Github user mgaido91 commented on a diff in the pull request: https://github.com/apache/spark/pull/21028#discussion_r187603861 --- Diff: sql/catalyst/src/main/scala/org/apache/spark/sql/catalyst/expressions/collectionOperations.scala --- @@ -529,6 +564,157 @@ case class ArrayContains(left: Expression, right: Expression) override def prettyName: String = "array_contains" } +/** + * Checks if the two arrays contain at least one common element. + */ +// scalastyle:off line.size.limit +@ExpressionDescription( + usage = "_FUNC_(a1, a2) - Returns true if a1 contains at least an element present also in a2. If the arrays have no common element and either of them contains a null element null is returned, false otherwise.", + examples = """ + Examples: + > SELECT _FUNC_(array(1, 2, 3), array(3, 4, 5)); + true + """, since = "2.4.0") +// scalastyle:off line.size.limit +case class ArraysOverlap(left: Expression, right: Expression) + extends BinaryArrayExpressionWithImplicitCast { + + override def dataType: DataType = BooleanType + + override def nullable: Boolean = { + left.nullable || right.nullable || left.dataType.asInstanceOf[ArrayType].containsNull || + right.dataType.asInstanceOf[ArrayType].containsNull + } + + override def nullSafeEval(a1: Any, a2: Any): Any = { + var hasNull = false + val arr1 = a1.asInstanceOf[ArrayData] + val arr2 = a2.asInstanceOf[ArrayData] + val (bigger, smaller, biggerDt) = if (arr1.numElements() > arr2.numElements()) { + (arr1, arr2, left.dataType.asInstanceOf[ArrayType]) + } else { + (arr2, arr1, right.dataType.asInstanceOf[ArrayType]) + } + if (smaller.numElements() > 0) { + val smallestSet = new mutable.HashSet[Any] + smaller.foreach(elementType, (_, v) => + if (v == null) { + hasNull = true + } else { + smallestSet += v + }) + bigger.foreach(elementType, (_, v1) => + if (v1 == null) { + hasNull = true + } else if (smallestSet.contains(v1)) { --- End diff -- actually it was not working also with `ArrayType`, so I addressed the problem in a more general way which supports both these cases. Thanks.
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