OK. I may completely misunderstand. If you are happy with what Rui and/or
others have given you, **read no further**, as it will just be noise.
Otherwise, I don't think tapply()/ave() etc. will do quite what you want
splitting by the list of separate factors -- or at least not easily. I
think it's simplest just to start by creating a single factor with just the
combinations of levels you have. It turns out here that because you are
ordering lexicographically -- which is what the factor() function will do
by default -- this will make it easy to get back your results in exactly
the form you want. It could be a bit of a hassle if this were not the case.
So first, starting from your already sorted DF3> fac <- with(DF3, factor(paste(State, name, day, sep = '.')))
## which gives:> fac
[1] CA.A.1 CA.A.2 CA.A.2 CA.A.2 CA.A.2 FL.B.3 FL.B.3 FL.B.3 FL.B.3 FL.B.3
[11] FL.B.4 FL.B.4
Levels: CA.A.1 CA.A.2 FL.B.3 FL.B.4 ## ordering the same as in DF3
Next I wrote a little function that I think applies the logic you
specified, yielding TRUE for when you want to raise the flag and FALSE if
not:
yourfun <- function(text, day){
len <- length(text)
if(len == 1) FALSE ## only 1 record in the group
else c(FALSE, (diff(day) < 50 & text[-1] == text[-len]))
## first record gets FALSE as there are none previous
}
Then I use the by() function to apply this groupwise as you specified:
> flag <-with(DF3,
+ by(DF3, fac, function(x)foo(x$text,x$day))
+ )
## Here's what you get> flag
fac: CA.A.1
[1] FALSE
-------------------------------------------------------
fac: CA.A.2
[1] FALSE TRUE FALSE FALSE
-------------------------------------------------------
fac: FL.B.3
[1] FALSE FALSE FALSE FALSE FALSE
-------------------------------------------------------
fac: FL.B.4
[1] FALSE TRUE
This is class "by", essentially a list. So one can use do.call() and
an
implicit cast to numeric to get the x,y flags that you specified:
> flag <- c("y", "x")[do.call(c, flag) + 1]
## yielding> flag
[1] "y" "y" "x" "y" "y"
"y" "y" "y" "y" "y"
"y" "x"
(you could also use the within() function to do this within DF3 and return
the modified DF)
HTH,
Bert
On Tue, Apr 26, 2022 at 10:53 PM Rui Barradas <ruipbarradas at sapo.pt>
wrote:
> Hello,
>
> Maybe something like the following will do it.
> In the ave function, don't forget that diff returns a vector of a
> different length, one less element. So combine with an initial zero.
> Then 1 + FALSE/TRUE equals 1/2 and subset the target vector
c("Y", "X")
> with these indices.
>
>
> i_ddiff <- with(DF3, ave(as.numeric(ddate), State, name, day, FUN = \(x)
> c(0L, diff(x))) < 50)
> DF3$ddiff <- c("Y", "X")[1L + i_ddiff]
>
>
> An alternative is to assign a default "Y" to the new column and
then
> assign "X" where the condition is TRUE. This is easier to read.
>
>
> DF3$ddiff <- "Y"
> DF3$ddiff[i_ddiff] <- "X"
>
>
> Hope this helps,
>
> Rui Barradas
>
> ?s 23:17 de 26/04/2022, Val escreveu:
> > Hi All,
> >
> > I want to flag a record based on the following condition.
> > The variables in the sample data are
> > State, name, day, text, ddate
> >
> > Sort the data by State, name, day ddate,
> >
> > Within State, name, day
> > assign consecutive number for each row
> > find the date difference between consecutive rows,
> > if the difference is less than 50 days and the text string in
> > previous and current rows are the same then flag the record as X,
> > otherwise Y.
> >
> > Here is sample data and my attempt,
> >
> > DF<-read.table(text="State name day text ddate
> > CA A 1 xch 2014/09/16
> > CA A 2 xck 2015/5/29
> > CA A 2 xck 2015/6/18
> > CA A 2 xcm 2015/8/3
> > CA A 2 xcj 2015/8/26
> > FL B 3 xcu 2017/7/23
> > FL B 3 xcl 2017/7/03
> > FL B 3 xmc 2017/7/26
> > FL B 3 xca 2017/3/17
> > FL B 3 xcb 2017/4/8
> > FL B 4 xhh 2017/3/17
> > FL B 4 xhh 2017/1/29",header=TRUE)
> >
> > DF$ddate <- as.Date (as.Date(DF$ddate),
format="%Y/%m/%d" )
> > DF3 <-
DF[order(DF$State,DF$name,DF$day,xtfrm(DF$ddate)), ]
> > DF3$C <- with(DF3, ave(State, name, day, FUN = seq_along))
> > DF3$diff <- with(DF3, ave(as.integer(ddate), State, name,
day,
> > FUN = function(x) x - x[1]))
> >
> > I stopped here, how do I evaluate the previous and the current rows
> > text string and date difference?
> >
> > Desired result,
> >
> >
> > State name day text ddate C diff flag
> > 1 CA A 1 xch 2014-09-16 1 0 y
> > 2 CA A 2 xck 2015-05-29 1 0 y
> > 3 CA A 2 xck 2015-06-18 2 20 x
> > 4 CA A 2 xcm 2015-08-03 3 66 y
> > 5 CA A 2 xcj 2015-08-26 4 89 y
> > 9 FL B 3 xca 2017-03-17 1 0 y
> > 10 FL B 3 xcb 2017-04-08 2 22 y
> > 7 FL B 3 xcl 2017-07-03 3 108 y
> > 6 FL B 3 xcu 2017-07-23 4 128 y
> > 8 FL B 3 xmc 2017-07-26 5 131 y
> > 12 FL B 4 xhh 2017-01-29 1 0 y
> > 11 FL B 4 xhh 2017-03-17 2 47 x
> >
> >
> >
> > Thank you,
> >
> > ______________________________________________
> > R-help at r-project.org mailing list -- To UNSUBSCRIBE and more, see
> > https://stat.ethz.ch/mailman/listinfo/r-help
> > PLEASE do read the posting guide
> http://www.R-project.org/posting-guide.html
> > and provide commented, minimal, self-contained, reproducible code.
>
> ______________________________________________
> R-help at r-project.org mailing list -- To UNSUBSCRIBE and more, see
> https://stat.ethz.ch/mailman/listinfo/r-help
> PLEASE do read the posting guide
> http://www.R-project.org/posting-guide.html
> and provide commented, minimal, self-contained, reproducible code.
>
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