Hi Ben,
I would look into ?findInterval() or ?cut() for an easier solution.
indx<- match(df[,1],as.POSIXct(dt))
indx2<- unique(df[,2])
lst1<- lapply(split(indx,((seq_along(indx)-1)%/%2)+1),function(x) seq(x[1],
x[2]))
res <- unlist(lapply(seq_along(lst1),function(i) {
val<-rep(indx2[i],length(lst1[[i]]))
names(val)<-lst1[[i]]
val
}))
res1<-res[match(seq_along(dt),names(res))]
res1[is.na(res1)]<- 0
names(res1)<- NULL
res1
# [1] 1 1 1 2 2 2 3 3 3 0 0 4 4 4 4 4 4 4 4 4 4 4 4 0
identical(id,res1)
#[1] TRUE
On Thursday, October 10, 2013 8:10 PM, Benjamin Gillespie <gy...@leeds.ac.uk>
wrote:
Hi all,
I hope you can help with this one!
I have a dataframe: 'df' that consists of a vector of times: 'dt2' and a vector
of group id's: 'group':
dates2=rep("01/02/13",times=8)
times2=c("12:00:00","12:30:00","12:45:00","13:15:00","13:30:00","14:00:00","14:45:00","17:30:00")
y =paste(dates2, times2)
dt2=strptime(y, "%m/%d/%y %H:%M:%S")
group=c(1,1,2,2,3,3,4,4)
df=data.frame(dt2,group)
I also have a vector: 'dt' which is a series of times:
dates=rep("01/02/13",times=20)
times=c("12:00:00","12:15:00","12:30:00","12:45:00","13:00:00","13:15:00","13:30:00","13:45:00","14:00:00","14:15:00","14:30:00","14:45:00","15:00:00","15:15:00","15:30:00","15:45:00","16:00:00","16:15:00","16:30:00","16:45:00","17:00:00","17:15:00","17:30:00","17:45:00")
x =paste(dates, times)
dt=strptime(x, "%m/%d/%y %H:%M:%S")
I wish to create a vector which looks like 'id':
id=c(1,1,1,2,2,2,3,3,3,0,0,4,4,4,4,4,4,4,4,4,4,4,4,0)
The rules I wish to follow to create 'id' are:
1. If a value in 'dt' is either equal to, or, within the range of times within
group x in dataframe 'df', then, the value in 'id' will equal x.
So, for example, in 'df', group 4 is between the times of "14:45:00" and
"17:30:00" on the "01/02/13". Thus, the 12th to 23rd value in 'id' equals 4 as
these values correspond to times within 'dt' that are equal to and within the
range of "14:45:00" and "17:30:00" on the "01/02/13".
If this doesn't make sense, please ask,
I'm not sure where to even start with this... possibly the 'cut' function?
Many thanks in advance,
Ben Gillespie, Research Postgraduate
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