[R] Expressions in lattice conditional variables
Peter Ehlers
ehlers at ucalgary.ca
Sat Mar 2 21:29:40 CET 2013
Frank,
As you probably realize, the problem is not with lattice but with
unique (actually: .Internal(unique, ...)):
unique(expression('a','b','b'))
generates your error.
Lattice needs a factor variable for the panels, and it tries to
use unique() on your 'vn' to do that.
For example, if you replace '|vn' with '|as.character(vn)', you'll
get the plot, but of course the strip labels are a mess.
I'm not sure that I understand your need correctly, but if it's
a matter of creating the strip labels, then I would explore using
the factor.levels argument to strip.default().
I hope that this isn't totally out to lunch.
Peter Ehlers
On 2013-03-02 09:20, Frank Harrell wrote:
> Whoops - these 2 lines should have been omitted from the program:
>
> n <- sprintf('%s (n%s=%g, n%s=%g)', v, nam[1],n[1], nam[2],n[2])
> vn[var == v] <- n
>
>
> Frank Harrell wrote
>> I would like to have a lattice conditioning ( | var ) variable have
>> expression() as values because I want panel labels to be able to use
>> plotmath notation for subscripts, etc. lattice barks at this. Does
>> anyone know of a trick workaround? An attempted example program is below.
>> Thanks -Frank
>>
>> require(lattice)
>> set.seed(1)
>> var <- c(rep('A', 100), rep('B', 100))
>> trt <- sample(c('T1','T2'), 200, TRUE)
>> x <- c(runif(100), 10*runif(100))
>> y <- x + c(runif(100)/10, runif(100))
>> N <- tapply(x, llist(var, trt), function(x) sum(!is.na(x)))
>> print(N)
>>
>> vn <- vector('expression', length(var))
>> for(v in unique(var)) {
>> i <- var == v
>> n <- tapply(!is.na(x[i]), trt[i], sum)
>> nam <- names(n)
>> w <- sprintf('paste(%s," (", n[%s]==%g,~~n[%s]==%g,")")',
>> v, nam[1], n[1], nam[2], n[2])
>> cat(w, '\n')
>> vn[var == v] <- parse(text=w)
>> n <- sprintf('%s (n%s=%g, n%s=%g)', v, nam[1],n[1], nam[2],n[2])
>> vn[var == v] <- n
>> }
>> trt <- factor(trt)
>>
>> xyplot(as.integer(trt) ~ x | vn, panel=panel.bpplot, ylim=c(0,3),
>> scale=list(y=list(at=1:2, labels=levels(trt)),
>> x=list(relation='free', limits=list(c(0,1),c(0,13)))),
>> ylab='Treatment', layout=c(1,2))
>>
>> Error in unique.default(x) :
>> unimplemented type 'expression' in 'HashTableSetup'
>
>
>
>
>
> -----
> Frank Harrell
> Department of Biostatistics, Vanderbilt University
> --
> View this message in context: http://r.789695.n4.nabble.com/Expressions-in-lattice-conditional-variables-tp4660089p4660090.html
> Sent from the R help mailing list archive at Nabble.com.
>
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