similar to: R object as a function

hi R folks, I need read a file from hardisk or www web. Then I need to define some new functions according to the contents of the read file. For example, i need write a package name "mypackage" like this: >library(mypackage) >read(some_file_on_web) #to see its content, suppose it contains: eat.drink.sleep then 3 new functions need to be created and usable. the problem is, how

I'm running an LSODA to generate some graphs, but I need to stop at a certain point and use those values to generate another LSODA output. This is working fine, but when I try to run the second LSODA, I get the "Error in eval(expr, envir, enclos) : object 'N' not found". Any ideas what can be causing this? I have no object 'N' anywhere in the script. I made an

Dear all, I have written a function for calculating the volume of a tree (=trad) or snag (=h?gst). The included volume regreesion model includes ten parameter values, which are tree species specific. bj?rk.formh?jd.pars is an object which includes the parameter values (parameter set) for birch (=bj?rk). There is one row per tree in the data object. > relev.kols[1:5,

hello I'm learning mgcv and would like to obtain numerical output corresponding to plot.gam. I can do so when seWithMean=FALSE (the default) but only approximately when seWithMean=TRUE. Can anyone show how to obtain the exact values? Alternatively, can you clarify the explanation in the manual "Note that, if seWithMean=TRUE, the confidence bands include the uncertainty about the

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),

Is there a way to call a function, and explicitly set an argument to 'not specified'? My situation is the following. I have a function which passes on most of its arguments to another function. The second function, myfun2, serializes all arguments and is out of my control. myfun <- function(...){ return(myfun2(...)); } now, the value for arguments of myfun are stored in variables.

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

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

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.

> 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)) >

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

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'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

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 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 =

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",

Dear R developer, I am not quite sure, if I should post my concern as a wish to r-bugs at r-project.org. Thus, as recommended, I first send an email to you. My request is the following: I would appreciate, if it was possible to obtain a plot of a 'stepfun' with a strict interpretation of xlim. What I mean: sf <- stepfun(1:4, 1:5) plot(sf, xlim=c(0,10)) does not bound the function to

hi y'all, I am wondering if there is any special command, function, package, etc to help me doing a cumulative distribution function, with y-scale - probability scale. I tried the help in R and i got the following answers: cumsum(base) Cumulative Sums, Products, and Extremes ecdf(stepfun) Empirical Cumulative Distribution Function cpgram(ts) Plot

Hello, My specific system is > version _ platform i386-pc-mingw32 arch i386 os mingw32 system i386, mingw32 status major 1 minor 5.1 year 2002 month 06 day 17 language R Each of the hmisc and stepfun packages has a function

Dear members, I would like a help for extracting the values from a step function (stepfun). >From help(stepfun) we have the following example: Y0<-c(1.,2.,4.,3.) y0<-c(1.,2.,3.,4.) sfun<-stepfun(1:3,y0,f=0) plot(sfun) Now, suppose instead I was given the object (*sfun*, say) from which I wanted to extract the values generated by the function *stepfun*. More precisely, I want to