On May 22, 2013, at 5:00 AM, catalin roibu wrote:
> Hello all!
> I have a problem to transform this list in a data frame. I try
> this command as.data.frame, but unsuccessfully.
> Please help me!
>
> thank you very much!
>
> structure(list(fns = list(structure(list(r = c(0, 0.048828125,
> 0.097
Thanks again,
A.K.
- Original Message -
From: David L Carlson
To: 'arun' ; 'R help'
Cc:
Sent: Sunday, July 1, 2012 5:09 PM
Subject: RE: [R] list to dataframe conversion-testing for identical
Yes it does have something to do with the representation of floating point
numbers.
HI All,
Thanks for your replies.
A.K.
- Original Message -
From: David Winsemius
To: arun
Cc: R help
Sent: Sunday, July 1, 2012 6:31 PM
Subject: Re: [R] list to dataframe conversion-testing for identical
On Jul 1, 2012, at 5:09 PM, David L Carlson wrote:
> Yes it does h
:3)
dat1 <- data.frame(listdat1)
--
David L Carlson
Associate Professor of Anthropology
Texas A&M University
College Station, TX 77843-4352
-Original Message-
From: r-help-boun...@r-project.org [mailto:r-help-bounces@r-
project.org] On
Hello,
But
> all.equal(dat1,dat2)
[1] TRUE
So I guess it does have to do with floating-point equality, all.equal
uses .Machine$double.eps. (Which could return FALSE on ocasions we would
expect TRUE, when, for instance, the tolerance could/should be
.Machine$double.eps^0.5.)
Rui Barradas
E
ation, TX 77843-4352
> -----Original Message-
> From: r-help-boun...@r-project.org [mailto:r-help-bounces@r-
> project.org] On Behalf Of arun
> Sent: Sunday, July 01, 2012 12:56 PM
> To: R help
> Subject: [R] list to dataframe conversion-testing for identical
>
> HI
HI R help,
I was trying to get identical data frame from a list using two methods.
#Suppose my list is:
listdat1<-list(rnorm(10,20),rep(LETTERS[1:2],5),rep(1:5,2))
#Creating dataframe using cbind
dat1<-data.frame(do.call("cbind",listdat1))
colnames(dat1)<-c("Var1","Var2","Var3")
#Second datafram
Does this do what you want:
> list <- list(A=1:4, B=1:6, C=1:9)
> result <- lapply(names(list), function(x){
+ data.frame(name = x
+ , length = length(list[[x]])
+ , gt5 = sum(list[[x]] > 5)
+ , lt5 = sum(list[[x]] < 5)
+ )
+ })
> do.call(rbind, result)
name l
Hi,
> Try
>
> list <- list(1:4, 1:6, 1:9)
> t(sapply(list, function(x) c(length(x), sum(x > 5), sum(x < 5
>
thank you...the sapply approach seems straight forward, although I don't get
the names into an own column... When the list elements are named the name is
used for the rownames. I'd
Try
list <- list(1:4, 1:6, 1:9)
t(sapply(list, function(x) c(length(x), sum(x > 5), sum(x < 5
HTH,
Jorge.-
On Wed, Feb 8, 2012 at 8:50 AM, Johannes Radinger <> wrote:
> Hi,
> I want to "melt" my list and get certain deskriptive factors (length of a
> vector etc.) into a dataframe. Best to
Hi,
I want to "melt" my list and get certain deskriptive factors (length of a
vector etc.) into a dataframe. Best to describe it with an example:
A <- seq(4)
B <- seq(6)
C <- seq(9)
ls <- list(A,B,C) # this is my list with vectors of different length
# thats the dataframe how it should look lik
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