Re: [R] Decomposing a List
You might add vapply() to you repertoire, as it is quicker than sapply but also does some error checking on the your input data. E.g., your f2 returns a matrix whose columns are the elements of the list l and you assume that there each element of l contains 2 character strings. f2 - function(l)matrix(unlist(l),nr=2) Here is a function based on vapply() the returns the same thing but also verifies that element of l is really a 2-long character vector. f2v - function (l) vapply(l, function(x) x, FUN.VALUE = character(2)) and a function to generate datasets of various sizes makeL - function(n)strsplit(paste(sample(LETTERS,n,rep=TRUE),sample(1:10,n,rep=TRUE),sep=+),+,fix=TRUE) Timing the functions on a million-long list I get l - makeL(n=10^6) system.time( r2 - f2(l) ) user system elapsed 0.088 0.000 0.090 system.time( r2v - f2v(l) ) user system elapsed 0.920.000.92 identical(r2, r2v) [1] TRUE vapply() is ten times slower than unlist() but three times faster than sapply(x,function(x)x). However, when you give it data that doesn't meet your expectations, which is common when using strsplit(), f2v tells you about the problem and f2 gives you an incorrect result: l[[10]] - c(a,b,c,d) system.time( r2v - f2v(l) ) Error in vapply(l, function(x) x, FUN.VALUE = character(2)) : values must be length 2, but FUN(X[[10]]) result is length 4 Timing stopped at: 0.004 0 0.002 system.time( rv - f2(l) ) user system elapsed 0.088 0.008 0.095 dim(rv) # you will have alignment problems later [1] 2 101 Bill Dunlap Spotfire, TIBCO Software wdunlap tibco.com -Original Message- From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org] On Behalf Of Bert Gunter Sent: Thursday, April 25, 2013 7:54 AM To: ted.hard...@wlandres.net Cc: R mailing list Subject: Re: [R] Decomposing a List Well, what you really want to do is convert the list to a matrix, and it can be done directly and considerably faster than with the (implicit) looping of sapply: f1 - function(l)sapply(l,[,1) f2 - function(l)matrix(unlist(l),nr=2) l - strsplit(paste(sample(LETTERS,1e6,rep=TRUE),sample(1:10,1e6,rep=TRUE),sep=+),+,f ix=TRUE) ## Then you get these results: system.time(x1 - f1(l)) user system elapsed 1.920.011.95 system.time(x2 - f2(l)) user system elapsed 0.060.020.08 system.time(x2 - f2(l)[1,]) user system elapsed 0.1 0.0 0.1 identical(x1,x2) [1] TRUE Cheers, Bert On Thu, Apr 25, 2013 at 3:32 AM, Ted Harding ted.hard...@wlandres.net wrote: Thanks, Jorge, that seems to work beautifully! (Now to try to understand why ... but that's for later). Ted. On 25-Apr-2013 10:21:29 Jorge I Velez wrote: Dear Dr. Harding, Try sapply(L, [, 1) sapply(L, [, 2) HTH, Jorge.- On Thu, Apr 25, 2013 at 8:16 PM, Ted Harding ted.hard...@wlandres.netwrote: Greetings! For some reason I am not managing to work out how to do this (in principle) simple task! As a result of applying strsplit() to a vector of character strings, I have a long list L (N elements), where each element is a vector of two character strings, like: L[1] = c(A1,B1) L[2] = c(A2,B2) L[3] = c(A3,B3) [etc.] From L, I wish to obtain (as directly as possible, e.g. avoiding a loop) two vectors each of length N where one contains the strings that are first in the pair, and the other contains the strings which are second, i.e. from L (as above) I would want to extract: V1 = c(A1,A2,A3,...) V2 = c(B1,B2,B3,...) Suggestions? With thanks, Ted. - E-Mail: (Ted Harding) ted.hard...@wlandres.net Date: 25-Apr-2013 Time: 11:16:46 This message was sent by XFMail __ 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. - E-Mail: (Ted Harding) ted.hard...@wlandres.net Date: 25-Apr-2013 Time: 11:31:57 This message was sent by XFMail __ 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. -- Bert Gunter Genentech Nonclinical Biostatistics Internal Contact Info: Phone: 467-7374 Website: http://pharmadevelopment.roche.com/index/pdb/pdb-functional-groups/pdb- biostatistics/pdb-ncb-home.htm __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r
