The grouping solutions offered seem to be the obvious way to do this and
may even be more efficient in R then what follows below. However, note
that they are to some extent doing unnecessary work, since the ordering in
the data frame already implicitly provides the grouping, and the hashing or
wha
In addition to many good solutions already provided, this solution uses
data.table package.
library(data.table)
mydf <- 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)))
setDT(mydf)
mydf[, `:=`(firstdate = with(.S
Very similar to what Oliver posted:
library(dplyr)
newdata <- olddata |>
group_by(ID) |>
mutate(firstdate = first(date))
newdata
1) I attached dplyr to the entire program. Oliver used dplyr::group_by() and
dplyr::mutate() to do the same thing.
2) I used the base R |> pipe while Oliver used
On 11/27/24 08:30, Sorkin, John wrote:
> I am an old, long time SAS programmer. I need to produce R code that
> processes a dataframe in a manner that is equivalent to that produced by
> using a by statement in SAS and an if first.day statement and a retain
> statement:
>
> I want to take data
Oh and don't forget:
#first line of code, bring dplyr into memory after that package has been
installed.
library(dplyr)
On Wednesday, November 27th, 2024 at 12:05 PM, Tom Woolman
wrote:
>
>
> Check out the dplyr package, specifically the mutate function.
>
> # Create new column based o
Dear John,
Considering that you've got the following dataframe:
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))
df <- data.frame(ID, date)
I would suggest to go this way:
newdf <- df %>% dplyr::group_by(ID) %>%
Check out the dplyr package, specifically the mutate function.
# Create new column based on existing column value
df <- df %>% mutate(FirstDay = if(ID = 2, 5))
df
Repeat as needed to capture all of the day/firstday combinations you want to
account for.
Like everything else in R, there are
I would use base R.
newdata <- cbind(olddata, FirstDay=olddata$date)
newdata$FirstDay <- with(newdata, {
for (thisID in unique(ID))
FirstDay[ID==thisID] <- FirstDay[ID==thisID][1]
FirstDay}
)
newdata
note that both my solution and Olivier have newdata$FirstDay[17:18] == 10
which is wha
Was wondering when this would be suggested. But the question was about getting
the final dataframe...
newdta <- olddta
newdta$FirstDay <- ave(newdata$date, newdata$ID, FUN = \(x) x[1L])
On November 27, 2024 11:13:49 AM PST, Rui Barradas wrote:
>Às 16:30 de 27/11/2024, Sorkin, John escreveu:
>>
I am an old, long time SAS programmer. I need to produce R code that processes
a dataframe in a manner that is equivalent to that produced by using a by
statement in SAS and an if first.day statement and a retain statement:
I want to take data (olddata) that looks like this
ID Day
1 1
Às 16:30 de 27/11/2024, Sorkin, John escreveu:
I am an old, long time SAS programmer. I need to produce R code that processes
a dataframe in a manner that is equivalent to that produced by using a by
statement in SAS and an if first.day statement and a retain statement:
I want to take data (ol
On 11/27/24 08:30, Sorkin, John wrote:
> I am an old, long time SAS programmer. I need to produce R code that
> processes a dataframe in a manner that is equivalent to that produced by
> using a by statement in SAS and an if first.day statement and a retain
> statement:
>
> I want to take data
Here is another version using for loops.
newdata3 <- olddata |>
dplyr::arrange(ID, date)
newdata3$firstday <- NA
for (i in 1:nrow(newdata3)) {
if (i == 1) {
dayz <- newdata3$date[i]
} else {
if (newdata3$ID[i] != newdata3$ID[i - 1]) {
dayz <- newdata3$date[i]
}
}
newdat
On 11/27/24 09:44, David Winsemius via R-help wrote:
On 11/27/24 08:30, Sorkin, John wrote:
I am an old, long time SAS programmer. I need to produce R code that processes
a dataframe in a manner that is equivalent to that produced by using a by
statement in SAS and an if first.day statement
John,
If I understood you, you want to take the minimum value of Day for each
grouping by ID and add a new column to contain that. Right?
There are likely many ways to do this in base R, but I prefer the
dplyr/tidyverse package in which you can use group_by(ID) piped to
mutate(FirstDay = min(Day
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