Friedericksen Hope
2010-Nov-11 08:05 UTC
[R] How to get a specific named element in a nested list
Hello,
I have a nested named list structure, like the following:
x <- list(
list(
list(df1,df2)
list(df3,
list(df4,df5))
list(df6,df7)))
with df1...d7 as data frames. Every data frame is named.
Is there a way to get a specific named element in x?
so, for example,
x[[c("df5")]] gives me the data frame 5?
Thank you in advance!
Best,
Friedericksen
Ivan Calandra
2010-Nov-11 09:23 UTC
[R] How to get a specific named element in a nested list
Hi,
A reproducible example would have been nice, but a correct code even
more (you forgot some commas)!
So if you meant this:
x <- list(
list(
list(df1,df2),
list(df3,
list(df4,df5)),
list(df6,df7)))
check str(x): (I removes the details of each df)
List of 1
$ :List of 3
..$ :List of 2
.. ..$ :'data.frame': 10 obs. of 2 variables: ## df1
.. ..$ :'data.frame': 10 obs. of 2 variables: ## df2
..$ :List of 2
.. ..$ :'data.frame': 10 obs. of 2 variables: ## df3
.. ..$ :List of 2
.. .. ..$ :'data.frame': 10 obs. of 2 variables: ## df4
.. .. ..$ :'data.frame': 10 obs. of 2 variables: ## df5
..$ :List of 2
.. ..$ :'data.frame': 10 obs. of 2 variables: ## df6
.. ..$ :'data.frame': 10 obs. of 2 variables: ## df7
And check x: (I removed the details again)
[[1]]
[[1]][[1]]
[[1]][[1]][[1]]
df1
[[1]][[1]][[2]]
df2
[[1]][[2]]
[[1]][[2]][[1]]
df3
[[1]][[2]][[2]]
[[1]][[2]][[2]][[1]]
df4
[[1]][[2]][[2]][[2]]
df5
[[1]][[3]]
[[1]][[3]][[1]]
df6
[[1]][[3]][[2]]
df7
If you want to access them, you need to follow the structure of your
list, e.g.
for df1:
x[[1]][[1]][[1]]
for df5:
x[[1]][[2]][[2]][[2]]
You need to access them with indexes because the list elements are unnamed.
Now if you do:
x <- list(uppermost=list(
medium1=list(df1=df1,df2=df2),
medium2=list(df3=df3,
lowermost=list(df4=df4,df5=df5)),
medium3=list(df6=df6,df7=df7)))
You can access them with names (but you still need to follow the
structure, i.e. x[[df5]] cannot work):
For df1:
x$uppermost$medium1$df1 or
x[["uppermost"]][["medium1"]][["df1"]]
for df5:
x$uppermost$medium2$lowermost$df5 or
x[["uppermost"]][["medium2"]][["lowermost"]][["df5"]]
But I guess you can write a function that would "search" through the
structure of x for a given df (that has to be named then) and return it.
Maybe you have your reasons to have such a complicated list, but if it
were me, I would make it easier to understand and to access (i.e. less
nested, max 2 levels)
HTH,
Ivan
Le 11/11/2010 09:05, Friedericksen Hope a ?crit :> Hello,
>
> I have a nested named list structure, like the following:
>
> x <- list(
> list(
> list(df1,df2)
> list(df3,
> list(df4,df5))
> list(df6,df7)))
>
> with df1...d7 as data frames. Every data frame is named.
>
> Is there a way to get a specific named element in x?
>
> so, for example,
>
> x[[c("df5")]] gives me the data frame 5?
>
> Thank you in advance!
>
> Best,
> Friedericksen
>
> ______________________________________________
> R-help at 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, self-contained, reproducible code.
>
--
Ivan CALANDRA
PhD Student
University of Hamburg
Biozentrum Grindel und Zoologisches Museum
Abt. S?ugetiere
Martin-Luther-King-Platz 3
D-20146 Hamburg, GERMANY
+49(0)40 42838 6231
ivan.calandra at uni-hamburg.de
**********
http://www.for771.uni-bonn.de
http://webapp5.rrz.uni-hamburg.de/mammals/eng/mitarbeiter.php
Michael Bedward
2010-Nov-11 09:55 UTC
[R] How to get a specific named element in a nested list
Hi Friedericksen,
This function will do it. No doubt there are more elegant ways :)
rmatch <- function(x, name) {
pos <- match(name, names(x))
if (!is.na(pos))
return(x[[pos]])
for (el in x) {
if (class(el) == "list") {
out <- getEl(el, name)
if (!is.null(out)) return(out)
}
}
}
Michael
On 11 November 2010 19:05, Friedericksen Hope
<friedericksen.hope at gmail.com> wrote:> Hello,
>
> I have a nested named list structure, like the following:
>
> x <- list(
> ? ? ? ?list(
> ? ? ? ? ? list(df1,df2)
> ? ? ? ? ? list(df3,
> ? ? ? ? ? ? ? ?list(df4,df5))
> ? ? ? ?list(df6,df7)))
>
> with df1...d7 as data frames. Every data frame is named.