Re: [R] Decomposing a List
Dear Dr. Harding, Try sapply(L, [, 1) sapply(L, [, 2) HTH, Jorge.- On Thu, Apr 25, 2013 at 8:16 PM, Ted Harding ted.hard...@wlandres.netwrote: Greetings! For some reason I am not managing to work out how to do this (in principle) simple task! As a result of applying strsplit() to a vector of character strings, I have a long list L (N elements), where each element is a vector of two character strings, like: L[1] = c(A1,B1) L[2] = c(A2,B2) L[3] = c(A3,B3) [etc.] From L, I wish to obtain (as directly as possible, e.g. avoiding a loop) two vectors each of length N where one contains the strings that are first in the pair, and the other contains the strings which are second, i.e. from L (as above) I would want to extract: V1 = c(A1,A2,A3,...) V2 = c(B1,B2,B3,...) Suggestions? With thanks, Ted. - E-Mail: (Ted Harding) ted.hard...@wlandres.net Date: 25-Apr-2013 Time: 11:16:46 This message was sent by XFMail __ 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. [[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] Decomposing a List
Thanks, Jorge, that seems to work beautifully! (Now to try to understand why ... but that's for later). Ted. On 25-Apr-2013 10:21:29 Jorge I Velez wrote: Dear Dr. Harding, Try sapply(L, [, 1) sapply(L, [, 2) HTH, Jorge.- On Thu, Apr 25, 2013 at 8:16 PM, Ted Harding ted.hard...@wlandres.netwrote: Greetings! For some reason I am not managing to work out how to do this (in principle) simple task! As a result of applying strsplit() to a vector of character strings, I have a long list L (N elements), where each element is a vector of two character strings, like: L[1] = c(A1,B1) L[2] = c(A2,B2) L[3] = c(A3,B3) [etc.] From L, I wish to obtain (as directly as possible, e.g. avoiding a loop) two vectors each of length N where one contains the strings that are first in the pair, and the other contains the strings which are second, i.e. from L (as above) I would want to extract: V1 = c(A1,A2,A3,...) V2 = c(B1,B2,B3,...) Suggestions? With thanks, Ted. - E-Mail: (Ted Harding) ted.hard...@wlandres.net Date: 25-Apr-2013 Time: 11:16:46 This message was sent by XFMail __ 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. - E-Mail: (Ted Harding) ted.hard...@wlandres.net Date: 25-Apr-2013 Time: 11:31:57 This message was sent by XFMail __ 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] Decomposing a List
Hi, May be this helps. L- list(c(A1,B1),c(A2,B2),c(A3,B3)) simplify2array(L)[1,] #[1] A1 A2 A3 simplify2array(L)[2,] #[1] B1 B2 B3 #or library(stringr) word(sapply(L,paste,collapse= ),1) #[1] A1 A2 A3 A.K. - Original Message - From: ted.hard...@wlandres.net ted.hard...@wlandres.net To: r-help@r-project.org Cc: Sent: Thursday, April 25, 2013 6:16 AM Subject: [R] Decomposing a List Greetings! For some reason I am not managing to work out how to do this (in principle) simple task! As a result of applying strsplit() to a vector of character strings, I have a long list L (N elements), where each element is a vector of two character strings, like: L[1] = c(A1,B1) L[2] = c(A2,B2) L[3] = c(A3,B3) [etc.] From L, I wish to obtain (as directly as possible, e.g. avoiding a loop) two vectors each of length N where one contains the strings that are first in the pair, and the other contains the strings which are second, i.e. from L (as above) I would want to extract: V1 = c(A1,A2,A3,...) V2 = c(B1,B2,B3,...) Suggestions? With thanks, Ted. - E-Mail: (Ted Harding) ted.hard...@wlandres.net Date: 25-Apr-2013 Time: 11:16:46 This message was sent by XFMail __ 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.
