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copied!<p>Here is a general solution, similar in spirit to that of @Ramnath, but avoiding the use of <code>union()</code> which is a binary function. The trick is to note that <code>union()</code> is implemented as:</p> <pre><code>unique(c(as.vector(x), as.vector(y))) </code></pre> <p>and the bit inside <code>unique()</code> can be achieved by unlisting the <code>n</code>th component of each list.</p> <p>The full solution then is:</p> <pre><code>unionFun <- function(n, obj) { unique(unlist(lapply(obj, `[[`, n))) } lapply(seq_along(lst[[1]]), FUN = unionFun, obj = lst) </code></pre> <p>which gives:</p> <pre><code>[[1]] [1] 1 2 3 4 5 6 7 8 9 10 11 12 [[2]] [1] 6 7 8 9 10 11 1 2 3 4 5 12 </code></pre> <p>on the data you showed.</p> <p>A couple of useful features of this are:</p> <ul> <li>we use <code>`[[`</code> to subset <code>obj</code> in <code>unionFun</code>. This is similar to <code>function(x) x$a</code> in @Ramnath's Answer. However, we don't need an anonymous function (we use <code>`[[`</code> instead). The equivalent to @Ramnath's Answer is: <code>lapply(lst, `[[`, 1)</code></li> <li>to generalise the above, we replace the <code>1</code> above with <code>n</code> in <code>unionFun()</code>, and allow our list to be passed in as argument <code>obj</code>.</li> </ul> <p>Now that we have a function that will provide the union of the <code>n</code>th elements of a given list, we can <code>lapply()</code> over the indices <code>k</code>, applying our <code>unionFun()</code> to each sub-element of <code>lst</code>, using the fact that the length of <code>lst[[1]]</code> is the same as <code>length(lst[[k]])</code> for all <code>k</code>.</p> <p>If it helps to have the names of the <code>n</code>th elements in the returned object, we can do:</p> <pre><code>> unions <- lapply(seq_along(lst[[1]]), FUN = unionFun, obj = lst) > names(unions) <- names(lst[[1]]) > unions $a [1] 1 2 3 4 5 6 7 8 9 10 11 12 $b [1] 6 7 8 9 10 11 1 2 3 4 5 12 </code></pre>

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