>
> Is there a way to get a specific named element in x?
>
> so, for example,
>
> x[[c("df5")]] gives me the data frame 5?
>
> Thank you in advance!
>
> Best,
> Friedericksen
>
> ______________________________________________
> R-help at 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, self-contained, reproducible code.
>
Janko Thyson
2010-Nov-11 10:06 UTC
[R] How to get a specific named element in a nested list
What you want is some sort of indexing nested lists based on names (as we
are used to for vectors, for example). As Ivan pointed out, I don't think
there's an "out-of-the-box" function in R that supports such
indexing as it
requires some sort of mapping of the nested list's hierarchical structure.
At first I thought one could use the information of 'as.relistable()'
and
'relist()' in some way, but I couldn't really make use of it.
So this is my own solution for retrieving all "branch names" of an
arbitrary
deeply nested list together with their recursive indexes which you then can
use to index/access a branch of your choice. I'm sure there are more elegant
ways, but at least it does the trick ;-). Currently requires that all
branches are named and names at a branch are unique(!). E.g., this is fine:
my.list=list(a=list(a.1=list(...), a.2=list(...)), b=list(...)); something
like this is not supported yet: my.list=list(a=list(a.1=list(...),
a.1=list(...)), a=list(...))). One could use regular expressions to handle
"stubs" of names. Right now you must use the "absolute path
name" (e.g.
"a$a.1$a.1.1) of a branch to access it (you get this info via
'listnames.get()', though). But it should be easy to handle
"stubs" (e.g.
"a.1.1" only) as well.
The two function defs and an example:
##### FUNCTION DEFS #####
listnames.get <- function(
list.obj,
do.basename=FALSE,
do.name.chain=TRUE,
...
)
{
# VALIDATE
if(!is.list(list.obj)) stop("Argument 'list.obj' must be a
list.")
# /
#---------------------------------------------------------------------------
# CORE FUNCTION
#---------------------------------------------------------------------------
listnames.get.core <- function(
# CUSTOM:
list.obj,
do.basename=FALSE,
do.name.chain=TRUE,
buffer,
...
)
{
if(!exists("index", buffer))
{
buffer$index <- new.env(parent=emptyenv())
buffer$index <- NULL
buffer$name <- NULL
}
#x=1
jnk <- sapply(1:length(list.obj), function(x)
{
list.branch <- list.obj[x]
list.branch.nme <- names(list.branch)
if(do.basename) list.branch.nme <-
basename(list.branch.nme)
list.obj.updt <- list.branch[[1]]
# UPDATE BUFFER
buffer$run <- c(buffer$run, x)
if(do.name.chain)
{
# buffer$name <-
paste(buffer$name, list.branch.nme, sep="$")
buffer$name <- c(buffer$name,
list.branch.nme)
} else
{
buffer$name <- list.branch.nme
}
# /
index.crnt <-
paste(as.character(buffer$run), collapse="-")
index.crnt <- data.frame(
name=paste(buffer$name, collapse="$"),
index=index.crnt,
stringsAsFactors=FALSE
)
index.updt <- rbind(buffer$index,
index.crnt)
buffer$index <- index.updt
if(is.list(list.obj.updt))
{
listnames.get.core(
list.obj=list.obj.updt,
do.basename=do.basename,
do.name.chain=do.name.chain,
buffer=buffer
)
}
# UPDATE BUFFER
buffer$run <- buffer$run[-length(buffer$run)]
buffer$name <- buffer$name[-length(buffer$name)]
# /
return(NULL)
})
return(TRUE)
}
# /CORE FUNCTION ----------
#---------------------------------------------------------------------------
# APPLICATION
#---------------------------------------------------------------------------
assign("buffer", new.env(parent=emptyenv()), envir=environment())
listnames.get.core(
list.obj=list.obj,
do.basename=do.basename,
buffer=buffer
)
# /APPLICATION ----------
return(buffer$index)
}
listbranch.get <- function(
list.obj,
query,
do.strict=TRUE,
do.rtn.val=TRUE,
msg.error=NULL,
...
)
{
# VALIDATE
if(!is.list(list.obj)) stop("Argument 'list.obj' must be a
list.")
# /
# ESTABLISH LIST INDEX
list.index <- listnames.get(
list.obj=list.obj,
do.basename=TRUE,
do.name.chain=TRUE
)
list.index.nms <- list.index$name
# /
query.0 <- query
# SEARCH FOR QUERY
if(do.strict)
{
query <- gsub("\\$", "\\\\$", query)
query <- gsub("\\.", "\\\\.", query)
query <- paste("^", query, "$", sep="")
} else
{
stop("'do.strict = FALSE not supported yet as it may result
in multiple results.")
