Dear all, A few months ago, I asked for your help on the following problem: I have a list with three (named) numeric vectors:> lst = list(a=c(A=1,B=8) , b=c(A=2,B=3,C=0), c=c(B=2,D=0) ) > lst$a A B 1 8 $b A B C 2 3 0 $c B D 2 0 Now, I'd love to use this list to create the following data frame:> dtf = data.frame(a=c(A=1,B=8,C=NA,D=NA),+ b=c(A=2,B=3,C=0,D=NA), + c=c(A=NA,B=2,C=NA,D=0) )> dtfa b c A 1 2 NA B 8 3 2 C NA 0 NA D NA NA 0 That is, I wish to "merge" the three vectors in the list into a data frame by their "(row)"names. And I got the following answer: library(zoo) z <- do.call(merge, lapply(lst, function(x) zoo(x, names(x)))) rownames(z) <- time(z) coredata(z) However, it does not seem to be working. Here's what I get when I try it:> lst = list(a=c(A=1,B=8) , b=c(A=2,B=3,C=0), c=c(B=2,D=0) ) > library(zoo) > z <- do.call(merge, lapply(lst, function(x) zoo(x, names(x))))Error in if (freq > 1 && identical(all.equal(freq, round(freq)), TRUE)) freq <- round(freq) : missing value where TRUE/FALSE needed In addition: Warning message: NAs introduced by coercion and z was not created. Any ideas on what is going on here? Thank you, Dimitri
Dmitri:
As you apparently have not received a reply....
IMHO, one of the glories of R is the ease with which you can create de novo
solutions for little problems like this yourself. While there may be more
efficient,robust, and elegant solutions already available, it can frequently
be considerably more time consuming to find and figure them out, as you
appear to have experienced. (And once outside base R and standard packages,
documentation can be problematic).
Anyway, whether you agree with that propoganda or not, here is a little
function (no claim for elegance or efficiency!) that does what you want, I
think:
makeFrame<-function(xlist)
{
allnames <- sort(unique(unlist(sapply(xlist,names))))
data.frame(lapply(xlist,function(y,an)structure(y[match(an,names(y))],
names=NULL),
an=allnames),row.names=allnames)
}
##test it
> lst
$a
A B
1 8
$b
A B C
2 3 0
$c
B D
2 0
> makeFrame(lst)
a b c
A 1 2 NA
B 8 3 2
C NA 0 NA
D NA NA 0
Cheers,
Bert Gunter
Genentech Nonclinical Statistics
South San Francisco, CA 94404
650-467-7374
-----Original Message-----
From: r-help-bounces at stat.math.ethz.ch
[mailto:r-help-bounces at stat.math.ethz.ch] On Behalf Of Dimitri Szerman
Sent: Thursday, April 05, 2007 11:58 AM
To: R-Help
Subject: [R] creating a data frame from a list
Dear all,
A few months ago, I asked for your help on the following problem:
I have a list with three (named) numeric vectors:
> lst = list(a=c(A=1,B=8) , b=c(A=2,B=3,C=0), c=c(B=2,D=0) )
> lst
$a
A B
1 8
$b
A B C
2 3 0
$c
B D
2 0
Now, I'd love to use this list to create the following data frame:
> dtf = data.frame(a=c(A=1,B=8,C=NA,D=NA),
+ b=c(A=2,B=3,C=0,D=NA),
+ c=c(A=NA,B=2,C=NA,D=0) )
> dtf
a b c
A 1 2 NA
B 8 3 2
C NA 0 NA
D NA NA 0
That is, I wish to "merge" the three vectors in the list into a data
frame
by their "(row)"names.
And I got the following answer:
library(zoo)
z <- do.call(merge, lapply(lst, function(x) zoo(x, names(x))))
rownames(z) <- time(z)
coredata(z)
However, it does not seem to be working. Here's what I get when I try it:
> lst = list(a=c(A=1,B=8) , b=c(A=2,B=3,C=0), c=c(B=2,D=0) )
> library(zoo)
> z <- do.call(merge, lapply(lst, function(x) zoo(x, names(x))))
Error in if (freq > 1 && identical(all.equal(freq, round(freq)),
TRUE)) freq <- round(freq) :
missing value where TRUE/FALSE needed
In addition: Warning message:
NAs introduced by coercion
and z was not created.
Any ideas on what is going on here?
Thank you,
Dimitri
______________________________________________
R-help at stat.math.ethz.ch 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.
Hi Dimitri,
You can try this one if you'd like:
lst = list(a=c(A=1,B=8) , b=c(A=2,B=3,C=0), c=c(B=2,D=0))
# get unique names
nms <- unique(rapply(lst,function(x) names(x)))
# create a vector of NA's and then fill it according
# to matching names for each element of list
doit <- function(x,nms) {
y <- rep(NA,length(nms)); names(y) <- nms
y[match(names(x),names(y))] <- x
return(y)
}
# apply it to the data
dtf <- as.data.frame(sapply(lst,doit,nms))
--- Dimitri Szerman <dimitrijoe at gmail.com> wrote:
> Dear all,
>
> A few months ago, I asked for your help on the following problem:
>
> I have a list with three (named) numeric vectors:
>
> > lst = list(a=c(A=1,B=8) , b=c(A=2,B=3,C=0), c=c(B=2,D=0) )
> > lst
> $a
> A B
> 1 8
>
> $b
> A B C
> 2 3 0
>
> $c
> B D
> 2 0
>
> Now, I'd love to use this list to create the following data frame:
>
> > dtf = data.frame(a=c(A=1,B=8,C=NA,D=NA),
> + b=c(A=2,B=3,C=0,D=NA),
> + c=c(A=NA,B=2,C=NA,D=0) )
>
> > dtf
> a b c
> A 1 2 NA
> B 8 3 2
> C NA 0 NA
> D NA NA 0
>
> That is, I wish to "merge" the three vectors in the list into a
data frame
> by their "(row)"names.
>
> And I got the following answer:
>
> library(zoo)
> z <- do.call(merge, lapply(lst, function(x) zoo(x, names(x))))
> rownames(z) <- time(z)
> coredata(z)
>
> However, it does not seem to be working. Here's what I get when I try
it:
>
> > lst = list(a=c(A=1,B=8) , b=c(A=2,B=3,C=0), c=c(B=2,D=0) )
> > library(zoo)
> > z <- do.call(merge, lapply(lst, function(x) zoo(x, names(x))))
> Error in if (freq > 1 && identical(all.equal(freq, round(freq)),
> TRUE)) freq <- round(freq) :
> missing value where TRUE/FALSE needed
> In addition: Warning message:
> NAs introduced by coercion
>
> and z was not created.
>
> Any ideas on what is going on here?
> Thank you,
> Dimitri
>
> ______________________________________________
> R-help at stat.math.ethz.ch 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.
>
____________________________________________________________________________________
Be a PS3 game guru.