r large "list" over 1 million rows. It is possible to
work with a matrix in ggplot2?
Jeff
-Original Message-
From: Duncan Murdoch
Sent: Saturday, July 24, 2021 12:03 PM
To: reichm...@sbcglobal.net; R-help@r-project.org
Subject: Re: [R] Converting a list to a data frame
Oth
Thanks for the tips
-Original Message-
From: Rui Barradas
Sent: Saturday, July 24, 2021 11:40 AM
To: reichm...@sbcglobal.net; R-help@r-project.org
Subject: Re: [R] Converting a list to a data frame
Hello,
This should do it:
as.data.frame(weight_chains$mcmc)
The only list member
over 1 million rows. It is possible to
work with a matrix in ggplot2?
Jeff
-Original Message-
From: Duncan Murdoch
Sent: Saturday, July 24, 2021 12:03 PM
To: reichm...@sbcglobal.net; R-help@r-project.org
Subject: Re: [R] Converting a list to a data frame
Others have shown you how to
uly 24, 2021 12:03 PM
To: reichm...@sbcglobal.net; R-help@r-project.org
Subject: Re: [R] Converting a list to a data frame
Others have shown you how to extract the matrix and convert it to a dataframe.
My only addition is to suggest that you don't do this:
matrix methods are often much
Others have shown you how to extract the matrix and convert it to a
dataframe. My only addition is to suggest that you don't do this:
matrix methods are often much more efficient than dataframe methods, so
if you can work with the matrix without conversion, you'll often find
things run a lot f
Hello,
This should do it:
as.data.frame(weight_chains$mcmc)
The only list member already has a dim attribute of length 2 and
dimnames' 2nd member are the colnames, just coerce to df.
Hope this helps,
Rui Barradas
Às 14:18 de 24/07/21, Jeff Reichman escreveu:
How does one convert a list
Here is a reprex that does what I think you want:
ex <- list(mcmc = matrix(1:12, ncol = 3, byrow=TRUE),
NULL, c("a","b", "s"))
dex <- data.frame(ex$mcmc)
names(dex) <- ex[[3]]
> dex
a b s
1 1 2 3
2 4 5 6
3 7 8 9
4 10 11 12
If this is not correct, you should provide a **
How does one convert a list into a data frame?
> str(weight_chains)
List of 1
$ : 'mcmc' num [1:10, 1:3] -105 -105 -105 -104 -103 ...
..- attr(*, "dimnames")=List of 2
.. ..$ : NULL
.. ..$ : chr [1:3] "a" "b" "s"
..- attr(*, "mcpar")= num [1:3] 1001 101000 1
- attr(*, "class"
Oh, how funny, hence the term Novice usR. UGH!
Thank you Sir.
WHP
From: Kevin E. Thorpe [mailto:kevin.tho...@utoronto.ca]
Sent: Friday, May 04, 2018 9:08 AM
To: Bill Poling ; Huzefa Khalil
Cc: R Help Mailing List
Subject: Re: [R] Converting a list to a data frame
It looks like you made a
"ata.frame"?
William H. Poling, Ph.D.
From: R-help [mailto:r-help-boun...@r-project.org] On Behalf Of Huzefa Khalil
Sent: Wednesday, May 02, 2018 1:24 PM
To: Kevin E. Thorpe
Cc: R Help Mailing List
Subject: Re: [R] Converting a list to a data frame
Hi Kevin,
There is probably a better
May 02, 2018 1:24 PM
*To:* Kevin E. Thorpe
*Cc:* R Help Mailing List
*Subject:* Re: [R] Converting a list to a data frame
Hi Kevin,
There is probably a better way, but it can be done in two steps like this
temp <- list(A=data.frame(x=1:2, y=3:4),B=data.frame(x=5:6,y=7:8))
temp <- lapply
If you require that the 'type' column be a factor with a level for each
element of the input list, then you need to do that after calling
dplyr::bind_rows(), just as with the base-R solutions.
> l <- list( A = data.frame(X=1:2, Y=11:12), B = data.frame(X=integer(),
Y=integer()), C = data.frame(X=3
On 05/03/2018 01:28 PM, Hadley Wickham wrote:
On Wed, May 2, 2018 at 11:53 AM, Jeff Newmiller
wrote:
Another approach:
library(tidyr)
L <- list( A = data.frame( x=1:2, y=3:4 )
, B = data.frame( x=5:6, y=7:8 )
)
D <- data.frame( Type = names( L )
, s
On Wed, May 2, 2018 at 11:53 AM, Jeff Newmiller
wrote:
> Another approach:
>
>
> library(tidyr)
> L <- list( A = data.frame( x=1:2, y=3:4 )
> , B = data.frame( x=5:6, y=7:8 )
> )
> D <- data.frame( Type = names( L )
>, stringsAsFactors = FALSE
>
A NA ...
--
David, the other.
>
> David C
>
> -Original Message-
> From: R-help On Behalf Of David L Carlson
> Sent: Wednesday, May 2, 2018 3:51 PM
> To: William Dunlap ; Kevin E. Thorpe
>
> Cc: r-help mailing list
> Subject: Re: [R] Converting a list to
p
and David Carlson is beautifully elegant.
