You can always add those names to the list: is this what you are after?
> example.names <- c("con1-1-masked-bottom-green.tsv",
"con1-1-masked-bottom-red.tsv"
+ , "con1-1-masked-top-green.tsv","con1-1-masked-top-red.tsv")
> example.list <- strsplit(example.names, "-")
> names(example.l
They aren't being stored, they are being generated on the fly. You can
create the same names using make.names()
example.names <- c("con1-1-masked-bottom-green.tsv",
"con1-1-masked-bottom-red.tsv", "con1-1-masked-top-green.tsv",
"con1-1-masked-top-red.tsv")
example.list <- strsplit(example.names,
I'm doing some string manipulation on a vector of file names, and noticed
something curious. When I strsplit the vector, I get a list of
character vectors.
The list is numbered, as lists are. When I cast that list as a data
frame with 'as.data.frame()', the resulting columns have names derived
fr
On 22/01/2016 2:29 AM, TJUN KIAT TEO wrote:
I am trying to populate an array of lists in R . Here is my code
TunePar<-matrix(list(Null),2,2)
TunePar[1,1]=list(G=2)
But when I type TunePar[1,1,], all I get is 2. The G has disappeared. why?
If I do this
Test=list(G=2)
Test
$G
[1] 2
Matric
Hi,
Provide a list of a list in the second assignment:
--
TunePar <- matrix(list(NULL), 2, 2)
TunePar[2,1] <- list(list(G = 2))
TunePar[2,1]
TunePar[2,1][[1]]$G
TunePar[[2]]$G
---
The point is that "[" returns the list element of the same level as the
original object (TunePar in the present ex
Hi,
Not sure this is what you wanted.
lstNew <- list(Spans, lapply(lapply(Spans,`[`,1),as.character) )
str(lstNew)
#List of 2
$ :List of 3
..$ : num [1:2] 8.37e+08 8.42e+08
..$ : num [1:2] 8.32e+08 8.37e+08
..$ : num [1:2] 9.30e+08 9.35e+08
$ :List of 3
..$ : chr "8.37e+08"
..$ : chr "8
Dear all,
I have a list that is created like that
Spans<-list( c(837e6,842e6),
c(832e6,837e6),
c(930.1e6,935.1e6)
)
I would like to include a second list that will contain the string that would
correspond to the numbers at the left side.
I would like thus insi
A matrix may only contain one data type. By not specifying when you
created m, it was filled with logical values of NA.
A logical value can't hold a list.
You can see that with
str(m)
which returns:
> str(m)
logi [1:3, 1:2] NA NA NA NA NA NA
- attr(*, "dimnames")=List of 2
..$ : chr [1:3] "Ro
Dear R experts,
since more or less half a year I am using R.
In many of my computations I construct huge matrices. Often I do so using
'cbind' on named lists:
do.call( 'cbind',
list(
"Column_A"=list("Row_one"=1.0, "Row_two"=2.0, "Row_three"=3.0),
"Column_B"=list("Row_one"=4.0, "Row_two
linear
algebra (or aviation or physics or almost anywhere else).
Bill Dunlap
Spotfire, TIBCO Software
wdunlap tibco.com
> -Original Message-
> From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org] On
> Behalf Of Kehl Dániel
> Sent: Monday, December 05,
Dear list members,
I have a really simple problem.
I connected to a DB and have the following query
adat <- dbGetQuery(con, paste("select * from kmdata where SzeAZ='",
szeazok[i], "' order by datum", sep=""))
now I have the data in the adat variable which is a list. In fact the
elements of t
Hello R users,
I am dealing with some resonably big data sets that I have split up
into lists based on various factors. In the code below, I have got my
code producing 100 values between point1x and point1y for the first
matrix in my list.
for (k in 1:length(point1x[[1]][, 1])) {
linex[[k]] = seq
-help@r-project.org Help
Subject: Re: [R] Lists of tables and conditional statements
On Mar 30, 2011, at 7:27 PM, Henrique Dallazuanna wrote:
> Try this:
>
> lapply(l, function(x)x[x[,'Sum'] == 3,])
If this is the right answer, you should send a "solved" message. The
On Mar 30, 2011, at 7:27 PM, Henrique Dallazuanna wrote:
Try this:
lapply(l, function(x)x[x[,'Sum'] == 3,])
If this is the right answer, you should send a "solved" message. The
dput extract was incomplete.
