similar to: Assignment in environment

> On 07 Feb 2016, at 14:46 , Duncan Murdoch <murdoch.duncan at gmail.com> wrot8[e: > [snippage] > > but in fact, this doesn't work: > > getValue(fn)[[1]] <- 3 > Error in getValue(fn)[[1]] <- 3 : could not find function "getValue" > > I suspect this is a parser problem. Umm, no... The canonical semantics are that foo(x)[[....]] <- bar

I keep finding myself in a situation where I want to calculate a variable name and then use it on the left hand side of an assignment. For example iteration <- 1 varName <- paste("run",iteration,sep="") myList$parse(text=varName) <- aColumn I want to take some existing variable "aColumn" and use the name "varName" name for it and put it into a

What is the right way to assign a new variable into each a of list of data frames? Here is my failed attempt: mylist <- list(df1 = data.frame(A = runif(5), B = runif(5)), df2 = data.frame(A = runif(5), B= runif(5))) lapply(mylist, function(x){x$Y <- x$A * x$B}) $df1 [1] 0.25589928 0.03446026 0.94992362 0.21388326 0.08668821 $df2 [1] 0.08771839 0.05643553 0.09036894

Hello I tried to define replacement functions for the class "mylist". When I test them in an active R session, they work -- however, when I put them into a package, they don't. Why and how to fix? make_my_list <- function( x, y ) { return(structure(list(x, y, class="mylist"))) } mylist <- make_my_list(1:4, letters[3:7]) mylist mylist[['x']] <- 4:6

Hi, I am moving from MATLAB, where one can easily assign a number of output values from a function like this: [x,y] = myfun(a,b) Then variables x and y can be directly used in the caller workspace. I understand that R functions return a single argument, which could be a list. This in a way makes it possible to return multiple values with a single function call, but accessing the list variables

I can define a list containing NULL elements: > myList <- list("aaa",NULL,TRUE) > names(myList) <- c("first","second","third") > myList $first [1] "aaa" $second NULL $third [1] TRUE > length(myList) [1] 3 However, if I assign NULL to any of the list element then such element is deleted from the list: > myList$second <-

Hi, I'm pretty new to R. I have an object (say a list) and I I have a function that I call on various columns in that list (excuse terminology if it's wrong/ambiguous). Imagine its like this (actual values are unimportant) and called mylist: >mylist A B 1 5 2 5 3 6 4 8 5 0 I have a function: foo = function(param){ #modify list A or B values depending on

Hi, I'm have a list of integer vectors and I want to perform an outer() like operation on the list. As an example, take the following list: mylist <- list(1:5,3:9,8:12) A simple example of the kind of thing I want to do is to find the sum of the shared numbers between each vector to give a result like: result <- array(c(15,12,0,12,42,17,0,17,50), dim=c(3,3)) Two for() loops is the

Dear R users, Let's say I have a list with components being 'm' matrices (as exemplified in the "mylist" object below). Now, I'd like to subset this list based on an index vector, which will partition each matrix 'm' in 2 sub-matrices. My questions are: 1. Is there an elegant way to have the results shown in mylist2 for an arbitrary number of matrices in mylist?

Dear R-users, I would like to assign elements to a list in the following manner: mylist <- list(a = a, b = b, c = c) To do this I tried myexpr <- expression(a = a, b = b, c = c) mylist <- list( eval(myexpr) ) It ends up by overwriting a when b is assigned and b when c is assigned. Additionally the element of the list does not have a name. Could you tell me why this is the case? Thank

Hi I have a list : mylist <- list( a = NULL, b = 1, c = 2 ) > mylist[1] $a NULL > is.null(mylist[1]) [1] FALSE > is.null(mylist$a) [1] TRUE why? I need to use mylist[1]

I have this awkward problem with trellis (lattice). I am trying to generate some plots through loops but the .eps file is empty. When I generate them in a list and print them outside the loop all is fine. this is an example below:( nothing shows up in foo.eps, but all show up in foo1.eps) R vesion 2.0.1, lattice version 0.10-16, on a debian 2.6.8-1 kernel. X <- data.frame(x=rnorm(10000),

How do I parse an identifier of a list component, e.g. mylist$mycomponent or mylist[[1]] ? Parse does not do the job, e.g. parse(text="mylist$mycomponent") returns an expression with just one term, instead of "mylist", "$", "mycomponent". What I need is a way to extract the list name (e.g. "mylist"), given an identifier of a component.

I've been doing some consulting with students who seem to come to R from SAS. They are usually pre-occupied with do loops and it is tough to persuade them to trust R lists rather than keeping 100s of named matrices floating around. Often it happens that there is a list with lots of matrices or data frames in it and we need to "stack those together". I thought it would be a simple

Dear List, Let's say I have a list whose components are 2 matrices (as exemplified in the "mylist" object below). I'd like to create a list with components being 4 matrices based on an logical index vector. is there a way to simplify what I'm doing to obtain the results in "mylist2"? I'd like something that would work on an arbitrary number of elements in

Sorry My previous report is not detailed. In R, you will get this: > mylist <- list() > mylist[[1]] Error in mylist[[1]] : subscript out of bounds > mylist[[1]] <- c(1) Error: (list) object cannot be coerced to vector type 14 > mylist[[1]] <- c(1,2) > mylist[[1]] <- c(1) > mylist [[1]] [1] 1 I was trying to assigning c(1) to (mylist[[1]] <- c(1)) -- it seems

Excuse me for what I'm sure is a stupid beginner's question, but I've given up trying to find the answer to this question from the help, RSiteSearch, or any of the usual places. I have a list that looks like this: >myList $first [1] "--" "18" "8" "32" $second [1] "--" "--" "40" "54" I want a

Hi, Try > is.null(mylist[[1]]) [1] TRUE Notice the double square brackets. From: ?`[` "The most important distinction between [, [[ and $ is that the [ can select more than one element whereas the other two select a single element." On Thu, Jun 15, 2017 at 11:33 AM, ce <zadig_1 at excite.com> wrote: > Hi > > I have a list : > > mylist <- list( a = NULL, b

Here i have a variable MyVar <- data.frame(read.csv("D:\\Doc.csv")) And now i am storing this variable name into a list. MyList <- list() MyList [length(MyList )+1]<- "MyVar" Now what is the requirement is, i need to call the variable name "MyVar" from the list "MyList " and get the data.

I find that the str function is more helpful for understanding the difference between a null list and a list containing a null list than the implicit print function call that the interpreter invokes when you enter an expression at the console. str( mylist[1] ) -- Sent from my phone. Please excuse my brevity. On June 15, 2017 8:39:47 AM PDT, Huzefa Khalil <huzefa.khalil at umich.edu>