In keeping with the theme of reducing unnecessary overhead
(and using William's example data)
> system.time( vAve <- ave(a, i, FUN=cummax) )
user system elapsed
0.125 0.003 0.127
> system.time( b <- unlist( lapply( split(a,i) , cummax) ) )
user system elapsed
0.320 0.007 0.327
Hi John,
I may be misunderstanding what you want, but this seems to produce the
output you specify:
A<-sample(-10:100,100)
i<-rep(1:10,c(5:13,19))
# replace the last value of x with the maximum
max_last<-function(x) return(c(x[-length(x)],max(x)))
as.vector(unlist(by(A,i,max_last)))
and this is w
Sorry, I can't help, but wanted to note that "apply methods" are
essentially just loops and rarely much faster or slower than explicit
loops. (Those of us who use them do so for clarity and maintainability
of code reasons).
Cheers,
Bert
Bert Gunter
"The trouble with having an open mind is that p
ave(A, i, FUN=cummax) loops but is faster than your aggregate-based
solution. E.g.,
> i <- rep(1:1, sample(0:210, replace=TRUE, size=1))
> length(i)
[1] 1056119
> a <- sample(-50:50, replace=TRUE, size=length(i))
> system.time( vAve <- ave(a, i, FUN=cummax) )
user system elapsed
0.
Folks
I have some very long vectors - typically 1 million long - which are
indexed by another vector, same length, with values from 1 to a few
thousand, sp each sub part of the vector may be a few hundred values long.
I want to calculate the cumulative maximum of each sub part the main
vector by
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