[R] Faster way to do it??...using apply?
Hi, I have a simple task, but I am looking for a clever and fast way to do it: I have a vector x with 0,1 or 2 and I want to create another vector y with the same length following the rules: If the element in x is equal to 0, the element in y is equal to 0 If the element in x is equal to 2, the element in y is equal to 1 If the element in x is equal to 1, the element in y is either 0 or 1 (sample from c(0,1)) thus the vector x [,1] [1,]0 [2,]2 [3,]1 [4,]2 [5,]0 [6,]1 [7,]2 could produce the vector y (this is one of the possibilities since y|x=1 is either 0 or 1 y [,1] [1,]0 [2,]1 [3,]1 [4,]1 [5,]0 [6,]0 [7,]1 I know how to do this using for loops but I was wondering if you guys could suggest a better way Thanks -- View this message in context: http://r.789695.n4.nabble.com/Faster-way-to-do-it-using-apply-tp3166161p3166161.html Sent from the R help mailing list archive at Nabble.com. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Faster way to do it??...using apply?
Try this: replace(replace(x, x == 1, sample(0:1, 1)), x == 2, 1) On Tue, Dec 28, 2010 at 2:43 PM, M.Ribeiro mresende...@yahoo.com.br wrote: Hi, I have a simple task, but I am looking for a clever and fast way to do it: I have a vector x with 0,1 or 2 and I want to create another vector y with the same length following the rules: If the element in x is equal to 0, the element in y is equal to 0 If the element in x is equal to 2, the element in y is equal to 1 If the element in x is equal to 1, the element in y is either 0 or 1 (sample from c(0,1)) thus the vector x [,1] [1,]0 [2,]2 [3,]1 [4,]2 [5,]0 [6,]1 [7,]2 could produce the vector y (this is one of the possibilities since y|x=1 is either 0 or 1 y [,1] [1,]0 [2,]1 [3,]1 [4,]1 [5,]0 [6,]0 [7,]1 I know how to do this using for loops but I was wondering if you guys could suggest a better way Thanks -- View this message in context: http://r.789695.n4.nabble.com/Faster-way-to-do-it-using-apply-tp3166161p3166161.html Sent from the R help mailing list archive at Nabble.com. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. -- Henrique Dallazuanna Curitiba-Paraná-Brasil 25° 25' 40 S 49° 16' 22 O [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Faster way to do it??...using apply?
Hi Henrique, Thanks for the fast answer, The only problem in your code, which I think I didn't mention in my message is that I would like one different random sampling procedure for each 1 in my vector The way it was written, it samples only once and replace by every 1: x = as.matrix(c(1,1,1,1,1,1,1,1,1,1,1,1,1,1,1)) replace(replace(x, x == 1, sample(0:1, 1)), x == 2, 1) [,1] [1,]1 [2,]1 [3,]1 [4,]1 [5,]1 [6,]1 [7,]1 [8,]1 [9,]1 [10,]1 [11,]1 [12,]1 [13,]1 [14,]1 [15,]1 Thanks -- View this message in context: http://r.789695.n4.nabble.com/Faster-way-to-do-it-using-apply-tp3166161p3166203.html Sent from the R help mailing list archive at Nabble.com. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Faster way to do it??...using apply?
Try this indeed replace(replace(x, x == 1, sample(0:1, sum(x == 1), rep = TRUE)), x == 2, 1) On Tue, Dec 28, 2010 at 3:14 PM, M.Ribeiro mresende...@yahoo.com.br wrote: Hi Henrique, Thanks for the fast answer, The only problem in your code, which I think I didn't mention in my message is that I would like one different random sampling procedure for each 1 in my vector The way it was written, it samples only once and replace by every 1: x = as.matrix(c(1,1,1,1,1,1,1,1,1,1,1,1,1,1,1)) replace(replace(x, x == 1, sample(0:1, 1)), x == 2, 1) [,1] [1,]1 [2,]1 [3,]1 [4,]1 [5,]1 [6,]1 [7,]1 [8,]1 [9,]1 [10,]1 [11,]1 [12,]1 [13,]1 [14,]1 [15,]1 Thanks -- View this message in context: http://r.789695.n4.nabble.com/Faster-way-to-do-it-using-apply-tp3166161p3166203.html Sent from the R help mailing list archive at Nabble.com. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. -- Henrique Dallazuanna Curitiba-Paraná-Brasil 25° 25' 40 S 49° 16' 22 O [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Faster way to do it??...using apply?
I don't know if it's any faster, but it is also possible this way: y - ifelse(x ==1, round(runif(x)), sign(x)) -- Jonathan P. Daily Technician - USGS Leetown Science Center 11649 Leetown Road Kearneysville WV, 25430 (304) 724-4480 Is the room still a room when its empty? Does the room, the thing itself have purpose? Or do we, what's the word... imbue it. - Jubal Early, Firefly r-help-boun...@r-project.org wrote on 12/28/2010 12:48:04 PM: [image removed] Re: [R] Faster way to do it??...using apply? Henrique Dallazuanna to: M.Ribeiro 12/28/2010 12:51 PM Sent by: r-help-boun...@r-project.org Cc: r-help Try this indeed replace(replace(x, x == 1, sample(0:1, sum(x == 1), rep = TRUE)), x == 2, 1) On Tue, Dec 28, 2010 at 3:14 PM, M.Ribeiro mresende...@yahoo.com.br wrote: Hi Henrique, Thanks for the fast answer, The only problem in your code, which I think I didn't mention in my message is that I would like one different random sampling procedure for each 1 in my vector The way it was written, it samples only once and replace by every 1: x = as.matrix(c(1,1,1,1,1,1,1,1,1,1,1,1,1,1,1)) replace(replace(x, x == 1, sample(0:1, 1)), x == 2, 1) [,1] [1,]1 [2,]1 [3,]1 [4,]1 [5,]1 [6,]1 [7,]1 [8,]1 [9,]1 [10,]1 [11,]1 [12,]1 [13,]1 [14,]1 [15,]1 Thanks -- View this message in context: http://r.789695.n4.nabble.com/Faster-way-to-do-it-using-apply- tp3166161p3166203.html Sent from the R help mailing list archive at Nabble.com. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. -- Henrique Dallazuanna Curitiba-Paraná-Brasil 25° 25' 40 S 49° 16' 22 O [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
