[R] Confused: Inconsistent result?
Petr PIKAL
petr.pikal at precheza.cz
Mon Feb 20 16:21:50 CET 2012
Hi
>
> This is copy & paste from my session:
>
> > xyz<-as.vector(c(ls(),as.matrix(lapply(ls(),class))))
> > dim(xyz)<-c(length(xyz)/2,2)
> >
> > allobj<-function(){
> + xyz<-as.vector(c(ls(),as.matrix(lapply(ls(),class))));
> + dim(xyz)<-c(length(xyz)/2,2);
> + return(xyz)
> + }
> > xyz
> [,1] [,2]
> [1,] "a" "character"
> [2,] "aa" "character"
> [3,] "abc" "character"
> [4,] "AirPassengers" "character"
> [5,] "allobj" "character"
> [6,] "allObjects" "character"
> [7,] "allObjects2" "character"
> [8,] "arrayFromAPL" "character"
> [9,] "classes" "character"
> [10,] "myCharVector" "character"
> [11,] "myDateVector" "character"
> [12,] "myNumericVector" "character"
> [13,] "newArrayFromAPL" "character"
> [14,] "obj" "character"
> [15,] "objClass" "character"
> [16,] "x" "character"
> [17,] "xyz" "character"
> [18,] "y" "character"
> > allobj()
> [,1] [,2]
> >
>
> As far as I can see, the function allobj has the same expressions as
those
> executed from the command line. Why are the results different?
Probably due to environment handling.
Do you really want to check if ls behaves as is intended and that it
produces character vector? Or your intention is a little bit more
ambitious and you want to know what objects do you have?
If the later, I recommend to use this function:
function (pos = 1, pattern, order.by)
{
napply <- function(names, fn) sapply(names, function(x) fn(get(x,
pos = pos)))
names <- ls(pos = pos, pattern = pattern)
obj.class <- napply(names, function(x) as.character(class(x))[1])
obj.mode <- napply(names, mode)
obj.type <- ifelse(is.na(obj.class), obj.mode, obj.class)
obj.size <- napply(names, object.size)
obj.dim <- t(napply(names, function(x) as.numeric(dim(x))[1:2]))
vec <- is.na(obj.dim)[, 1] & (obj.type != "function")
obj.dim[vec, 1] <- napply(names, length)[vec]
out <- data.frame(obj.type, obj.size, obj.dim)
names(out) <- c("Type", "Size", "Rows", "Columns")
if (!missing(order.by))
out <- out[order(out[[order.by]]), ]
out
}
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