search for: subianto

Thank you very much for your useful suggestions. These are exactly what I was looking for. foo <- list(foo1, foo2, foo3) lapply(foo, function(x) matrix(unlist(x), nrow = length(x), byrow = TRUE)) or lapply(foo, function(x) do.call('rbind', x)) Best, Muhammad Subianto On 4/11/06, Muhammad Subianto <msubianto at gmail.com> wrote: > Dear all, > I have a result my experiment like this below (here my toy example): > > foo1 <- list() > foo1[[1]] <- c(10, 20, 30) > foo1[[2]] <- c(11, 21, 31) > > foo2 <- list() > foo2[[1]]...

...1 5 -1 [6,] 1 1 1 4 6 0 1 6 -1 [7,] 1 1 1 4 6 0 1 7 -1 [8,] 1 1 1 4 6 0 1 8 -1 > How can I do to check columns 2,3,4,6,7 and 9 have the same value, and columns 1,5 and 8 have different values. Best, Muhammad Subianto

...> as.character(d[3,]) [1] "1" "3" "1" > I think as.character the result something like [3] "XYZ" "C" "A" I don't know how to convert it. Any help gratefully received. Thank you very much in advance. Kindly regards, Muhammad Subianto

...x1200 0 0 0 1 1 0 0 1 1 1 0 0 1 1 0 0 1 1 0 1 0 1 1 0 0 1 1 1 1 0 1 1 0 0 1 1 ... How can I change automatically 0=no and 1=yes. Thank you very much in advance. Kindly regards, Muhammad Subianto

...ear R-helper, Is there possible to make this array: > a <- array(1:12, c(4, 3)) > a [,1] [,2] [,3] [1,] 1 5 9 [2,] 2 6 10 [3,] 3 7 11 [4,] 4 8 12 > like: c(1,5,9) c(2,6,10) c(3,7,11) c(4,8,12) Thank you very much in advance. Regards, Muhammad Subianto

...[1,] 1 0 1 1 [3,] 3 5 2 -1 [4,] 4 1 0 1 I have tried with matrix(train %in% test, dim(train)) apply(train, 1, paste, collapse="") %in% apply(test, 1, paste, collapse="") It doesn't work. How can I do. Thanks for any help. Best, Muhammad Subianto

...13 18 1 [4,] 4 9 14 19 1 [5,] 5 10 15 20 1 [[2]][[2]] [,1] [,2] [,3] [,4] [5] [1,] 1 6 11 16 1 [2,] 2 7 12 17 1 [3,] 3 8 13 18 -1 [4,] 4 9 14 19 1 [5,] 5 10 15 20 -1 Thanks you for any help. Best wishes, Muhammad Subianto #Gabor Grothendieck ggrothendieck at gmail.com #Mon Aug 28 13:53:52 CEST 2006 Here are two ways: 1. use indexes: lapply(seq(along = x), function(i) cbind(x[[i]], y[[i]])) 2. use mapply: mapply(cbind, x, y, SIMPLIFY = FALSE) On 8/28/06, Muhammad Subianto <msubianto at gmail.com> wro...

...22/75 1 10 40 24/75 25/75 26/75 45 61 117 28/75 62 Any suggestions would be appreciated. Thanks you. Best, Muhammad Subianto PS. I found this website: http://www.mindspring.com/~alanh/fracs.html

...maybe like | | | | | | | | | | | | | |pos|neg| |pos|neg| |pos|neg| | | | | | | | | | --------- --------- --------- v1 v2 v3 .... v7 v8 Thanks you for any helps. Regards, Muhammad Subianto

...table, Y X A B C A 0.950933 0.143600 0.956133 B 0.000533 0.986467 0.032066 C 0.005333 0.000000 0.009266 I cannot use table() or ftable() functions because the 3rd column (Z) is probability. Are there any function to make a table as I want? Kind regards, Muhammad Subianto

