Dear Eduardo,
try:
f <- function(x){}
s <- 0.2
body(f) <- substitute({x^2 + s}, list(s = s))
Best,
Matthias
On 15.02.2011 16:53, Eduardo de Oliveira Horta wrote:
Thanks... but I guess I didn't make myself clear. What I was trying to
do was precisely to "store" inside the function the number associated
to s[i] rather than the call to s[i], such that I wouldn't need to
keep that object in subsequent function calls.
In other words, I wanted to use lapply to get functions equivalent to:
s<- c( 0.2, 0.45, 0.38, 0.9)
f<-list()
f[[1]]<- function(x) x^2+0.2
f[[2]]<- function(x) x^2+0.45
f[[3]]<- function(x) x^2+0.38
f[[4]]<- function(x) x^2+0.9
Best regards,
Eduardo
On Tue, Feb 15, 2011 at 7:20 AM, Dennis Murphy<djmu...@gmail.com> wrote:
Hi:
If you look at the error message, you'll see that you removed s before
evaluating f, and since an element of s is called in the function....
Try
s<- c( 0.2, 0.45, 0.38, 0.9)
f<- lapply(1:10, function(i)local({ force(i) ; function(x)x^2+s[i]}))
f[[1]](s)
[1] 0.2400 0.4025 0.3444 1.0100
f is a list with 10 components, the first of which is
[[1]]
function (x)
x^2 + s[i]
<environment: 0x0000000002a26d48>
Each component occupies a different environment. To see what you get,
f[[1]](0.1)
[1] 0.21
for(i in 1:10) print(f[[i]](i))
[1] 1.2
[1] 4.45
[1] 9.38
[1] 16.9
[1] NA
[1] NA
[1] NA
[1] NA
[1] NA
[1] NA
for(i in 1:10) print(f[[i]](1))
[1] 1.2
[1] 1.45
[1] 1.38
[1] 1.9
[1] NA
[1] NA
[1] NA
[1] NA
[1] NA
[1] NA
HTH,
Dennis
On Mon, Feb 14, 2011 at 9:50 PM, Eduardo de Oliveira Horta
<eduardo.oliveiraho...@gmail.com> wrote:
Hello again.
Let me try something a little more intricate. Let's say instead of
forcing evaluation of 'i' I'd want to force evaluation of a vector;
for example:
s<- c( 0.2, 0.45, 0.38, 0.9)
f<- lapply(1:10, function(i)local({ force(i) ; function(x)x^2+s[i]}))
rm(s)
f[[1]](0.1)
Error in f[[1]](0.1) : object 's' not found
Any thoughts?
Best regards,
Eduardo
sessionInfo()
R version 2.11.1 (2010-05-31)
x86_64-pc-mingw32
locale:
[1] LC_COLLATE=Portuguese_Brazil.1252 LC_CTYPE=Portuguese_Brazil.1252
[3] LC_MONETARY=Portuguese_Brazil.1252 LC_NUMERIC=C
[5] LC_TIME=Portuguese_Brazil.1252
attached base packages:
[1] stats graphics grDevices utils datasets methods base
other attached packages:
[1] Revobase_4.2.0 RevoScaleR_1.1-1 lattice_0.19-13
loaded via a namespace (and not attached):
[1] grid_2.11.1 pkgXMLBuilder_1.0 revoIpe_1.0 tools_2.11.1
[5] XML_3.1-0
On Mon, Nov 15, 2010 at 7:10 PM, William Dunlap<wdun...@tibco.com>
wrote:
You could make f[[i]] be function(t)t^2+i for i in 1:10
with
f<- lapply(1:10, function(i)local({ force(i) ; function(x)x^2+i}))
After that we get the correct results
> f[[7]](100:103)
[1] 10007 10208 10411 10616
but looking at the function doesn't immdiately tell you
what 'i' is in the function
> f[[7]]
function (x)
x^2 + i
<environment: 0x19d7458>
You can find it in f[[7]]'s environment
> get("i", envir=environment(f[[7]]))
[1] 7
The call to force() in the call to local() is not
necessary in this case, although it can help in
other situations.
Bill Dunlap
Spotfire, TIBCO Software
wdunlap tibco.com
-----Original Message-----
From: r-help-boun...@r-project.org
[mailto:r-help-boun...@r-project.org] On Behalf Of Eduardo de
Oliveira Horta
Sent: Monday, November 15, 2010 12:50 PM
To: r-help@r-project.org
Subject: [R] Defining functions inside loops
Hello,
I was trying to define a set of functions inside a loop, with
the loop index
working as a parameter for each function. Below I post a
simpler example, as
to illustrate what I was intending:
f<-list()
for (i in 1:10){
f[[i]]<-function(t){
f[[i]]<-t^2+i
}
}
rm(i)
With that, I was expecting that f[[1]] would be a function
defined by t^2+1,
f[[2]] by t^2+2 and so on. However, the index i somehow
doesn't "get in" the
function definition on each loop, that is, the functions
f[[1]] through
f[[10]] are all defined by t^2+i. Thus, if I remove the
object i from the
workspace, I get an error when evaluating these functions.
Otherwise, if
don't remove the object i, it ends the loop with value equal
to 10 and then
f[[1]](t)=f[[2]](t)=...=f[[10]](t)=t^2+10.
I am aware that I could simply put
f<-function(u,i){
f<-t^2+i
}
but that's really not what I want.
Any help would be appreciated. Thanks in advance,
Eduardo Horta
[[alternative HTML version deleted]]
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PLEASE do read the posting guide
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and provide commented, minimal, self-contained, reproducible code.
______________________________________________
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PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
--
Dr. Matthias Kohl
www.stamats.de
______________________________________________
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PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
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