Re: [R] Simple - Finding vector in a vector

ounces@r- >> project.org] On Behalf Of Jessica Streicher >> Sent: Monday, October 08, 2012 9:19 AM >> To: Mike Spam >> Cc: r-help@r-project.org >> Subject: Re: [R] Simple - Finding vector in a vector >> >> > x<-c(NA , 1 ,NA, 1 , 1 , 1 , 1 , 1 ,1 ,N

Re: [R] Simple - Finding vector in a vector

be something like, 1 1 1 3 3 3 3 1 1 but anyway i can work with rle, if there is no suitable function. thanks, Nico 2012/10/8 Mike Spam : > Hey Rui, > > Perfect! Thanks!! :) > > Nico > > 2012/10/8 Rui Barradas : >> Hello, >> >> See ?rle >> >&

Re: [R] Simple - Finding vector in a vector

Hey Rui, Perfect! Thanks!! :) Nico 2012/10/8 Rui Barradas : > Hello, > > See ?rle > > Hope this helps, > > Rui Barradas > Em 08-10-2012 13:55, Mike Spam escreveu: >> >> Hi, >> >> just a simple question. >> Assumed i have a vector, >&

[R] Simple - Finding vector in a vector

Hi, just a simple question. Assumed i have a vector, FALSE TRUE TRUE TRUE FALSE TRUE FALSE TRUE FALSE or NA 1 1 1 NA 1 NA 1 NA what i need is the position where an element is the same - three (or in general multiple) times in a row. in this case: i want to get the position where it is

Re: [R] effective way to return only the first argument of "which()"

gt; } >> >> return(NULL) >> } >> >>> system.time(for(i in 1:100)which(x>0.999)[1]) >> utilisateur système écoulé >> 9.740 5.795 15.890 >>> system.time(for(i in 1:100)which.max(x>0.999)) >> utilisateur système écoul

Re: [R] effective way to return only the first argument of "which()"

Hi, Thanks Michael, but i think this is even slower. x <-sample(2000) which(x < 5)[1] which.max(x < 5) system.time(for(i in 1:100) which.max(x < 5)) User System verstrichen 60.84 13.70 86.33 system.time(for(i in 1:100) which(x < 5)[1]) User Syste

[R] effective way to return only the first argument of "which()"

Hi, I was looking for a function like "which()" but only returns the first argument. Compare: x <- c(1,2,3,4,5,6) y <- 4 which(x>y) returns: 5,6 which(x>y)[1] returns: 5 which(x>y)[1] is exactly what i need. I did use this but the dataset is too big (~18 mio. Points). That's why i need a more