similar to: call a function with explicitly not setting an argument

Hello all, I'm trying to use the apply function on a data frame, by applying a function that takes a one row data.frame as argument . Here's the example : myfun = function(x) paste(x$f1 , x$f2) df = data.frame(f1 = c(1,4,10),f2 = "hello") apply(df,1,myfun) ==> Does not work (I get "character(0)" ) Though : myfun(df[1,]) works, and myfun(df) works as well. So if

I've seen a number of problems like this over the years. The fact that the singular values of the Jacobian have a ration larger than the usual convergence tolerances can mean the codes stop well before the best fit. That is the "numerical analyst" view. David and Jeff have given geometric and statistical arguments. All views are useful, but it takes some time to sort them all out and

I am not as expert as John, but I thought it worth pointing out that the variable substitution technique gives up one set of constraints for another (b=0 in this case). I also find that plots help me see what is going on, so here is my reproducible example (note inclusion of library calls for completeness). Note that NONE of the optimizers mentioned so far appear to be finding the true best

> On Jun 18, 2017, at 6:24 AM, Manoranjan Muthusamy <ranjanmano167 at gmail.com> wrote: > > I am using nlsLM {minpack.lm} to find the values of parameters a and b of > function myfun which give the best fit for the data set, mydata. > > mydata=data.frame(x=c(0,5,9,13,17,20),y = c(0,11,20,29,38,45)) > > myfun=function(a,b,r,t){ > prd=a*b*(1-exp(-b*r*t)) >

Hi All I have a dataframe: myframe<-data.frame(ID=c("first","second"),x=c(1,2),y=c(3,4)) And I have a function myfun: myfun<-function(x,y) x+y I would like to write a function myfun2 that takes myframe and myfun as parameters and returns a list as below: mylist $first [1] 4 $second [2] 6 Could you please help me with this? Doesn't seem like the

I ran the following script. I satisfied the constraint by making a*b a single parameter, which isn't always possible. I also ran nlxb() from nlsr package, and this gives singular values of the Jacobian. In the unconstrained case, the svs are pretty awful, and I wouldn't trust the results as a model, though the minimum is probably OK. The constrained result has a much larger sum of squares.

https://cran.r-project.org/web/views/Optimization.html (Cran's optimization task view -- as always, you should search before posting) In general, nonlinear optimization with nonlinear constraints is hard, and the strategy used here (multiplying by a*b < 1000) may not work -- it introduces a discontinuity into the objective function, so gradient based methods may in particular be

Dear R experts. I might be missing something obvious. I have been trying to fix this problem for some weeks. Please help. #data ped <- c(rep(1, 4), rep(2, 3), rep(3, 3)) y <- rnorm(10, 8, 2) # variable set 1 M1a <- sample (c(1, 2,3), 10, replace= T) M1b <- sample (c(1, 2,3), 10, replace= T) M1aP1 <- sample (c(1, 2,3), 10, replace= T) M1bP2 <- sample (c(1, 2,3), 10, replace= T)

I am using nlsLM {minpack.lm} to find the values of parameters a and b of function myfun which give the best fit for the data set, mydata. mydata=data.frame(x=c(0,5,9,13,17,20),y = c(0,11,20,29,38,45)) myfun=function(a,b,r,t){ prd=a*b*(1-exp(-b*r*t)) return(prd)} and using nlsLM myfit=nlsLM(y~myfun(a,b,r=2,t=x),data=mydata,start=list(a=2000,b=0.05), lower = c(1000,0),

Dear all, I'd appreciate any help to rectify what must be a misconception of mine how environments work: ########################## myEnv <- new.env() myEnv$a.env <- 1 myEnv$symbols.env <- "a.env" a.global <- 2 symbols.global <- "a.global" myFun <- function(symbols){do.call("print", lapply(symbols, FUN=as.name))} do.call("myFun",

I am trying to write a function, which would allow to call various methods and would pass to them extra arbitrary parameters. My first attempt was to use call() as illustrated below, but apparently '...' cannot be used in such context. How can this be achieved ? Best regards, Ryszard > myfun <- function(method, x, ...) { + v <- eval(call(method, x, ...)) + } > method =

Hello folks, I have a small piece of C code written like this: typedef struct { char a; int b; int c; } foo; foo myFoo[5] = {{0}}; With llvm-gcc 2.5, I get the following IR: %struct.foo = type { i8, i32, i32 } @myFoo = global [5 x %struct.foo] zeroinitializer, align 32 With the current 2.6, I get this: %0 = type { i8, [11 x i8] }

Hi We've found that when using parse_args(..., positional_arguments=FALSE), it is permissible to invoke our script with either "--myfoo=bar" or "--myfoo bar"; that is, whether or not the equals sign is present makes no difference, and in fact both usage forms are demonstrated in the optparse vignette. However, we've found that when using parse_args(...,

How can I right justify the right-axis tick values? They appear in the example below as left-justified. I have tried several different ways and all fail in different ways. The example below creates the right axis tick value with no attempt at adjustment. alternates I have tried are 1. formatting the values. This doesn't work because they are in a proportional font and the blanks are too

Dear List, just out of pure curiosity: is it possible to define a function via 'substitute()' such that the function's formal arguments are specified by a "wildcard" that is substituted when the expression is evaluated? Simple example: x.args <- formals("data.frame") x.body <- expression( out <- myArg + 100, return(out) ) expr <-

Hello Julien, I think the reason for the change was because there is processor context information stored in the type in 2.6. The reason it's there is to support multicore JIT architecture. --Sam ----- Original Message ---- > From: Julien Lerouge <jlerouge at apple.com> > To: LLVM Developers Mailing List <llvmdev at cs.uiuc.edu> > Sent: Thursday, September 10, 2009

Hi, Suppose I have created something like this in R: foo <- "myfoo" and I want to find out the number of character in foo (in other words, R should return 5 since "myfoo" has 5 charactors. How can I do it? I tried: length(foo) but it returned 1. Cheers, Kevin ------------------------------------------------------------------------------ Ko-Kang Kevin Wang

Dear R-help, I have a home-made package works perfectly under R 1.9.1. Now I'm trying to port it to R > 2.0. So I rebuild the package under R 2.1.1. It installs and loads OK. But when I try to call some functions "myfoo" in this package, it returns error message: "Couldn't find function myfoo". I checked all available manuals, and didn't find any specific

On Sat, Feb 26, 2011 at 10:25 PM, Chris Lattner <clattner at apple.com> wrote: > Several people have been proding me to write up my thoughts on how to fix the IR  type system for LLVM 3.0.  Here are some (fairly stream of conscious) thoughts on the matter: > http://nondot.org/sabre/LLVMNotes/TypeSystemRewrite.txt > > Comments welcome! So struct types would unique by name. How

Hi, suppose that I have the following data.frame: cnae4 cnpj 1995 1996 1997 1998 1999 2000 2001 2002 2003 2004 Y 24996 10020470 1 1 2 12 16 21 17 51 43 19 183 24996 10020470 69 91 79 92 91 77 90 96 98 108 891 36145 10020470 0 0 0 0 2 83 112 97 91 144 529 44444 10023333 5 20 60 0 0 0 0 5 20 1000 1110 I would like to create a new variable X that indicates which