Displaying 20 results from an estimated 800 matches similar to: ""R CMD INSTALL in R 2.5.1 (2007-06-27)""
2004 Dec 28
1
editing a package directly from R
Dear R,
I have made a package 'myfuncs' of some home made functions and
installed it without problems. It contains no C or fortran code and no
dynamic link library. Is there a way to edit this package directly from
R (or Xemacs with ESS) or do I always have to edit the source and then
reinstall the package? I have a weak memory that such editing could be
done in S-Plus 2000 or 6.1?
2001 Sep 07
1
"load" for text file with functions?
Hi!
I'm writing down the methods that I used to
migrate from Splus to R and expand a bit the
R docs for this subject. Prior to send these
notes to the FAQ curator I would like to make sure
on the following:
Given an ascii text file with several
functions (i.e., myfuncs.txt), is there any equivalent to
load(), so that the functions become
usable in R? Or is it necessary
to install it as a
2008 Oct 21
1
R CMD INSTALL problem
Dear list members,
I've run into a problem with R CMD INSTALL under Windows Vista and R 2.8.0:
--------- snip -----------
C:\Users\John Fox\workspace>c:\R\R-2.8.0\bin\R CMD INSTALL car
installing to ''
---------- Making package car ------------
adding build stamp to DESCRIPTION
installing NAMESPACE file and metadata
installing R files
installing inst files
installing
2008 Oct 30
3
Save a function existing library
Hello all,
I am fairly new to R and I have created a function that I use quit
frequently. I was wondering if there is away to save this function in
an existing library so I can call it by using the function name once the
library is loaded?
Thanks,
Doug
--
---------------------------------
Douglas M. Hultstrand, MS
Senior Hydrometeorologist
Metstat, Inc. Windsor, Colorado
voice:
2004 Aug 30
3
Multiple lapply get-around
I am faced with a situation wherein I have to use multiple lapply's. The
pseudo-code could be approximated to something as below:
For each X from i=1 to n
For each Y based on j=1 to m
For each F from 1 to f
Do some calculation based on Fij
Store Xi,Yj = Fij
End For F
End for Y
End for X
Is there anyway to optimize the processing logic further? I *guess*
using the multiple lapply
2007 Jun 28
2
using self-written functions
Hi, I am pretty new to R, so I apologize for the obvious question.
I
have worked with R for a few months now and in the process have written
several functions that I frequently use in various data analysis
projects. I tend to give each project a directory of its own and set
the working directory to that.
Since there are several tasks that
need to be accomplished in many of my projects, I
2010 Mar 31
2
Simplifying particular piece of code
Hello, everyone
I have a piece of code that looks like this:
mrets <- merge(mrets, BMM.SR=apply(mrets, 1, MyFunc, ret="BMM.AV120",
stdev="BMM.SD120"))
mrets <- merge(mrets, GM1.SR=apply(mrets, 1, MyFunc, ret="GM1.AV120",
stdev="GM1.SD120"))
mrets <- merge(mrets, IYC.SR=apply(mrets, 1, MyFunc, ret="IYC.AV120",
2011 Feb 18
1
debugger() fails if "..." in function arguments
Dear all,
I'm having a problem with debugger() in both R 2.8.0 and R 2.12.0.
Probably also versions in-between.
I don't see it logged in the bug database, but it's hard for me to
imagine that no-one else has encountered it. So my question is whether
it's a known problem with a workaround, or do I log it as a new problem?
The situation is that if I use
2004 Sep 09
4
scoping rules
Can someone help me with this simple example?
sq <- function() {
y <- x^2
y
}
myfunc <- function() {
x <- 10
sq()
}
myfunc()
executing the above in R yields:
> myfunc()
Error in sq() : Object "x" not found
I understand that R's scoping rules cause it to look for "x" in the
environment in which "sq" was defined (the global environment in
2002 Aug 06
1
re| `By reference'
David Brahm <brahm at alum.mit.edu> wrote:
>VBMorozov at lbl.gov wrote:
>> I would like to pass variables to a function in R in "by reference"...
