Dear Richard, Martin, Dimitris and Domenico
thank you very much for your help.
I must say that the fastest procedure appears to be the one suggested by
Richard
This runs pretty quickly:
unSpells <- nrow(Atr) - apply(Atr,2,function(x) max(which(x==1)))
#c(4,0,7,0)
If I may abuse of your kin
Hi Mario --
This function
f <- function(m) {
## next 2 lines due to Bill Dunlap
## http://tolstoy.newcastle.edu.au/R/e4/devel/08/04/1206.html
csum <- cumsum(!m)
crun <- csum - cummax(m * csum)
matrix(ifelse(crun > 0, (crun-1) %% nrow(m) + 1, 0),
nrow=nrow(m))
}
ret
> >> It works, but the for (i in ...) loop slows down the simulation a
lot.
> >>
> >> Any suggestion on how to avoid this loop? (or in general, to speed up
> >> this part of the simulation)
> Actually, I have not specified the following: i want to consider only
> the "most recent" sequence of
then try the following:
Atr <- cbind(rep(1:0, each = 4), 1, c(1, rep(0, 7)), 1)
Atr <- rbind(c(0, 1, 0, 1), Atr)
apply(Atr, 2, function (x) {
rr <- rle(x)
if (tail(rr$values, 1) == 0) tail(rr$length, 1) else 0
})
I hope this what you're looking for.
Best,
Dimitris
Mario Lavezzi wrote
Hi Dimitris, thank you very much.
Actually, I have not specified the following: i want to consider only
the "most recent" sequence of zeros, that is the last part of the time
series.
That is, If I have:
[,1] [,2] [,3] [,4]
[1,]0101
[2,]1111
[3,]11
Try this:
unSpells[tail(Atr,1)==0] <-
apply(Atr,2,function(x)sum(x==0))[tail(Atr,1)==0]
Or (if you don't have to preserve the value in the unSpells vector):
unSpells <- apply(Atr,2,function(x)sum(x==0))
But in this case you have 0 instead of 1 in the second and fourth position.
Ciao,
domenico
it's not totally clear to me what exactly do you need in this case, but
have a look at the following:
Atr <- cbind(rep(1:0, each = 4), 1, c(1, rep(0, 7)), 1)
unSpells <- colSums(Atr == 0)
unSpells[unSpells == 0] <- 1
unSpells
I hope it helps.
Best,
Dimitris
Mario Lavezzi wrote:
Hello,
I ha
Hello,
I have the following problem.
I am running simulations on possible states of a set of agents
(1=employed, 0=unemployed).
I store these simulated time series in a matrix like the following,
where rows indicates time periods, columns the number of agents (4
agents and 8 periods in this
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