}
idx <- grep(query, list.index.nms, perl=TRUE)
if(!length(idx))
{
if(is.null(msg.error))
{
msg.error <- paste("Query not successful: '",
query.0, "' ('", query, "')", sep="")
}
stop(cat(msg.error, sep="\n"))
}
# /
# BUILDING RECURSIVE INDEX
idx <- list.index$index[idx]
idx <- as.numeric(unlist(strsplit(idx, split="-")))
# /
if(do.rtn.val)
{
# RECURSIVE INDEXING
rtn <- list.obj[[idx]]
# /
} else
{
rtn <- idx
}
return(rtn)
}
##### EXAMPLE #####
my.list <- list(
a=list(a.1="a", a.2=list(a.2.1="a", a.2.2="b"),
a.3=list(a.3.1="a"),
b=list(b.1=list(b.1.1="a"), b.2="b"),
c="a"
))
# RETRIEVE 'COMPLETE' INDEX (A DATA FRAME; NAMES AND INDEX)
listnames.get(list.obj=my.list, do.basename=TRUE, do.name.chain=TRUE)
# GET RECURSIVE INDEX ONLY
idx <- listbranch.get(list.obj=my.list, query="a$a.2$a.2.2",
do.strict=TRUE, do.rtn.val=FALSE)
my.list[[idx]]
# GET RECURSIVELY INDEXED 'BRANCH CONTENT' DIRECTLY
my.list.sub <- listbranch.get(list.obj=my.list,
query="a$a.2$a.2.2",
do.strict=TRUE, do.rtn.val=TRUE)
my.list.sub
Hope this helps,
Janko
> -----Urspr?ngliche Nachricht-----
> Von: r-help-bounces at r-project.org [mailto:r-help-bounces at
r-project.org] Im
> Auftrag von Friedericksen Hope
> Gesendet: Donnerstag, 11. November 2010 09:05
> An: r-help at stat.math.ethz.ch
> Betreff: [R] How to get a specific named element in a nested list
>
> Hello,
>
> I have a nested named list structure, like the following:
>
> x <- list(
> list(
> list(df1,df2)
> list(df3,
> list(df4,df5))
> list(df6,df7)))
>
> with df1...d7 as data frames. Every data frame is named.
>
> Is there a way to get a specific named element in x?
>
> so, for example,
>
> x[[c("df5")]] gives me the data frame 5?
>
> Thank you in advance!
>
> Best,
> Friedericksen
>
> ______________________________________________
> R-help at 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, self-contained, reproducible code.
Michael Bedward
2010-Nov-11 10:45 UTC
[R] How to get a specific named element in a nested list
On 11 November 2010 21:08, Janko Thyson <janko.thyson at ku-eichstaett.de> wrote:> > Could it be that you forgot to supply 'getEL()'? Or do I have to use some > package to make it available? >Oops - no. The problem was me stupidly renaming the function without modifying the code. Try this instead... rmatch <- function(x, name) { pos <- match(name, names(x)) if (!is.na(pos)) return(x[[pos]]) for (el in x) { if (class(el) == "list") { out <- rmatch(el, name) if (!is.null(out)) return(out) } } } Sorry about that. Michael>> -----Urspr?ngliche Nachricht----- >> Von: r-help-bounces at r-project.org [mailto:r-help-bounces at r-project.org] Im >> Auftrag von Michael Bedward >> Gesendet: Donnerstag, 11. November 2010 10:56 >> An: friedericksen.hope at gmail.com >> Cc: r-help at stat.math.ethz.ch >> Betreff: Re: [R] How to get a specific named element in a nested list >> >> Hi Friedericksen, >> >> This function will do it. No doubt there are more elegant ways :) >> >> rmatch <- function(x, name) { >> ? pos <- match(name, names(x)) >> ? if (!is.na(pos)) >> ? ? return(x[[pos]]) >> >> ? for (el in x) { >> ? ? if (class(el) == "list") { >> ? ? ? out <- getEl(el, name) >> ? ? ? if (!is.null(out)) return(out) >> ? ? } >> ? } >> } >> >> >> Michael >> >> On 11 November 2010 19:05, Friedericksen Hope >> <friedericksen.hope at gmail.com> wrote: >> > Hello, >> > >> > I have a nested named list structure, like the following: >> > >> > x <- list( >> > ? ? ? ?list( >> > ? ? ? ? ? list(df1,df2) >> > ? ? ? ? ? list(df3, >> > ? ? ? ? ? ? ? ?list(df4,df5)) >> > ? ? ? ?list(df6,df7))) >> > >> > with df1...d7 as data frames. Every data frame is named. >> > >> > Is there a way to get a specific named element in x? >> > >> > so, for example, >> > >> > x[[c("df5")]] gives me the data frame 5? >> > >> > Thank you in advance! >> > >> > Best, >> > Friedericksen >> > >> > ______________________________________________ >> > R-help at 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, self-contained, reproducible code. >> > >> >> ______________________________________________ >> R-help at 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, self-contained, reproducible code. > >
Possibly Parallel Threads
- Reshaping data with xtabs giving me 'extra' data
- create data frame(s) from a list with different numbers of rows
- Retrieve all names of nested list and index list based on these names
- splitting into multiple dataframes and then create a loop to work
- How to remove rows representing concurrent sessions from data.frame?