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). This way I want to create at least 10 tables (10 random
> tables) and save them separately in Excel.
>
>
> Thanks
>
> [[alternative HTML version deleted]]
>
> ______________________________________________
> R-help@r-project.org mailing list -- To UNSUBSCRIBE and more, see
> 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.
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