Dear David and Petr, It worked and I think now I can modify the code little bit to fit my requirement. Thank you so much for your help.
Thanks, MW On Wed, Oct 18, 2017 at 7:15 AM, David L Carlson <dcarl...@tamu.edu> wrote: > Building on Petr's suggestion, you could modify his code to get all 10 > samples at once in a compact format: > > > Samples <- lapply(lll, function(x) replicate(10, sample(x, rep=TRUE))) > # Samples is a list containing 3 matrices, one for each group > # Each column gives the index (row) numbers for a particular sample > > str(Samples) > List of 3 > $ A: int [1:3, 1:10] 2 1 3 1 1 3 2 3 1 2 ... > $ B: int [1:7, 1:10] 8 5 8 6 5 8 10 6 4 7 ... > $ C: int [1:9, 1:10] 14 15 19 13 18 19 19 12 13 16 ... > # Row numbers for all 10 samples of 3 for group A > > Samples$A > [,1] [,2] [,3] [,4] [,5] [,6] [,7] [,8] [,9] [,10] > [1,] 2 1 2 2 3 3 2 3 2 3 > [2,] 1 1 3 1 3 3 2 3 3 3 > [3,] 3 3 1 2 2 1 2 3 3 1 > # The first sample from group A > > dat1[Samples$A[ , 1], ] > RegionA site temp group > 2 Ra s2 21 A > 1 Ra s1 23 A > 3 Rb s3 10 A > > # Save the first sample from group A as a .csv file > > write.csv(dat1[Samples$A[ , 1], ], row.names=FALSE, file="Test.csv") > > # Save all ten group A samples as a single .csv file > # Faster but you have to split them into separate tables by hand > > write.csv(dat1[Samples$A[ , 1:10], ], row.names=FALSE, file="Test.csv") > > ---------------------------------------- > David L Carlson > Department of Anthropology > Texas A&M University > College Station, TX 77843-4352 > > > -----Original Message----- > From: R-help [mailto:r-help-boun...@r-project.org] On Behalf Of PIKAL Petr > Sent: Wednesday, October 18, 2017 3:38 AM > To: Marna Wagley <marna.wag...@gmail.com>; r-help mailing list < > r-help@r-project.org> > Subject: Re: [R] creating tables with replacement > > Hi > > maybe there is another more elegant solution but something like this > > > idx <- 1:nrow(dat1) > > lll <- split(idx, dat1$group) > > dat1[unlist(lapply(lll, sample, rep=TRUE)),] > > gives you selected rows. > > You could use for cycle or save those data frames manually > > Cheers > > Petr > > -----Original Message----- > > From: R-help [mailto:r-help-boun...@r-project.org] On Behalf Of Marna > > Wagley > > Sent: Wednesday, October 18, 2017 9:46 AM > > To: r-help mailing list <r-help@r-project.org> > > Subject: [R] creating tables with replacement > > > > Hi R User, > > I am new in R and trying to create tables with selecting rows randomly > > (but with replacement) for each group but each group should have same > > number as original. Is it possible to create it using the following > example data set? > > > > Your help is highly appreciated. > > > > dat1<-structure(list(RegionA = structure(c(1L, 1L, 2L, 3L, 3L, 4L, > > > > 5L, 5L, 6L, 6L, 6L, 6L, 6L, 6L, 6L, 6L, 6L, 6L, 6L), .Label = c("Ra", > > > > "Rb", "Rc", "Rd", "Re", "Rf"), class = "factor"), site = > > structure(c(1L, > > > > 12L, 13L, 14L, 15L, 16L, 17L, 18L, 19L, 2L, 3L, 4L, 5L, 6L, 7L, > > > > 8L, 9L, 10L, 11L), .Label = c("s1", "s10", "s11", "s12", "s13", > > > > "s14", "s15", "s16", "s17", "s18", "s19", "s2", "s3", "s4", "s5", > > > > "s6", "s7", "s8", "s9"), class = "factor"), temp = c(23L, 21L, > > > > 10L, 15L, 16L, 8L, 13L, 1L, 23L, 19L, 25L, 19L, 12L, 16L, 19L, > > > > 21L, 12L, 5L, 7L), group = structure(c(1L, 1L, 1L, 2L, 2L, 2L, > > > > 2L, 2L, 2L, 2L, 3L, 3L, 3L, 3L, 3L, 3L, 3L, 3L, 3L), .Label = c("A", > > > > "B", "C"), class = "factor")), .Names = c("RegionA", "site", > > > > "temp", "group"), class = "data.frame", row.names = c(NA, -19L > > > > )) > > > > > > Here group A has 3 rows, > > > > B had 7 rows, and > > > > C has 9 rows > > > > > > I want to select rows randomly (with replacement) for each group but > > each group should have same number of rows as above (group A=3 rows, > > B=7 rows, and > > c=9 rows). 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