Re: [R] Decomposing a List
Well, what you really want to do is convert the list to a matrix, and it can be done directly and considerably faster than with the (implicit) looping of sapply: f1 - function(l)sapply(l,[,1) f2 - function(l)matrix(unlist(l),nr=2) l - strsplit(paste(sample(LETTERS,1e6,rep=TRUE),sample(1:10,1e6,rep=TRUE),sep=+),+,fix=TRUE) ## Then you get these results: system.time(x1 - f1(l)) user system elapsed 1.920.011.95 system.time(x2 - f2(l)) user system elapsed 0.060.020.08 system.time(x2 - f2(l)[1,]) user system elapsed 0.1 0.0 0.1 identical(x1,x2) [1] TRUE Cheers, Bert On Thu, Apr 25, 2013 at 3:32 AM, Ted Harding ted.hard...@wlandres.net wrote: Thanks, Jorge, that seems to work beautifully! (Now to try to understand why ... but that's for later). Ted. On 25-Apr-2013 10:21:29 Jorge I Velez wrote: Dear Dr. Harding, Try sapply(L, [, 1) sapply(L, [, 2) HTH, Jorge.- On Thu, Apr 25, 2013 at 8:16 PM, Ted Harding ted.hard...@wlandres.netwrote: Greetings! For some reason I am not managing to work out how to do this (in principle) simple task! As a result of applying strsplit() to a vector of character strings, I have a long list L (N elements), where each element is a vector of two character strings, like: L[1] = c(A1,B1) L[2] = c(A2,B2) L[3] = c(A3,B3) [etc.] From L, I wish to obtain (as directly as possible, e.g. avoiding a loop) two vectors each of length N where one contains the strings that are first in the pair, and the other contains the strings which are second, i.e. from L (as above) I would want to extract: V1 = c(A1,A2,A3,...) V2 = c(B1,B2,B3,...) Suggestions? With thanks, Ted. - E-Mail: (Ted Harding) ted.hard...@wlandres.net Date: 25-Apr-2013 Time: 11:16:46 This message was sent by XFMail __ 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. - E-Mail: (Ted Harding) ted.hard...@wlandres.net Date: 25-Apr-2013 Time: 11:31:57 This message was sent by XFMail __ 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. -- Bert Gunter Genentech Nonclinical Biostatistics Internal Contact Info: Phone: 467-7374 Website: http://pharmadevelopment.roche.com/index/pdb/pdb-functional-groups/pdb-biostatistics/pdb-ncb-home.htm __ 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] Decomposing a List
On Apr 25, 2013, at 7:53 AM, Bert Gunter wrote: Well, what you really want to do is convert the list to a matrix, and it can be done directly and considerably faster than with the (implicit) looping of sapply: f1 - function(l)sapply(l,[,1) f2 - function(l)matrix(unlist(l),nr=2) l - strsplit(paste(sample(LETTERS,1e6,rep=TRUE),sample(1:10,1e6,rep=TRUE),sep=+),+,fix=TRUE) Consider this alternative: L = list( c(A1,B1), c(A2,B2), c(A3,B3) ) simplify2array(L) [,1] [,2] [,3] [1,] A1 A2 A3 [2,] B1 B2 B3 -- David. ## Then you get these results: system.time(x1 - f1(l)) user system elapsed 1.920.011.95 system.time(x2 - f2(l)) user system elapsed 0.060.020.08 system.time(x2 - f2(l)[1,]) user system elapsed 0.1 0.0 0.1 identical(x1,x2) [1] TRUE Cheers, Bert On Thu, Apr 25, 2013 at 3:32 AM, Ted Harding ted.hard...@wlandres.net wrote: Thanks, Jorge, that seems to work beautifully! (Now to try to understand why ... but that's for later). Ted. On 25-Apr-2013 10:21:29 Jorge I Velez wrote: Dear Dr. Harding, Try sapply(L, [, 1) sapply(L, [, 2) HTH, Jorge.- On Thu, Apr 25, 2013 at 8:16 PM, Ted Harding ted.hard...@wlandres.netwrote: Greetings! For some reason I am not managing to work out how to do this (in principle) simple task! As a result of applying strsplit() to a vector of character strings, I have a long list L (N elements), where each element is a vector of two character strings, like: L[1] = c(A1,B1) L[2] = c(A2,B2) L[3] = c(A3,B3) [etc.] From L, I wish to obtain (as directly as possible, e.g. avoiding a loop) two vectors each of length N where one contains the strings that are first in the pair, and the other contains the strings which are second, i.e. from L (as above) I would want to extract: V1 = c(A1,A2,A3,...) V2 = c(B1,B2,B3,...) Suggestions? With thanks, Ted. - E-Mail: (Ted Harding) ted.hard...