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Would you mind posting the solution just for the record in case somebody else finds this topic while looking for it?

I learnt that the hard way - an old post http://newlisper.wordpress.com/2006/09/18/my-mistake-2/ describes my experience in tedious detail... To be fair I found it a bit weird when I first read about it as well. I think having h and o prefixes would have been clearer, but I guess as long as one kno...

See here in the documentation The string must begin with '0x' for hexadecimal strings or '0' (zero) for octal strings. If exp is invalid, int returns nil as a default value if not otherwise specified. If you prefix it with a 0, it will be evaluated as octal. There is no 9 or 8 in octal, since 8 dec ...

My candidate: (define lst '("A" "B" "C")) (define (permuations lst i) (if (= i 1) ;; if i = 1, then just return the list lst ;; otherwise permutate the list with the permutations of i-1 (apply append (map (fn (x) (map (fn (y) (append x y)) lst)) (permuations lst (- i 1))) 2))) (println (permuations ...

From the manual it seems like you are using find-all wrong. If the input is a list, then you cannot use a regex, but must use a list-matching-pattern. If you want to match the elements in the list, then try filter or map. What I think you want to do can be done like this: (filter (fn (x) (find-all {...

I'm not 100% sure what you wanted, since your description and the example seem to contradict, so let me see: You want all duplicates in a list, sorted by how often they occur? ;; reverse inner list, so that only the value is an item again (println (map (fn (x) (first x)) ;; filter out those elements...

Ah oh, thanks for the publicity I guess :)