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copied!<p>Here is a solution using <code>plyr</code> with a simulated dataset.</p> <pre><code>library(plyr) set.seed(1001) dat = data.frame(matrix(rnorm(1000), ncol = 10), treatment = sample(c("control", "control", "treatment"), 100, replace = T) ) # divide data set into training and test sets tr_prop = 0.5 # proportion of full dataset to use for training training_set = ddply(dat, .(treatment), function(., seed) { set.seed(seed); .[sample(1:nrow(.), trunc(nrow(.) * tr_prop)), ] }, seed = 101) test_set = ddply(dat, .(treatment), function(., seed) { set.seed(seed); .[-sample(1:nrow(.), trunc(nrow(.) * tr_prop)), ] }, seed = 101) # check that proportions are equal across datasets ddply(dat, .(treatment), function(.) nrow(.)/nrow(dat) ) ddply(training_set, .(treatment), function(.) nrow(.)/nrow(training_set) ) ddply(test_set, .(treatment), function(.) nrow(.)/nrow(test_set) ) c(nrow(training_set), nrow(test_set), nrow(dat)) # lengths of sets </code></pre> <p>Here, I use <code>set.seed()</code> to ensure identical behavior of <code>sample()</code> when constructing the training/test sets with <code>ddply</code>. This strikes me as a bit of a hack; perhaps there is another way to achieve the same result using a single call to <code>**ply</code> (but returning two dataframes). Another option (without egregious use of <code>set.seed</code>) would be to use <code>dlply</code> and then piece together elements of the resulting list into training/test sets:</p> <pre><code>set.seed(101) # for consistancy with 'ddply' above split_set = dlply(dat, .(treatment), function(.) { s = sample(1:nrow(.), trunc(nrow(.) * tr_prop)); list(.[s, ], .[-s,]) } ) # join together with ldply() training_set = ldply(split_set, function(.) .[[1]]) test_set = ldply(split_set, function(.) .[[2]]) </code></pre>

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