1. Please do not post in HTML. This is a plain text list and it can get mangled.
2. Thank you for the example data and code. The problem is -- see the
Help for strip.custom() -- that strip.custom() wants a character
vector or expression for the factor.levels argument. But...
> str(dist.names)
Factor w/ 3 levels "normal","t","uniform": 1 1 2 3
So dist.names is a factor, not a character vector. (I presume you know
the difference! If not, you need to do your homework and read an R
tutorial or two). Thus, this works:
histogram(~x | name, data = my.df,
strip = strip.custom(factor.levels=as.character(dist.names)))
Cheers,
Bert
Bert Gunter
"Data is not information. Information is not knowledge. And knowledge
is certainly not wisdom."
-- Clifford Stoll
On Sat, Jul 4, 2015 at 6:53 PM, Naresh Gurbuxani
<naresh_gurbux...@hotmail.com> wrote:
> I would like to use a mapping to name panel strips. Directly using mapping
> table does not work. What is the problem here?
> my.df <- data.frame(x = rnorm(100, mean = 0, sd = 2), name = "A")
> my.df <- rbind(my.df, data.frame(x = rnorm(100, mean = 0, sd = 4), name =
> "B"))
> my.df <- rbind(my.df, data.frame(x = rt(100, 3), name = "C"))
> my.df <- rbind(my.df, data.frame(x = runif(100, min = -3, max = 3), name =
> "D"))
>
>
> library(lattice)
> histogram(~x | name, data = my.df)
> name.dist <- data.frame(name = c("A", "B", "C", "D"), dist = c("normal",
> "normal", "t", "uniform"))
> dist.names <- name.dist$dist
>
> # Below code workshistogram(~x | name, data = my.df, strip =
> strip.custom(factor.levels = c("normal", "normal", "t", "uniform")))
> # Below code does not workhistogram(~x | name, data = my.df, strip =
> strip.custom(factor.levels = dist.names))
> [[alternative HTML version deleted]]
>
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