Thanks again,
Kevin
-Original Message-
From: R-help On Behalf Of David L Carlson
Sent: Wednesday, May 2, 2018 3:51 PM
To: William Dunlap ; Kevin E. Thorpe
Cc: r-help mailing list
Subject: Re: [R] Converting a list to a data frame
Or add th
William Dunlap ; Kevin E. Thorpe
>
> Cc: r-help mailing list
> Subject: Re: [R] Converting a list to a data frame
>
> Or add the type column first and then rbind:
>
> x <- list(A=data.frame(x=1:2, y=3:4),B=data.frame(x=5:6,y=7:8))
> x2 <- do.call(rbind, lapply(names(
On 05/03/2018 07:58 AM, Jeff Newmiller wrote:
This is very nice to learn about, Denis, but it seems only fair to point out
that the result of rbindlist is not a data frame. You can convert it to a data
frame easily, but the copy and indexing semantics of data tables are quite
different than
This is very nice to learn about, Denis, but it seems only fair to point out
that the result of rbindlist is not a data frame. You can convert it to a data
frame easily, but the copy and indexing semantics of data tables are quite
different than data tables, which could be a real headache for s
To: William Dunlap ; Kevin E. Thorpe
Cc: r-help mailing list
Subject: Re: [R] Converting a list to a data frame
Or add the type column first and then rbind:
x <- list(A=data.frame(x=1:2, y=3:4),B=data.frame(x=5:6,y=7:8))
x2 <- do.call(rbind, lapply(names(x), function(z)
data.fra
as A&M University
College Station, TX 77843-4352
-Original Message-
From: R-help On Behalf Of William Dunlap via
R-help
Sent: Wednesday, May 2, 2018 12:28 PM
To: Kevin E. Thorpe
Cc: R Help Mailing List
Subject: Re: [R] Converting a list to a data frame
> x1 <- do.call(
On 05/02/2018 07:11 PM, Kevin E. Thorpe wrote:
I suspect this is pretty easy, but I'm having trouble figuring it out.
Basically, I have a list of data frames such as the following example:
list(A=data.frame(x=1:2, y=3:4),B=data.frame(x=5:6,y=7:8))
I would like to turn this into data frame w
Another approach:
library(tidyr)
L <- list( A = data.frame( x=1:2, y=3:4 )
, B = data.frame( x=5:6, y=7:8 )
)
D <- data.frame( Type = names( L )
, stringsAsFactors = FALSE
)
D$data <- L
unnest(D, data)
#> Type x y
#> 1A 1 3
#> 2A
On Wed, 2 May 2018, Kevin E. Thorpe wrote:
I suspect this is pretty easy, but I'm having trouble figuring it out.
Basically, I have a list of data frames such as the following example:
list(A=data.frame(x=1:2, y=3:4),B=data.frame(x=5:6,y=7:8))
I would like to turn this into data frame where
> x1 <- do.call(rbind, c(x, list(make.row.names=FALSE)))
> x2 <- cbind(type=rep(names(x), vapply(x, nrow, 0)), x1)
> str(x2)
'data.frame': 4 obs. of 3 variables:
$ type: Factor w/ 2 levels "A","B": 1 1 2 2
$ x : int 1 2 5 6
$ y : int 3 4 7 8
Bill Dunlap
TIBCO Software
wdunlap tibco.co
Hi Kevin,
There is probably a better way, but it can be done in two steps like this
temp <- list(A=data.frame(x=1:2, y=3:4),B=data.frame(x=5:6,y=7:8))
temp <- lapply(names(temp), function(n, temp) {
temp[[n]]$type <- n
return(temp[[n]])
}, temp = temp)
do.call(rbind, temp)
On Wed, May 2,
I suspect this is pretty easy, but I'm having trouble figuring it out.
Basically, I have a list of data frames such as the following example:
list(A=data.frame(x=1:2, y=3:4),B=data.frame(x=5:6,y=7:8))
I would like to turn this into data frame where the list elements are
essentially rbind'ed t
I believe that a slightly more efficient way of doing this without
leaving base R is:
cbind(do.call(rbind,x), set = rep(seq_along(x), vapply(x,nrow,1)) )
Cheers,
Bert
Bert Gunter
"The trouble with having an open mind is that people keep coming along
and sticking things into it."
-- Opus (aka
Hi,
You have also "rbindlist()" function in package "data.table" that does
exactly what you need.
Kind Regards,
Carlos Ortega
www.qualityexcellence.es
2016-11-04 13:37 GMT+01:00 Kevin E. Thorpe :
> There is probably a very simple elegant way to do this, but I have been
> unable to find it. He
Hi Kevin,
There may be a more elegant way but the following do.call and lapply should
solve your problem.
do.call(rbind, lapply(seq(length(x)), function(i) data.frame(set=i,
x[[i]])))
Regards,
Charles
On Fri, Nov 4, 2016 at 7:37 AM, Kevin E. Thorpe
wrote:
> There is probably a very simple ele
There is probably a very simple elegant way to do this, but I have been
unable to find it. Here is a toy example. Suppose I have a list of data
frames like this.
print(x <-
list('1'=data.frame(id=1:4,expand.grid(x1=0:1,x2=0:1)),'2'=data.frame(id=5:8,expand.grid(x1=2:3,x2=2:3
$`1`
id x1
Farrel Buchinsky-3 wrote:
>
> I have a column in which dates and times are specified thus
> m/d/ HH:MM:SS
> Alas, some entries do not include the time and therefore are only
> m/d/
> so I used read.csv and specified that the relevant column should be read
> as
> is and it remained as a
I have a column in which dates and times are specified thus
m/d/ HH:MM:SS
Alas, some entries do not include the time and therefore are only
m/d/
so I used read.csv and specified that the relevant column should be read as
is and it remained as a character variable.
I then split the value on
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