Common list elements: Difference between revisions

</pre>

=={{header|Wren}}==

{{libheader|Wren-seq}}

As we're dealing here with lists rather than sets, some guidance is needed on how to deal with duplicates in each list in the general case. A drastic solution would be to remove all duplicates from the result. Instead, the following matches duplicates - so if List A contains 2 'a's and List B contains 3 'a's, there would be 2 'a's in the result.

<lang ecmascript>import "/seq" for Lst

var common2 = Fn.new { |l1, l2|

// minimimze number of lookups

var c1 = l1.count

var c2 = l2.count

var shortest = (c1 < c2) ? l1 : l2

var longest = (l1 == shortest) ? l2 : l1

longest = longest.toList // matching duplicates will be destructive

var res = []

for (e in shortest) {

var ix = Lst.indexOf(longest, e)

if (ix >= 0) {

res.add(e)

longest.removeAt(ix)

}

}

return res

}

var commonN = Fn.new { |ll|

var n = ll.count

if (n == 0) return ll

if (n == 1) return ll[0]

var first2 = common2.call(ll[0], ll[1])

if (n == 2) return first2

return ll.skip(2).reduce(first2) { |acc, l| common2.call(acc, l) }

}

var lls = [

[[2, 5, 1, 3, 8, 9, 4, 6], [3, 5, 6, 2, 9, 8, 4], [1, 3, 7, 6, 9]],

[[2, 2, 1, 3, 8, 9, 4, 6], [3, 5, 6, 2, 2, 2, 4], [2, 3, 7, 6, 2]]

]

for (ll in lls) {

System.print("Intersection of %(ll) is:")

System.print(commonN.call(ll))

System.print()

}</lang>

{{out}}

<pre>

Intersection of [[2, 5, 1, 3, 8, 9, 4, 6], [3, 5, 6, 2, 9, 8, 4], [1, 3, 7, 6, 9]] is:

[3, 6, 9]

Intersection of [[2, 2, 1, 3, 8, 9, 4, 6], [3, 5, 6, 2, 2, 2, 4], [2, 3, 7, 6, 2]] is:

[2, 3, 6, 2]