Thanks a lot. That worked perfectly.
Best,Farnoosh Sheikhi
Sent: Wednesday, August 28, 2013 10:37 AM
Subject: Re: list to data frame
If the expected result is data.frame()
as.data.frame(t(as.data.frame(lapply(dat1,sum# would be data.frame. But,
not
Hi,
set.seed(249)
dat1<- as.data.frame(matrix(sample(1:20,40*20,replace=TRUE),ncol=20))
#Based on your code:
vecNew<-sort(t(as.data.frame(lapply(dat1,sum))),decreasing=TRUE)[1:10]
#or
vec1<-sort(unlist(lapply(dat1,sum),use.names=FALSE),decreasing=TRUE)[1:10]
identical(vec1,vecNew)
#[1] TRUE
#
On Sun, Apr 10, 2011 at 06:01:39PM +, Franklin Tamborello II wrote:
> I need to make a data frame out of the data that I currently have in
> a list. This works, but is ugly:
> ineffData<-rbind(ineffFilesList[[1]], ineffFilesList[[2]],
> ineffFilesList[[3]], ineffFilesList[[4]], ineffFilesList[[
Hi Franklin,
Try
do.call(rbind, ineffFilesList)
See ?do.call for more details.
HTH,
Jorge
On Sun, Apr 10, 2011 at 2:01 PM, Franklin Tamborello II <> wrote:
> I need to make a data frame out of the data that I currently have in a
> list. This works, but is ugly:
> ineffData<-rbind(ineffFilesL
Thanks a lot!
This solves my problem!
Joh
On Monday 26 July 2010 17:06:37 Joshua Wiley wrote:
> Hi,
>
> Here is another option if you already have a list you want to convert.
> This will handle different elements of the list being different
> lengths.
>
> #Using your example data
> mydata <-
Hi,
Here is another option if you already have a list you want to convert.
This will handle different elements of the list being different
lengths.
#Using your example data
mydata <- list(c(1,2,3),c(4,5,6))
data.frame(
OriginalListIndex = rep(x = seq_along(mydata),
times = unlist(lapply(myd
Try this:
stack(data.frame(list('A' = c(1,2,3), 'B' = c(4,5,6
On Mon, Jul 26, 2010 at 11:46 AM, Johannes Graumann <
johannes_graum...@web.de> wrote:
> Hi,
>
> Any ideas on how to efficiently convert
>
> > list(c(1,2,3),c(4,5,6))
>
> to
>
> > data.frame(OriginalListIndex=c(1,1,1,2,2,2),Item=c
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