Dear Colleagues,
I am grateful to all of you for helping me with my question, how to write R
code that will identify the first row of each ID within a data frame, create a
variable first=1 for the first row and first=0 for all repeats of the ID.
WOW!!!
I just saw Boris Steipe's answer to my question:
olddata$first <- as.numeric(! duplicated(olddata$ID))
The solution is elegant, short, easy to understand, and it uses base R! All
important characteristics of a good solution, at least for me. While I want to
learn solutions using packages that extend base R, I believe that a good
programmer learns how to do something using the base language and once that is
learned, explores way to solve a programing problem using advanced packages.
Each and every one of you (I hope I did not miss anyone in my list of email
addresses) took the time to read my emails and respond to me. Your collective
help is invaluable, and I am in your collect debt.
Many, many thanks,
John
John David Sorkin M.D., Ph.D.
Professor of Medicine, University of Maryland School of Medicine;
Associate Director for Biostatistics and Informatics, Baltimore VA Medical
Center Geriatrics Research, Education, and Clinical Center;
PI Biostatistics and Informatics Core, University of Maryland School of
Medicine Claude D. Pepper Older Americans Independence Center;
Senior Statistician University of Maryland Center for Vascular Research;
Division of Gerontology and Paliative Care,
10 North Greene Street
GRECC (BT/18/GR)
Baltimore, MD 21201-1524
Cell phone 443-418-5382
________________________________________
From: Bert Gunter <bgunter.4...@gmail.com>
Sent: Sunday, December 1, 2024 11:30 AM
To: Rui Barradas
Cc: Sorkin, John; r-help@r-project.org (r-help@r-project.org)
Subject: Re: [R] Identify first row of each ID within a data frame, create a
variable first =1 for the first row and first=0 of all other rows
Rui:
"f these two, diff is faster. But of all the solutions posted so far,
Ben Bolker's is the fastest."
But the explicit version of diff is still considerably faster:
> D <- c(rep(1,10),rep(2,6),rep(3,2))
> microbenchmark(c(1L,diff(D)), times = 1000L)
Unit: microseconds
expr min lq mean median uq max neval
c(1L, diff(D)) 3.075 3.198 3.34396 3.28 3.362 29.684 1000
> microbenchmark( as.integer(!duplicated(D)), times =1000L)
Unit: microseconds
expr min lq mean median uq max neval
as.integer(!duplicated(D)) 1.476 1.558 1.644264 1.599 1.64 16.4 1000
> microbenchmark( D - c(0L, D[-length(D)]), times = 1000L)
Unit: nanoseconds ## note that unit is nanoseconds not microseconds
expr min lq mean median uq max neval
D - c(0L, D[-length(D)]) 369 410 489.335 492 533 9840 1000
Cheers,
Bert
On Sat, Nov 30, 2024 at 11:05 PM Rui Barradas <ruipbarra...@sapo.pt> wrote:
>
> Às 02:27 de 01/12/2024, Sorkin, John escreveu:
> > Dear R help folks,
> >
> > First my apologizes for sending several related questions to the list
> > server. I am trying to learn how to manipulate data in R . . . and am
> > having difficulty getting my program to work. I greatly appreciate the help
> > and support list member give!
> >
> > I am trying to write a program that will run through a data frame organized
> > by ID and for the first line of each new group of data lines that has the
> > same ID create a new variable first that will be 1 for the first line of
> > the group and 0 for all other lines.
