# [R] creating a data frame from a list

Stephen Tucker brown_emu at yahoo.com
Fri Apr 6 09:40:24 CEST 2007

```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
> NAs introduced by coercion
>
> and z was not created.
>
> Any ideas on what is going on here?
> Thank you,
> Dimitri
>
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> and provide commented, minimal, self-contained, reproducible code.
>

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