I have two arrays of strings, say
A = ('abc', 'joia', 'abas8', '09ma09', 'oiam0')
and
B = ('gfdg', '89jkjj', '09ma09', 'asda', '45645ghf', 'dgfdg', 'yui345gd', '6456ds', '456dfs3', 'abas8', 'sfgds').
What I want to do is simply to count the number of elements of every string in A that appears in B (if any). For example, the resulted array here should be: C = (0, 0, 1, 1, 0). How can I do that?
try this:
A.map( x => B.count(y => y == x)))
You can do it how idursun suggested, it will work, but may be not efficient as if you'll prepare intersection first. If B is much bigger than A it will give massive speedup. 'intersect' method has better 'big-O' complexity then doing linear search for each element of A in B.
val A = Array("abc", "joia", "abas8", "09ma09", "oiam0")
val B = Array("gfdg", "89jkjj", "09ma09", "asda", "45645ghf", "dgfdg", "yui345gd", "6456ds", "456dfs3", "abas8", "sfgds")
val intersectCounts: Map[String, Int] =
A.intersect(B).map(s => s -> B.count(_ == s)).toMap
val count = A.map(intersectCounts.getOrElse(_, 0))
println(count.toSeq)
Result
(0, 0, 1, 1, 0)
intersect and toMap. Just val intersectCounts = B.groupBy(identity).mapValues(_.size) should do the job. It's a bit faster too (still O(n+m), but less overhead).Use a foldLeft construction as the yield off of each element of A:
val A = List("a","b")
val B = List("b","b")
val C = for (a <- A)
yield B.foldLeft(0) { case (totalc : Int, w : String) =>
totalc + (if (w == a) 1 else 0)
}
And the result:
C: List[Int] = List(0, 2)