>Just in case the ensuing discussion got too esoteric, here's one simple
answer:
>R> x <- 1:10
>R> MyFunc <- function(x, zz) assign(deparse(substitute(zz)), sum(x), 1)
>R> MyFunc(x,y)
>R>
2009 Aug 06
1
Using 'field names' of a data.frame in a function
I may be doing this wrong! but I have a function which I have simplified a lot below. I want to pass some 'field names' of a data-frame to the function for it to then do some manipulation of.
Here's my code:
#build a simple dataset
mydataset = data.frame (
2002 Aug 03
2
variable scope
Dear R-guRus:
I would like to pass variables to a function in R in "by reference",
e.g Fortran style.
For example, suppose I have the following code
x<-c(1:10)
y<-1
MyFunc<-function(x,y) {y<-sum(x); return(NULL)}
MyFunc(x,y)
print(y)
in this case print(y) will produce "1" instead of 55 (which is sum(x)) -
how do I make sure that afte the function is run, y
2002 Aug 03
2
variable scope
Dear R-guRus:
I would like to pass variables to a function in R in "by reference",
e.g Fortran style.
For example, suppose I have the following code
x<-c(1:10)
y<-1
MyFunc<-function(x,y) {y<-sum(x); return(NULL)}
MyFunc(x,y)
print(y)
in this case print(y) will produce "1" instead of 55 (which is sum(x)) -
how do I make sure that afte the function is run, y
2011 Jan 10
2
Integration in R
Dear all,
It has been ages since I studied integration in college. Right now I
try to recover all this kind of knowledge and then try to understand how
integration works.
Thus I am doing some first 'experiments' and I would like to request your help and comments.
I have the function:
p2<-function(x){0.5*(3*x^2-1)}
# I found the square of p2 by using some pencil and
2006 Nov 09
3
function
R-help,
I am trying to create a function that i pass a data set to and have the
function return some calculations based on data.
Allow me to illustrate:
myfunc <- function(lst,mn,sd){
lst <- sort(lst)
mn <- mean(lst)
sd <- sqrt(var(lst))
return(lst,mn,sd)
}
data1 <-c (1,2,3,4,5)
data2 <- c(6,7,8,9,10)
myfunc(data1,data1mn,data1sd)
myfunc(data2,data2mn,data2sd)
2012 Aug 29
5
Extracting the name of a function (inverse of match.fun("myFun"))
Hi all,
is there a way to extract the name of a function, i.e. do the reverse
of match.fun applied to a character string? I would like to print out
the name of a function supplied to another function as an argument.
For example:
myFunc = function(x) { x+1 }
applyFunc = function(fnc, x)
{
fnc = match.fun(fnc)
fnc(x)
}
Is there a way to obtain "myFunc" from the argument fnc in
2008 Mar 27
1
A faster way to compute finite-difference gradient of a scalar function of a large number of variables
Hi All,
I would like to compute the simple finite-difference approximation to the
gradient of a scalar function of a large number of variables (on the order
of 1000). Although a one-time computation using the following function
grad() is fast and simple enough, the overhead for repeated evaluation of
gradient in iterative schemes is quite significant. I was wondering whether
there are
2012 Jul 20
3
Execute a function
Hi,
I would like to evaluate a function, with 3 arguments, for instance,
myfunc<-function(a,b,c) { sqrt(a)-exp(b)+4*c
}
How to execute myfunc(x,y,z), for all x, all y and all z, where x,y,z are
vectors?
Thank you very much in advance
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View this message in context: http://r.789695.n4.nabble.com/Execute-a-function-tp4637182.html
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2011 Aug 24
1
Passing a large amount of parameters to a function
Hello,
I have a function with a long list of parameters (of different types,
numeric and string)
myFunc <-function(p1, p2, p3, p4, p5...etc)
{
do.something(p1,p2,....)
}
I want to loop over this to provide a different set of parameters to the
list every time.
for (ii in 1:N)
{
myFunc(p1(ii), p2(ii),....etc)
}
I would like to simplify the notation and use some kind of structure, maybe
2009 Apr 07
2
Puzzled by an error with apply()
I've written a function, myFunc, that works fine with myFunc(data,
...), but when I use apply() to run it with an array of data
apply(myArray, 1, myFunc, ...)
I get a strange error:
Error in match.fun(FUN) : '1' is not a function, character or symbol
which really puzzles me because '1' is meant to be the margin of the
array I want to apply over, but how come does apply()