Dear Ivan,
No problem.
If you want it in a single plot:
matplot(do.call(cbind,g),ylab="value",pch=1:2,main="Some
plot",col=c("red","orange"),type="o")
legend("topleft",inset=.01,lty=c(1,1),title="Plot",col=c("red","orange"),names(g),horiz=TRUE)
A.K.
________________________________
From: Ivan Alves <papu...@me.com>
To: Duncan Murdoch <murdoch.dun...@gmail.com>; arun <smartpink...@yahoo.com>
Cc: "R-help@r-project.org" <R-help@r-project.org>
Sent: Wednesday, April 17, 2013 11:33 AM
Subject: Re: [R] use of names() within lapply()
Dear Duncan and A.K.
Many thanks for your super quick help. The modified lapply did the trick,
mapply died with a error "Error in dots[[2L]][[1L]] : object of type 'builtin'
is not subsettable".
Kind regards,
Ivan
On 17 Apr 2013, at 17:12, Duncan Murdoch <murdoch.dun...@gmail.com> wrote:
> On 17/04/2013 11:04 AM, Ivan Alves wrote:
>> Dear all,
>>
>> List g has 2 elements
>>
>> > names(g)
>> [1] "2009-10-07" "2012-02-29"
>>
>> and the list plot
>>
>> lapply(g, plot, main=names(g))
>>
>> results in equal plot titles with both list names, whereas distinct titles
>> names(g[1]) and names(g[2]) are sought. Clearly, lapply is passing 'g' in
>> stead of consecutively passing g[1] and then g[2] to process the additional
>> 'main' argument to plot. help(lapply) is mute as to what to element-wise
>> pass parameters. Any suggestion would be appreciated.
>
> I think you want mapply rather than lapply, or you could do lapply on a
> vector of indices. For example,
>
> mapply(plot, g, main=names)
>
> or
>
> lapply(1:2, function(i) plot(g[[i]], main=names(g)[i]))
>
> Duncan Murdoch
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