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
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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
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