A recursive solution to find k-subset permutations (in pseudo-code): kSubsetPermutations(partial, set, k) { for (each element in set) { if (k equals 1) { store partial + element } else { make copy of set remove element from copy of set recurse with (partial + element, copy of set, k – 1) } } } … Read more
First, if you intend to do random access on a list, you should pick a list implementation that supports that efficiently. From the javadoc on LinkedList: All of the operations perform as could be expected for a doubly-linked list. Operations that index into the list will traverse the list from the beginning or the end, … Read more
nzcoops is spot on with his suggestion. I posed this question in the R Chat a while back and Paul Teetor suggested defining a new function: `%notin%` <- function(x,y) !(x %in% y) Which can then be used as follows: foo <- letters[1:6] > foo[foo %notin% c(“a”, “c”, “e”)] [1] “b” “d” “f” Needless to say, … Read more
For now, a temporary solution could be, `..` <- function (…, .env = globalenv()) { get(deparse(substitute(…)), env = .env) } ..(a) ## [1] “b” dt[a==..(a)] ## a b c ## 1: b a 15 ## 2: b a 11 ## 3: b b 8 ## 4: b b 4 ## 5: b c 5 ## … Read more
Jon Bentley discusses this in either ‘Programming Pearls’ or ‘More Programming Pearls’. You need to be careful with your N of M selection process, but I think the code shown works correctly. Rather than randomly shuffle all the items, you can do the random shuffle only shuffling the first N positions – which is a … Read more