viirya commented on a change in pull request #33142:
URL: https://github.com/apache/spark/pull/33142#discussion_r661588066
##########
File path:
sql/catalyst/src/main/scala/org/apache/spark/sql/catalyst/expressions/EquivalentExpressions.scala
##########
@@ -93,25 +72,36 @@ class EquivalentExpressions {
*/
private def addCommonExprs(
exprs: Seq[Expression],
- addFunc: Expression => Boolean = addExpr): Unit = {
- val exprSetForAll = mutable.Set[Expr]()
- addExprTree(exprs.head, addExprToSet(_, exprSetForAll))
-
- val candidateExprs = exprs.tail.foldLeft(exprSetForAll) { (exprSet, expr)
=>
- val otherExprSet = mutable.Set[Expr]()
- addExprTree(expr, addExprToSet(_, otherExprSet))
- exprSet.intersect(otherExprSet)
+ map: mutable.HashMap[ExpressionEquals, ExpressionStats]): Unit = {
+ assert(exprs.length > 1)
+ var localEquivalenceMap = mutable.HashMap.empty[ExpressionEquals,
ExpressionStats]
+ addExprTree0(exprs.head, localEquivalenceMap)
+
+ exprs.tail.foreach { expr =>
+ val otherLocalEquivalenceMap = mutable.HashMap.empty[ExpressionEquals,
ExpressionStats]
+ addExprTree0(expr, otherLocalEquivalenceMap)
+ localEquivalenceMap = localEquivalenceMap.filter { case (key, _) =>
+ otherLocalEquivalenceMap.contains(key)
+ }
}
- // Not all expressions in the set should be added. We should filter out
the related
- // children nodes.
- val commonExprSet = candidateExprs.filter { candidateExpr =>
- candidateExprs.forall { expr =>
- expr == candidateExpr ||
expr.e.find(_.semanticEquals(candidateExpr.e)).isEmpty
+ localEquivalenceMap.foreach { case (commonExpr, state) =>
+ val possibleParents = localEquivalenceMap.filter { case (_, v) =>
v.height > state.height }
+ val notChild = possibleParents.forall { case (k, _) =>
+ k == commonExpr || k.e.find(_.semanticEquals(commonExpr.e)).isEmpty
+ }
+ if (notChild) {
+ // `commonExpr` appears in every expression in `exprs`, so we need to
call `addExprToMap`
+ // `exprs.length` times, to make `useCount` accurate.
+ exprs.indices.foreach(_ => addExprToMap(commonExpr.e, map))
Review comment:
`a + b + c` is not a common subexpression. You can think it as `(a + b >
1)` and `a + b + c` both run once. Only if there is another `a + b + c`, `a + b
+ c` will be considered a common subexpression.
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