similar to: get list element names within lapply / sapply call

|Hello, Given a function with several arguments, I would like to perform an lapply (or equivalent) while holding one or more arguments fixed to some common value, and I would like to do it in as elegant a fashion as possible, without resorting to wrapping a separate wrapper for the function if possible. Moreover I would also like it to work in cases where one or more arguments to the original

I am trying to use apply and a list to supply names to a set of tables I want to generate. Below is an example that I hope mimics the larger original problem. EXAMPLE aa <- c( 2,2,1,1,2) bb <- c(5,6,6,7,4) aan <- c("yes", "no") bbn <- c("a", "b", "c", "d") mynames <- c("abby", "billy") mylist <-

Wouldn't that change how simplify='array' is handled? > str(sapply(1:3, function(x)diag(x,5,2), simplify="array")) int [1:5, 1:2, 1:3] 1 0 0 0 0 0 1 0 0 0 ... > str(sapply(1:3, function(x)diag(x,5,2), simplify=TRUE)) int [1:10, 1:3] 1 0 0 0 0 0 1 0 0 0 ... > str(sapply(1:3, function(x)diag(x,5,2), simplify=FALSE)) List of 3 $ : int [1:5, 1:2] 1 0 0 0 0 0 1 0 0

Hi there During execution of sapply I want to extract the number of times the function given to supply has been executed. I came up with: mylist <- list(a=3,b=6,c=9) sapply(mylist,function(x)as.numeric(gsub("[^0-9]","",deparse(substitute(x))))) This works fine, but looks quite ugly. I'm sure that there's a more elegant way to do this. Any suggestion? Christian

You?re right, it sure does. My suggestion causes it to fail when simplify = ?array? From: William Dunlap [mailto:wdunlap at tibco.com] Sent: Tuesday, March 13, 2018 12:11 PM To: Doran, Harold <HDoran at air.org> Cc: r-help at r-project.org Subject: Re: [R] Possible Improvement to sapply Wouldn't that change how simplify='array' is handled? > str(sapply(1:3,

On 03/13/2018 09:23 AM, Doran, Harold wrote: > While working with sapply, the documentation states that the simplify argument will yield a vector, matrix etc "when possible". I was curious how the code actually defined "as possible" and see this within the function > > if (!identical(simplify, FALSE) && length(answer)) > > This seems superfluous to me,

Hi R users I like to extract (or collect) a numeric element with a name from a list. Is there any way to extract just a numeric element without the name attached to the element. For example, >mylist Mantel-Haenszel chi-squared test with continuity correction data: table(mydata[, x]) Mantel-Haenszel X-squared = 8.3832, df = 1, p-value = 0.003787 alternative hypothesis: true

I have a question regarding accessing the names of list components when applying a function with lapply. Here is an example that demonstrates what I'd like to do. I have a list like this one: mylist <- list(a=letters[1:10], b=letters[10:1], c=letters[1:3]) Now I would like to append the names of the list components to their corresponding vectors with the c() function. I thought this

Hello I am using R 2.2.0 with Windows XP. I've got a five element list object, each element containing two dataframes of equivalent size. > str(mylist) List of 1 $ data1:List of 2 ..$ data1a :`data.frame': 77 obs. of 63 variables: .. ..$ var1 : num [1:77] 0.41375 0.00056 1.43040 1.43528 0.61730 ... .. ..$ var2 : num [1:77] 1.154 1.686 0.673 0.800 0.760 ... ..

Hi r-users, I have a matrix B and a list of 3x3 matrices (mylist). I want to calculate the quantiles in the list using each of the value of B as probabilities. The codes I wrote are: B <- matrix (runif(12, 0, 1), 3, 4) mylist <- lapply(mylist, function(x) {matrix (rnorm(9), 3, 3)}) for (i in 1:length(B)) { quant <- lapply (mylist, quantile, probs=B[i]) } But quant

While working with sapply, the documentation states that the simplify argument will yield a vector, matrix etc "when possible". I was curious how the code actually defined "as possible" and see this within the function if (!identical(simplify, FALSE) && length(answer)) This seems superfluous to me, in particular this part: !identical(simplify, FALSE) The preceding

Hi - I have a list (call it 'mylist') with the following elements: (i) a function (call it 'myfunc' and expressed as 'mylist$myfunc') and (ii) a variable (call it 'myvar' and expressed as 'mylist$myvar'). Since I use mylist as a pseudo-class (I assign mylist to multiple different R objects), I would like to access the mylist R object from within the

Hi all, I'm trying to use lapply on a list with the following command: out<-lapply(mylist,myfun,par1=p,par2=d) (1) where myfun<-function(x,par1,par1) {.....} (2) now this function is in fact a wrapper for some Fortran code I have written so I think this might be the problem. When I call lapply() as in (1) I get the following message: Error in get(x,

I have split my original dataframe to generate a list of dataframes each of which has 3 columns of factors and a 4th column of numeric data. I would like to use lapply to apply the fitdistr() function to only the 4th column (x$isi) of the dataframes in the list. Is there a way to do this or am I misusing lapply? As a second solution I tried splitting only the numeric data column to yield a list

Martin In terms of context of the actual problem, sapply is called millions of times because the work involves scoring individual students who took a test. A score for student A is generated and then student B and such and there are millions of students. The psychometric process of scoring students is complex and our code makes use of sapply many times for each student. The toy example used

Hi I want to extract columns from a data frame using a vector with the desired column names. This short example uses the select argument in the subset function to accomplish what I am trying to do. Is there a better solution? #names of desired columns colnames <- c("col1","col3") #my data data <-

Hello: Forgive me, this is surely a simple question but I can't figure it out, having consulted the help archives and "Data Manipulation With R" (Spector). I have a list of 11 data frames with one common variable in each (prov). I'd like to use lapply to go through and recode one particular level of that common variable. I can get the recode to work, but it only returns the

Could your code use vapply instead of sapply? vapply forces you to declare the type and dimensions of FUN's output and stops if any call to FUN does not match the declaration. It can use much less memory and time than sapply because it fills in the output array as it goes instead of calling lapply() and seeing how it could be simplified. Bill Dunlap TIBCO Software wdunlap tibco.com On Tue,

Hi It seems, I just miss something. I defined treshold <- function(pred) { if (pred < 0.5) pred <- 0 else pred <- 1 return(pred) } and want to use apply it on a vector sapply(mylist[,,3],threshold) but I get: Error in match.fun(FUN) : Object "threshold" not found thanks for help cheers chris -- Christoph Lehmann <christoph.lehmann at gmx.ch>

Quite possibly, and I?ll look into that. Aside from the work I was doing, however, I wonder if there is a way such that sapply could avoid the overhead of having to call the identical function to determine the conditional path. From: William Dunlap [mailto:wdunlap at tibco.com] Sent: Tuesday, March 13, 2018 12:14 PM To: Doran, Harold <HDoran at air.org> Cc: Martin Morgan <martin.morgan