--
David.
On Wed, Mar 30, 2011 at 7:38 PM, Herbert, Alan G
wrote:
Hi R-users
Try this:
lapply(l, function(x)x[x[,'Sum'] == 3,])
On Wed, Mar 30, 2011 at 7:38 PM, Herbert, Alan G wrote:
> Hi R-users,
>
> I have a list containing numeric tables of differing row length. I want to
> make a new list that contains only rows from tables with a "Sum" greater than
> 3, plus the
On Mar 30, 2011, at 5:38 PM, Herbert, Alan G wrote:
Hi R-users,
I have a list containing numeric tables of differing row length. I
want to make a new list that contains only rows from tables with a
"Sum" greater than 3, plus the names of each table. I was wondering
whether there is an el
Hi R-users,
I have a list containing numeric tables of differing row length. I want to make
a new list that contains only rows from tables with a "Sum" greater than 3,
plus the names of each table. I was wondering whether there is an elegant way
to do this using apply of related functions as th
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On 12/08/2010 11:22 AM, jim holtman wrote:
> It sounds like you want to use a "list" instead of a dataframe,
No - I would like to have a data,frame. I am aware of the differences,
but as far as I understand, each column in a data.frame can have a
diff
It sounds like you want to use a "list" instead of a dataframe,
especially if the elements are a different length.
> d <- list() # initialize
> d[[length(d) + 1]] <- list() # extend
> d[[length(d)]]$fun <- sin # add a function
> d[[length(d) + 1]] <- list() # extend again
> d[[length(d)]]$fun
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Hi
I would like to have a data.frame, where one column contains functions,
and another one lists. i.e.:
FUN <- function(l) {return(l$a+l$b+l$c}
LIST <- list(a=1, b=2, c=3)
d <- data.frame(fun=FUN, no=LIST, value=2, b=TRUE)
FUN <- function(l) {retur
alf Of Peter Langfelder
> Sent: Monday, September 20, 2010 9:52 PM
> To: r-help
> Subject: Re: [R] Lists with NULL entries
>
> Hi Joshua,
>
> thanks, I came up with that solution myself after a bit of thinking.
> Normally I wouldn't worry about NULL components of lists
Hi Joshua,
thanks, I came up with that solution myself after a bit of thinking.
Normally I wouldn't worry about NULL components of lists, but dimnames
is a list and often some components are null and is therefore a bit
tricky to manipulate...
Peter
On Mon, Sep 20, 2010 at 7:39 PM, Joshua Wiley
Sorry, that was a really half-hearted reply. This will create a new
list that is the old list shifted down (and should be much faster than
the for loop too).
lst <- list(NULL,2)
lst2 <- vector("list", length(lst) + 1)
lst2[2:length(lst2)] <- lst
lst
lst2
If you really need to use a for loop, may
Hello Peter,
This is because assigning a value of NULL removes that element of the
list. I am not quite sure what the reference for that is. I remember
reading it in the documentation once though. I looked through ?list
to no avail. At any rate, to avoid it, you would have to assign
something
Hello,
I encountered a weird problem. Consider the following code that takes
a list "lst" and shifts all elements one index up (for example, to
make space for a new first element):
lst = list(1,2)
ll = length(lst);
for (i in ll:1)
lst[[i+1]] = lst[[i]];
lst
If you run it, you get the expected
r-help-boun...@r-project.org
> [mailto:r-help-boun...@r-project.org] On Behalf Of ewaters
> Sent: Tuesday, 23 February 2010 3:27 p.m.
> To: r-help@r-project.org
> Subject: [R] Lists into matrices within lists...again
>
>
> Related questions to this have been asked before, but
On Feb 22, 2010, at 8:27 PM, ewaters wrote:
Related questions to this have been asked before, but I have tried all
options they gave me unsuccessfully (do.call and unlist).
