Clojure - Splitting a vector

Question

If I have two arguments [[1 2] [3 4]] and [5 6], how can I get to [[1 5] [2 6] [3 5] [4 6]].

I thought I may have to use for so I tried,

(fo         


        

Answer 1

This is one (hmm - not particularly elegant) way to get your answer, without using for:

(defn f [x y]
  (let [x' (apply concat x)
        multiples (/ (count x') (count y))
        y' (apply concat (repeat multiples y))]
    (mapv vector x' y')))

(f [[1 2] [3 4]] [5 6])
;;=> [[1 5] [2 6] [3 5] [4 6]]

Answer 2

(mapcat #(map vector % [5 6]) [[1 2] [3 4]])

or using for:

(for [c [[1 2] [3 4]]
      p (map vector c [5 6])]
  p)

Answer 3

If I understood your question correctly, your solution is actually very close. You did not need to nest the for expressions explicitly, since doing so creates a new list at every level, instead, just use multiple bindings:

(for [x [[1 2][3 4]]
      xx x
      y [5 6]]
  [xx y])

Which will result in

([1 5] [1 6] [2 5] [2 6] [3 5] [3 6] [4 5] [4 6])

Edit

Now that I finally read your question carefully, I came up with the same answer as @Lee did (mapcat (fn[x] (map vector x [5 6])) [[1 2] [3 4]]), which is the right one and should probably be accepted.