Hi Ben,
It looks like the condition is not met in majority of the split elements. So,
when you create a dataframe with the a column with 0 element and another column
with an element, it shows the Error message.
data.frame(dt2=NULL,group=1)
#Error in data.frame(dt2 = NULL, group = 1) :
# argum
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From: arun [smartpink...@yahoo
Also,
df1 <- do.call(rbind,lapply(split(df,df$group),function(x)
data.frame(dt2=seq(x[1,1],x[nrow(x),1],by="15 min"),group=x[1,2])))
id1<- df1[,2][match(as.POSIXct(dt),df1[,1])]
id1[is.na(id1)]<- 0
identical(id1,id)
#[1] TRUE
A.K,
On Thursday, October 10, 2013 9:29 PM, arun wrote:
Hi Ben,
I
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) {
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, ti
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