Thank you, Henrique,
my example was simplified. In a more complexe
function I want to use the objects, not just
their names. In your solution, I have to adapt
the function itself, depending on the name of the
data.frame, which I would like to avoid.
Thanks,
Heinz
At 13:36 28.09.2009, Henrique Dallazuanna wrote:
You can use names insteed:
DF <- data.frame(a=1:3, b=2:4)
lapply(names(DF), function(x){
print(x)
DF[x]
})
On Mon, Sep 28, 2009 at 8:22 AM, Heinz Tuechler <tuech...@gmx.at> wrote:
> Dear All,
>
> to produce output of several columns of a data frame, I tried to use lapply
> and also l_ply. In both cases, I would like to print a header line
> containing also the name of the respective column in the data frame.
>
> For example, I would like the following
>
> lapply(data.frame(a=1:3, b=2:4), function(x) print(deparse(substitute(x))))
>
> to produce:
> [1] "a"
> [1] "b"
>
> and not, what it actually does:
> [1] "X[[1L]]"
> [1] "X[[2L]]"
> $a
> [1] "X[[1L]]"
>
> $b
> [1] "X[[2L]]"
>
> or with l_ply (plyr package)
> l_ply(data.frame(a=1:3, b=2:4), function(x) print(deparse(substitute(x))))
>
> to produce:
> [1] "a"
> [1] "b"
>
> and not, what it actually does:
> [1] ".data[[i]]"
> [1] ".data[[i]]"
>
> Is this possible?
>
> Thanks,
> Heinz
>
> ______________________________________________
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>
--
Henrique Dallazuanna
Curitiba-Paraná-Brasil
25° 25' 40" S 49° 16' 22" O
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