If you have an algorithm that only works on sorted data, it is easy to
write a function that sorts [a copy of] the data, applies the algorithm,
then puts the result back in the order of the original data. E.g.,
f <- function (data) {
ord <- with(data, order(TT, ID, FY)) # data[ord,] will be sorted in your
required order
data$EXCL3 <- 1 * duplicated(data[ord, 1:2], fromLast = TRUE)[order(ord)] #
[order(ord)] puts it back in original order
data
}
E.g.,
> i <- c(12, 5, 10, 6, 4, 2, 1, 3, 7, 11, 9, 8)
> scrambled <- HTDF[i,]
> f(scrambled)
FY ID TT EXCL3
12 FY13 2 TER 0
5 FY12 1 TER 0
10 FY12 2 HC 0
6 FY09 2 HC 0
4 FY12 1 HC 1
2 FY10 1 HC 0
1 FY09 1 HC 0
3 FY11 1 HC 0
7 FY10 2 HC 1
11 FY13 2 HC 1
9 FY11 2 HC 0
8 FY10 2 TER 0
Or is your dataset so large that this sorting and unsorting would take too
long or too much space?
(There are faster ways of doing this than duplicated(), but the details depend
on some details like whether or not there may be more than 2 FY/ID duplicates.]
Bill Dunlap
Spotfire, TIBCO Software
wdunlap tibco.com
> -----Original Message-----
> From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org] On
> Behalf
> Of Lopez, Dan
> Sent: Monday, November 11, 2013 12:50 PM
> To: R help (r-help@r-project.org)
> Subject: [R] How do I derive a logical variable in a dataframe based on
> another row in the
> same dataframe?
>
> Hi R Experts,
>
> How do I mark rows in dataframe based on a condition that's based off another
> row in
> the same dataframe?
>
> I want to mark any combination of FY,ID, TT=='HC' rows that have a
> FY,ID,TT=='TER' row
> with a 1. In my example below this is rows 4, 7 and 11.
> My data looks something like this:
> FY ID TT
> 1 FY09 1 HC
> 2 FY10 1 HC
> 3 FY11 1 HC
> 4 FY12 1 HC
> 5 FY12 1 TER
> 6 FY09 2 HC
> 7 FY10 2 HC
> 8 FY10 2 TER
> 9 FY11 2 HC
> 10 FY12 2 HC
> 11 FY13 2 HC
> 12 FY13 2 TER
>
> I know for this specific example I can use:
> HTDF$EXCL3<-1*duplicated(HTDF[,1:2],fromLast=T)
>
> However my actual data set is NOT sorted by FY, ID and TT. TT is a binary
> factor variable.
> I want to know if there is another way of doing the same thing without
> sorting the data.
> I tried the last line of code below but it gave me unexpected results. It
> marks the first
> three rows with 0 and everything else with 1. Based on the warning messages
> looks like
> it has something to do with longer object length is not a multiple of shorter
> object
> length. But I am now stumped.
>
> #REPRODUCIBLE EXAMPLE
> FY<-
> factor(c("FY09","FY10","FY11","FY12","FY12","FY09","FY10","FY10","FY11","FY12","FY13
> ","FY13"))
> ID<-c(rep(1,5),rep(2,7))
> TT<-factor(c(rep("HC",4),"TER","HC","HC","TER","HC","HC","HC","TER"))
> HTDF<-data.frame(FY,ID,TT)
>
> #Summarize data and get max TT. TT is a binary factor variable
> library(sqldf)
> HTDF.MAX<-sqldf('SELECT ID,FY,Max(TT) "MAXTT" FROM HTDF GROUP BY ID,FY')
>
> # Initiate new variable and assign 0 or 1
> HTDF$EXCL<-0
>
> # THIS IS WHERE I AM GETTING UNEXPECTE RESULTS
> HTDF$EXCL<-
> ifelse(HTDF$FY==HTDF.MAX$FY&HTDF$ID==HTDF.MAX$ID&HTDF$TT==HTDF.MAX$MAX
> TT,0,1)
>
>
> Dan Lopez
> Workforce Analyst
> LLNL
> HRIM - Workforce Analytics & Metrics
>
> [[alternative HTML version deleted]]
>
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