How do I multiply all elements in one collection with all the elements in another collection?

Question

Say I have:

(def s1 [1 2 3 4 5])
(def s2 [1 2 3 4 5])

For every x in s1, I want to multiply it with every y in s2.

---

To clarify, I basically want the Cartesian product, so I don't think map works here.

Answer 1

(for [x1 s1
      x2 s2]
  (* x1 x2))

Answer 2

Here is the java 1.5 (or newer) equivalent of Michal's code:

List<Integer> numbers = new ArrayList<Integer>();    

for(int x1 : s1) {
  for(int x2 : s2) {
    numbers.add(x1 * x2);
  }
}

The difference is that for loops in java don't return a sequence like they do in clojure, so you need to use a mutable ArrayList to construct the result.

Definitely not as pretty as the clojure version, but much better than what you would have had to do in Java 1.4.

Answer 3

A simple, visual demonstration of the basic functionality of for:

user=> (pprint 
         (for [tens (range 10) 
               ones (range 10)]
           [tens ones]))
([0 0]
 [0 1]
 [0 2]
 [0 3]
 [0 4]
 [0 5]
 [0 6]
 [0 7]
 [0 8]
 [0 9]
 [1 0]
 [1 1]
 [1 2]
 [1 3]
 [1 4]
 [1 5]
 [1 6]
 [1 7]
 [1 8]
 [1 9]
 [2 0]
 [2 1]
 [2 2]
 [2 3]
 [2 4]
 [2 5]
 [2 6]
 [2 7]
 [2 8]
 [2 9]
 [3 0]
 [3 1]
 [3 2]
 [3 3]
 [3 4]
 [3 5]
 [3 6]
 [3 7]
 [3 8]
 [3 9]
 [4 0]
 [4 1]
 [4 2]
 [4 3]
 [4 4]
 [4 5]
 [4 6]
 [4 7]
 [4 8]
 [4 9]
 [5 0]
 [5 1]
 [5 2]
 [5 3]
 [5 4]
 [5 5]
 [5 6]
 [5 7]
 [5 8]
 [5 9]
 [6 0]
 [6 1]
 [6 2]
 [6 3]
 [6 4]
 [6 5]
 [6 6]
 [6 7]
 [6 8]
 [6 9]
 [7 0]
 [7 1]
 [7 2]
 [7 3]
 [7 4]
 [7 5]
 [7 6]
 [7 7]
 [7 8]
 [7 9]
 [8 0]
 [8 1]
 [8 2]
 [8 3]
 [8 4]
 [8 5]
 [8 6]
 [8 7]
 [8 8]
 [8 9]
 [9 0]
 [9 1]
 [9 2]
 [9 3]
 [9 4]
 [9 5]
 [9 6]
 [9 7]
 [9 8]
 [9 9])

Answer 4

While solution using for is nicer, here is a map-only version if you have troubles understanding for:

(map #(map (partial * %) s2) s1)

for above expands to something similar, except it would use another anonymous function instead of partial, something like this:

(map (fn [x] (map (fn [y] (* x y)) s2)) s1)

or, neatly formated:

(map
  (fn [x]
    (map
      (fn [y]
        (* x y))
      s2))
  s1)

Answer 5

As simple as it can get:

(map * '(1 2) '(3 4))

will yield:

(3 8)