> >
> > e.g. if my original data is
> > olddata
> > ID date
> > 1 1
> > 1 1
> > 1 2
> > 1 2
> > 1 3
> > 1 3
> > 1 4
> > 1 4
> > 1 5
> > 1 5
> > 2 5
> > 2 5
> > 2 5
> > 2 6
> > 2 6
> > 2 6
> > 3 10
> > 3 10
> >
> > the new data will be
> > newdata
> > ID date first
> > 1 1 1
> > 1 1 0
> > 1 2 0
> > 1 2 0
> > 1 3 0
> > 1 3 0
> > 1 4 0
> > 1 4 0
> > 1 5 0
> > 1 5 0
> > 2 5 1
> > 2 5 0
> > 2 5 0
> > 2 6 0
> > 2 6 0
> > 2 6 0
> > 3 10 1
> > 3 10 0
> >
> > When I run the program below, I receive the following error:
> > Error in df[, "ID"] : incorrect number of dimensions
> >
> > My code:
> > # Create data.frame
> > ID <- c(rep(1,10),rep(2,6),rep(3,2))
> > date <- c(rep(1,2),rep(2,2),rep(3,2),rep(4,2),rep(5,2),
> > rep(5,3),rep(6,3),rep(10,2))
> > olddata <- data.frame(ID=ID,date=date)
> > class(olddata)
> > cat("This is the original data frame","\n")
> > print(olddata)
> >
> > # This function is supposed to identify the first row
> > # within each level of ID and, for the first row, set
> > # the variable first to 1, and for all rows other than
> > # the first row set first to 0.
> > mydoit <- function(df){
> > value <- ifelse (first(df[,"ID"]),1,0)
> > cat("value=",value,"\n")
> > df[,"first"] <- value
> > }
> > newdata <- aggregate(olddata,list(olddata[,"ID"]),mydoit)
> >
> > Thank you,
> > John
> >
> >
> > John David Sorkin M.D., Ph.D.
> > Professor of Medicine, University of Maryland School of Medicine;
> > Associate Director for Biostatistics and Informatics, Baltimore VA Medical
> > Center Geriatrics Research, Education, and Clinical Center;
> > PI Biostatistics and Informatics Core, University of Maryland School of
> > Medicine Claude D. Pepper Older Americans Independence Center;
> > Senior Statistician University of Maryland Center for Vascular Research;
> >
> > Division of Gerontology and Paliative Care,
> > 10 North Greene Street
> > GRECC (BT/18/GR)
> > Baltimore, MD 21201-1524
> > Cell phone 443-418-5382
> >
> >
> >
> > ______________________________________________
> > R-help@r-project.org mailing list -- To UNSUBSCRIBE and more, see
> > https://stat.ethz.ch/mailman/listinfo/r-help
> > PLEASE do read the posting guide
> > https://www.r-project.org/posting-guide.html
> > and provide commented, minimal, self-contained, reproducible code.
> Hello,
>
> And here are two other solutions.
>
>
> olddata$first <- with(olddata, ave(seq_along(ID), ID, FUN = \(x) x ==
> x[1L]))
>
> olddata$first <- c(1L, diff(olddata$ID))
>
>
> Of these two, diff is faster. But of all the solutions posted so far,
> Ben Bolker's is the fastest. And it can be made a little faster if
> as.integer substitutes for as.numeric.
> And dplyr::mutate now has a .by argument, which avoids explicit the call
> to group_by, with a performance gain.
>
>
> library(microbenchmark)
>
> mb <- microbenchmark(
> ave = with(olddata, ave(seq_along(ID), ID, FUN = \(x) x == x[1L])),
> dup_num = as.numeric(! duplicated(olddata$ID)),
> dup_int = as.integer(! duplicated(olddata$ID)),
> diff = diff = c(1L, diff(olddata$ID)),
> dplyr_grp = olddata %>% group_by(ID) %>% mutate(first =
> as.integer(row_number() == 1)),
> dplyr = olddata %>% mutate(first = as.integer(row_number() == 1), .by
> = ID)
> )
> print(mb, order = "median")
>
>
>
> However, note that dplyr operates in entire data.frames and therefore is
> expected to be slower when tested against instructions that process one
> column only.
>
>
> Hope this helps,
>
> Rui Barradas
>
>
> --
> Este e-mail foi analisado pelo software antivírus AVG para verificar a
> presença de vírus.
> http://www.avg.com/
>
> ______________________________________________
> R-help@r-project.org mailing list -- To UNSUBSCRIBE and more, see
> https://stat.ethz.ch/mailman/listinfo/r-help
> PLEASE do read the posting guide https://www.r-project.org/posting-guide.html
> and provide commented, minimal, self-contained, reproducible code.
______________________________________________
R-help@r-project.org mailing list -- To UNSUBSCRIBE and more, see
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide https://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