I start with three lists of summary statistics, 100 elements each,
which I
bind together:
None of this represents
Related questions to this have been asked before, but I have tried all
options they gave me unsuccessfully (do.call and unlist).
I start with three lists of summary statistics, 100 elements each, which I
bind together:
statslist <- as.data.frame(cbind (means, vars, mcrs))
I then take 100 sampl
How about like this:
for (i in seq_along(a)) {
result <- as.list(a[1:i])
cat("iterator", i, ":\n")
print(result)
}
On Sat, Jul 25, 2009 at 6:48 AM, Alberto Lora M wrote:
> Hi Everybody
>
> I have the following problem
>
> suppose that we
>
> a<-c("uno","dos","tres")
>
> I am working with a
Here is one way:
> a<-c("uno","dos","tres")
> x <- list()
> a<-c("uno","dos","tres")
> x <- list()
> for (i in seq_along(a)){
+ # add to the list
+ x[[i]] <- a[i]
+ str(x)
+ }
List of 1
$ : chr "uno"
List of 2
$ : chr "uno"
$ : chr "dos"
List of 3
$ : chr "uno"
$ : chr "dos"
$ : chr "t
Hi Everybody
I have the following problem
suppose that we
a<-c("uno","dos","tres")
I am working with a while cycle and the idea is in each iteration adding an
item to a list
In the first iteration the resultshould be:
[[1]]
[1] "uno"
In the second
[[1]]
[1] "uno"
[[2]]
[1] "dos"
And the fina
Many thanks for all your suggestions - much appreciated!
Nikol
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal,
Petr PIKAL wrote:
> Hi
>
>
> r-help-boun...@r-project.org napsal dne 04.03.2009 13:18:30:
>
>
>> Hello
>> I am am new to R and any help with the following would be appreciated:
>> I have a list (example attached) and I would like to create a new list
>> which is a filtered version of this list.
Nikol Simecek wrote:
> Hello
> I am am new to R and any help with the following would be appreciated:
> I have a list (example attached) and I would like to create a new list
> which is a filtered version of this list. I.e I would like a list that
> only contains elements with this value:
>
> Chr 1
Dear Nikol,
Try this:
do.call(c,lapply(yourlist,function(x) x[2] ))
HTH,
Jorge
On Wed, Mar 4, 2009 at 7:18 AM, Nikol Simecek wrote:
> Hello
> I am am new to R and any help with the following would be appreciated:
> I have a list (example attached) and I would like to create a new list
> whic
Hi
r-help-boun...@r-project.org napsal dne 04.03.2009 13:18:30:
> Hello
> I am am new to R and any help with the following would be appreciated:
> I have a list (example attached) and I would like to create a new list
> which is a filtered version of this list. I.e I would like a list that
> o
Hello
I am am new to R and any help with the following would be appreciated:
I have a list (example attached) and I would like to create a new list
which is a filtered version of this list. I.e I would like a list that
only contains elements with this value:
Chr 10 : 21853562 - 21855482
Any p
Dear R experts,
I have a problem with a function I wrote. The fuction looks like this:
series<-function(x,s){
foo<-list();
ind3<-integer();
for (j in diff){
for (i in 1:(n-12)){
if
(!(x[i,j]==0)&!(x[i+1,j]==0)&!(x[i+2,j]==0)&!(x[i+3,j]==0)&!(x[i+4,j]==0)&!(x[i+5,j]==0)&!(x[i+6,j]==0)
&!(x[i+
I think this is what you want. You need to look at the help file for
?'[' and ?'[['. Also understand the differences between a data.frame
and a list. in the case of defining qn you need to use a list and not
a data.frame because here is what happens with a data.frame:
> q1 <- data.frame(q=d1, n
I cannot find out how to build data structures as lists of data structures.
I want to do...
r-help@r-project.org
d1 <- data.frame(x=1, y=2)
d2 <- data.frame(x=1, y=3)
d3 <- data.frame(x=21, y=3)
q1 <- data.frame(q=d1, n="a")
q2 <- data.frame(q=d2, n="a")
q3 <- data.frame(q=d3, n="a")
v <-
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