...> Yes, I get a matrix TRUE. But I need to change a column name like [,1] [,2] [,3]. I need the result like [,1] [,2] [,3] [1,] 10 1 1 [2,] 20 2 1 [3,] 10 3 1 [4,] 20 4 3 [5,] 10 5 2 [6,] 20 6 2 How can I do that? Thanks in advance, Muhammad Subianto

...t; "xyz.skin" "xyz.bmi" "xyz.ped" "xyz.age" "xyz.type" How can I make this (automatically). I don't want to make manual with more 100 variables. I would be very happy if anyone could help me. Thank you for your time. Kindly regards, Muhammad Subianto

Dear all, I tried to install rgl package on my linux machine fc5, I got an error. Here I run R as user, $ R > options(repos=c(CRAN="http://cran.at.r-project.org/")) > install.packages("rgl", lib="/home/subianto/local/lib/R/library/site-packages", dependencies=TRUE) trying URL 'http://cran.at.r-project.org/src/contrib/rgl_0.67-2.tar.gz' Content type 'application/x-tar' length 567592 bytes opened URL ================================================== downloaded 554Kb * Installing *sour...

...like: 24 2 3 3 1 1 2 3 0 1 45 1 3 10 1 1 3 4 0 1 43 2 3 7 1 1 3 4 0 1 42 3 2 9 1 1 3 3 0 1 36 3 3 8 1 1 3 2 0 1 19 4 4 0 1 1 3 3 0 1 38 2 3 6 1 1 3 2 0 1 21 3 3 1 1 0 3 2 0 1 27 2 3 3 1 1 3 4 0 1 45 1 1 8 1 1 2 2 1 1 ... How can I do it. Thanks you for your help. Best regards, Muhammad Subianto

...46 1 1 1 1 1 random number is 3 0 1 0 1 1 random number is 28 1 1 0 1 1 So, the result random tuples (order from 6 random number): 0 1 0 1 1 0 1 0 1 1 1 1 0 1 1 1 1 0 1 1 1 1 1 1 1 0 0 0 0 1 I would be very happy if anyone could help me. Thank you very much in advance. Kindly regards, Muhammad Subianto

...load("filename.zip") Error: bad restore file magic number (file may be corrupted) -- no data loaded > My question: How can I open/load "filename.zip"? Is there any function to open R workspace which it store in zip file? I hope some one can give me advices. Best, Muhammad Subianto > version _ platform i386-pc-mingw32 arch i386 os mingw32 system i386, mingw32 status major 2 minor 2.1 year 2005 month 12 day 20 svn rev 36812 language R >

...5,29),] Because I have very large dataset I can not make this manual. Then I need the ID of row did not change, I mean like, [3,] [5,] [29,] In dataset this is about ID our customers. I was wondering if anyone give me a trick to make simple. Thanks you very much for any suggestions. Best, Muhammad Subianto

...30 [2] 11 21 31 [[2]] [1] 100 200 300 [2] 110 210 310 [[3]] [1] 1000 2000 3000 [2] 1100 2100 3100 I saw on the R-help archives page similar like this but I can't find a solution. http://tolstoy.newcastle.edu.au/R/help/05/05/4678.html Thanks very much for any suggestions. Sincerely, Muhammad Subianto

...: Mon Oct 24, 2005 at 04:23:01 PM Finish time: Wed Oct 26, 2005 at 03:26:19 PM Run time: 1.960625 secs. This is about two seconds or one day and nine hours? Then, how could I convert to 1 day, 23 hours, ? minutes, ? seconds. Thanks you very much for any suggestions. Best wishes, Muhammad Subianto > ################################################################################ > # Begin of program and timestamp: > cat(format(begin.time <- Sys.time(), "%a %b %d %X %Y") ,"\n") Mon Oct 24 04:23:01 PM 2005 > cat("Start time:", secs <- f...

...0.2 setosa 5.4 3.9 1.7 0.4 setosa 4.6 3.4 1.4 0.3 setosa 5.0 3.4 1.5 0.2 setosa 4.4 2.9 1.4 0.2 setosa 4.9 3.1 1.5 0.1 setosa Sincerely, Muhammad Subianto