(x, error)
arrows(mat[, 1], max.bias, mat[, 2], max.bias, length = 0.05, angle = 90, code = 3)

Best,
Dimitris



Dimitris Rizopoulos
Ph.D. Student
Biostatistical Centre
School of Public Health
Catholic University of Leuven

Address: Kapucijnenvoer 35, Leuven, Belgium

Tel: +32/(0)16/336899
Fax: +32/(0)16/337015
Web: http://med.kuleuven.be/biostat/
     http://www.student.kuleuven.be/~m0390867/dimitris.htm


Quoting Gavin Simpson <gavin.simpson@ucl.ac.uk>:

> On Sat, 2006-05-20 at 17:39 +0800, Berwin A Turlach wrote:
> > G'day Gavin,
> >
> > >>>>> "GS" == Gavin Simpson <gavin.simpson@ucl.ac.uk> writes:
> >
> > GS> The problem is getting the range/interval for each group
> from
> > GS> (4,4.3], so I can automate this.
> > Most likely there is an easier way, but this seems to work:
> >
> > ## get the levels of groups:
> > > tmp <- levels(groups)
> > ## remove the opening "(" and closing "]" from the string:
> > > tmp1 <- sapply(tmp, function(x) substr(x, 2, nchar(x)-1))
> > ## split into two character strings:
> > > tmp2 <- strsplit(tmp1, ",")
> > ## turn into results into two numbers:
> > > tmp3 <- lapply(tmp2, as.numeric)
> >
> > ## Of course, we can do everything in one go:
> > > lapply(strsplit(sapply(levels(groups), function(x) substr(x, 2,
> nchar(x)-1)), ","), as.numeric)
>
> Many thanks Berwin. My brain wasn't in character string processing
> mode,
> but your solution works just fine. For the archives then, here is
> the
> full script:
>
> ## example data
> dat <- seq(4, 7, by = 0.05)
> x <- sample(dat, 30)
> y <- sample(dat, 30)
> ## residuals
> error <- x - y
> ## break range of x into 10 groups
> groups <- cut(x, breaks = 10)
> ##calculate bias (mean) per group
> max.bias <- aggregate(error, list(group = groups), mean)$x
> ## turn cut intervals into numeric
> interv <- lapply(strsplit(sapply(levels(groups),
> function(x) substr(x, 2,
> nchar(x)-1)),
> ","),
> as.numeric)
> ## reformat cut intervals as 2 col matrix for easy plotting
> interv <- matrix(unlist(interv), ncol = 2, byrow = TRUE)
> ## plot the residuals vs observed
> plot(x, error, type = "n")
> abline(h = 0, col = "grey")
> panel.smooth(x, error)
> ## add bias indicators per group
> arrows(interv[,1], max.bias, interv[,2], max.bias,
> length = 0.05, angle = 90, code = 3)
>
> All the best,
>
> G
>
> <snip />
> > Cheers,
> >
> > Berwin
> >
> > ========================== Full address
> ============================
> > Berwin A Turlach Tel.: +61 (8) 6488 3338
> (secr)
> > School of Mathematics and Statistics +61 (8) 6488 3383
> (self)
> > The University of Western Australia FAX : +61 (8) 6488 1028
> > 35 Stirling Highway
> > Crawley WA 6009 e-mail: berwin@maths.uwa.edu.au
> > Australia
> http://www.maths.uwa.edu.au/~berwin
> >
> --
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