@wlandres.net Date: 25-Apr-2013 Time: 11:16:46 This message was sent by XFMail __ 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. - E-Mail: (Ted Harding) ted.hard...@wlandres.net Date: 25-Apr-2013 Time: 11:31:57 This message was sent by XFMail __ 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. -- Bert Gunter Genentech Nonclinical Biostatistics Internal Contact Info: Phone: 467-7374 Website: http://pharmadevelopment.roche.com/index/pdb/pdb-functional-groups/pdb-biostatistics/pdb-ncb-home.htm __ 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. David Winsemius Alameda, CA, USA __ 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] Decomposing a List
Well... WIth the same list,l,as before: system.time(x3 - simplify2array(l)) user system elapsed 2.110.052.20 system.time(x2 - f2(l)) ## the matrix(unlist(...)) one user system elapsed 0.110.000.11 identical(x2,x3) [1] TRUE So kind of a big difference if you care about efficiency... (and I can't remember all those specialized functions, anyway!) -- Bert On Thu, Apr 25, 2013 at 8:53 PM, David Winsemius dwinsem...@comcast.net wrote: On Apr 25, 2013, at 7:53 AM, Bert Gunter wrote: Well, what you really want to do is convert the list to a matrix, and it can be done directly and considerably faster than with the (implicit) looping of sapply: f1 - function(l)sapply(l,[,1) f2 - function(l)matrix(unlist(l),nr=2) l - strsplit(paste(sample(LETTERS,1e6,rep=TRUE),sample(1:10,1e6,rep=TRUE),sep=+),+,fix=TRUE) Consider this alternative: L = list( c(A1,B1), c(A2,B2), c(A3,B3) ) simplify2array(L) [,1] [,2] [,3] [1,] A1 A2 A3 [2,] B1 B2 B3 -- David. ## Then you get these results: system.time(x1 - f1(l)) user system elapsed 1.920.011.95 system.time(x2 - f2(l)) user system elapsed 0.060.020.08 system.time(x2 - f2(l)[1,]) user system elapsed 0.1 0.0 0.1 identical(x1,x2) [1] TRUE Cheers, Bert On Thu, Apr 25, 2013 at 3:32 AM, Ted Harding ted.hard...@wlandres.net wrote: Thanks, Jorge, that seems to work beautifully! (Now to try to understand why ... but that's for later). Ted. On 25-Apr-2013 10:21:29 Jorge I Velez wrote: Dear Dr. Harding, Try sapply(L, [, 1) sapply(L, [, 2) HTH, Jorge.- On Thu, Apr 25, 2013 at 8:16 PM, Ted Harding ted.hard...@wlandres.netwrote: Greetings! For some reason I am not managing to work out how to do this (in principle) simple task! As a result of applying strsplit() to a vector of character strings, I have a long list L (N elements), where each element is a vector of two character strings, like: L[1] = c(A1,B1) L[2] = c(A2,B2) L[3] = c(A3,B3) [etc.] From L, I wish to obtain (as directly as possible, e.g. avoiding a loop) two vectors each of length N where one contains the strings that are first in the pair, and the other contains the strings which are second, i.e. from L (as above) I would want to extract: V1 = c(A1,A2,A3,...) V2 = c(B1,B2,B3,...) Suggestions? With thanks, Ted. - E-Mail: (Ted Harding) ted.hard...@wlandres.net Date: 25-Apr-2013 Time: 11:16:46 This message was sent by XFMail __ 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. - E-Mail: (Ted Harding) ted.hard...@wlandres.net Date: 25-Apr-2013 Time: 11:31:57 This message was sent by XFMail __ 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. -- Bert Gunter Genentech Nonclinical Biostatistics Internal Contact Info: Phone: 467-7374 Website: http://pharmadevelopment.roche.com/index/pdb/pdb-functional-groups/pdb-biostatistics/pdb-ncb-home.htm __ 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. David Winsemius Alameda, CA, USA -- Bert Gunter Genentech Nonclinical Biostatistics Internal Contact Info: Phone: 467-7374 Website: http://pharmadevelopment.roche.com/index/pdb/pdb-functional-groups/pdb-biostatistics/pdb-ncb-home.htm __ 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.
