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copied!<p>Your data</p> <pre><code>df <- list(structure(list(a = 1:5, b = 6:11), .Names = c("a", "b")), structure(list(a = 6:11, b = 1:5), .Names = c("a", "b")), structure(list(a = 12, b = 12), .Names = c("a", "b"))) </code></pre> <p>This gives you the unique values of the nested lists:</p> <pre><code>library(plyr) df.l <- llply(df, function(x) unlist(unique(x))) R> df.l [[1]] [1] 1 2 3 4 5 6 7 8 9 10 11 [[2]] [1] 6 7 8 9 10 11 1 2 3 4 5 [[3]] [1] 12 </code></pre> <p><strong>EDIT</strong></p> <p>Thanks to Ramnath I changed the code a bit and hope this answer fits the needs of your question. For illustration I keep the previous answer as well. The slightly changed data has now an additional list.</p> <pre><code>df <- list(structure(list(a = 1:5, b = 6:11), .Names = c("a", "b")), structure(list(a = 6:11, b = 1:5), .Names = c("a", "b")), structure(list(a = 12, b = 12, c = 10:14), .Names = c("a", "b", "c"))) f.x <- function(x.list) { x.names <- names(x.list) i <- combn(x.names, 2) l <- apply(i, 2, function(y) x.list[y]) llply(l, unlist) } </code></pre> <p>Now you can apply the function to your data.</p> <pre><code>all.l <- llply(df, f.x) llply(all.l, function(x) llply(x, unique)) R> [[1]] [[1]][[1]] [1] 1 2 3 4 5 6 7 8 9 10 11 [[2]] [[2]][[1]] [1] 6 7 8 9 10 11 1 2 3 4 5 [[3]] [[3]][[1]] [1] 12 [[3]][[2]] [1] 12 10 11 13 14 [[3]][[3]] [1] 12 10 11 13 14 </code></pre> <p>However, the nested structure is not very user friendly. That could be changed a bit...</p>

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