Displaying 20 results from an estimated 3000 matches similar to: "Parsing for list components"
2007 Nov 07
1
Shortcut to refer to an attached dataframe?
When I attach data frames I often want to be able to refer to the whole
data frame rather then one of its components. For example:
attach (my.data.frame)
summary(my.data.frame)
That's fine but often the frame has a very long name so I'd prefer some
shorthand way of referring to it by its position on the search list.
This applies especially to cases where I have a nested data frame
2001 Mar 02
1
Introduction to R (S) [in German]
Couse notes for a short introduction to the S language have been posted as
<http://www.ci.tuwien.ac.at/R/doc/contrib/s.pdf;I>
This is material for a 4-5 days course, assuming some background in
statistics (German style).
As a matter of fact it is a Trojan horse, claiming it is teaching the
S language, but under the shelf it is trying to convey some ideas
about statistics.
For those
2001 Mar 02
1
Introduction to R (S) [in German]
Couse notes for a short introduction to the S language have been posted as
<http://www.ci.tuwien.ac.at/R/doc/contrib/s.pdf;I>
This is material for a 4-5 days course, assuming some background in
statistics (German style).
As a matter of fact it is a Trojan horse, claiming it is teaching the
S language, but under the shelf it is trying to convey some ideas
about statistics.
For those
2006 Mar 30
3
access list component names with lapply
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
2015 May 04
2
Define replacement functions
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
2003 May 31
5
parse on left hand side of R assignment
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
2009 Oct 25
3
NULL elements in lists ... a nightmare
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 <-
2007 Oct 20
1
Getting at what a named object represents in a function...
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
2004 May 10
2
Lists and outer() like functionality?
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
2011 Apr 05
1
Help in splitting a list
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?
2017 Jun 15
4
is.null(mylist[1]) and is.null(mylist$a) returns different values
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]
2005 Jan 30
3
trellis graphics in loops
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),
2010 Sep 04
4
Please explain "do.call" in this context, or critique to "stack this list faster"
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
2011 Apr 03
1
Help in splitting ists into sub-lists
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
1999 May 09
1
subscripting in list() (PR#187)
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
2010 Mar 18
1
Substitute NAs in a data frame
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"
$second
[1] "--" "--" "40" "54"
I want a
2017 Jun 15
0
is.null(mylist[1]) and is.null(mylist$a) returns different values
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
2012 Aug 28
3
Get variable data Reading from the list
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.
2017 Jun 15
1
is.null(mylist[1]) and is.null(mylist$a) returns different values
I find that the str function is more helpful for understanding the difference between a null list and a list containing a null list than the implicit print function call that the interpreter invokes when you enter an expression at the console.
str( mylist[1] )
--
Sent from my phone. Please excuse my brevity.
On June 15, 2017 8:39:47 AM PDT, Huzefa Khalil <huzefa.khalil at umich.edu>
2016 Feb 07
3
Assignment in environment
Dear all,
I have a function "fn" with its own environment, i.e.
env <- environment(fn)
and env is not .GlobalEnv. And another function
getValue <- function(x) environment(x)$mylist
which returns the list object "mylist" which is in "env". If I want to
modify "mylist", I could write
'getValue<-' <- function(x, value) {