jayzhan211 commented on code in PR #9628:
URL: https://github.com/apache/arrow-datafusion/pull/9628#discussion_r1527168866
##########
datafusion/optimizer/src/simplify_expressions/expr_simplifier.rs:
##########
@@ -1405,12 +1409,116 @@ impl<'a, S: SimplifyInfo> TreeNodeRewriter for
Simplifier<'a, S> {
Transformed::yes(lit(false))
}
+ // expr IN () --> false
+ // expr NOT IN () --> true
+ Expr::InList(InList {
+ expr,
+ list,
+ negated,
+ }) if list.is_empty() && *expr != Expr::Literal(ScalarValue::Null)
=> {
+ Transformed::yes(lit(negated))
+ }
+
+ // null in (x, y, z) --> null
+ // null not in (x, y, z) --> null
+ Expr::InList(InList {
+ expr,
+ list: _,
+ negated: _,
+ }) if is_null(expr.as_ref()) => Transformed::yes(lit_bool_null()),
+
+ // expr IN ((subquery)) -> expr IN (subquery), see ##5529
+ Expr::InList(InList {
+ expr,
+ mut list,
+ negated,
+ }) if list.len() == 1
+ && matches!(list.first(), Some(Expr::ScalarSubquery { .. })) =>
+ {
+ let Expr::ScalarSubquery(subquery) = list.remove(0) else {
+ unreachable!()
+ };
+
+ Transformed::yes(Expr::InSubquery(InSubquery::new(
+ expr, subquery, negated,
+ )))
+ }
+
+ // Combine multiple OR expressions into a single IN list
expression if possible
+ //
+ // i.e. `a = 1 OR a = 2 OR a = 3` -> `a IN (1, 2, 3)`
+ Expr::BinaryExpr(BinaryExpr {
+ left,
+ op: Operator::Or,
+ right,
+ }) if are_inlist_and_eq(left.as_ref(), right.as_ref()) => {
+ let left = as_inlist(left.as_ref());
+ let right = as_inlist(right.as_ref());
+
+ let lhs = left.unwrap();
+ let rhs = right.unwrap();
+ let lhs = lhs.into_owned();
+ let rhs = rhs.into_owned();
+ let mut seen: HashSet<Expr> = HashSet::new();
+ let list = lhs
+ .list
+ .into_iter()
+ .chain(rhs.list)
+ .filter(|e| seen.insert(e.to_owned()))
+ .collect::<Vec<_>>();
+
+ let merged_inlist = InList {
+ expr: lhs.expr,
+ list,
+ negated: false,
+ };
+
+ return Ok(Transformed::yes(Expr::InList(merged_inlist)));
+ }
+
// no additional rewrites possible
expr => Transformed::no(expr),
})
}
}
+fn are_inlist_and_eq(left: &Expr, right: &Expr) -> bool {
+ let left = as_inlist(left);
+ let right = as_inlist(right);
+ if let (Some(lhs), Some(rhs)) = (left, right) {
+ lhs.expr.try_into_col().is_ok()
+ && rhs.expr.try_into_col().is_ok()
+ && lhs.expr == rhs.expr
+ && !lhs.negated
+ && !rhs.negated
+ } else {
+ false
+ }
+}
+
+/// Try to convert an expression to an in-list expression
+fn as_inlist(expr: &Expr) -> Option<Cow<InList>> {
+ match expr {
+ Expr::InList(inlist) => Some(Cow::Borrowed(inlist)),
+ Expr::BinaryExpr(BinaryExpr { left, op, right }) if *op ==
Operator::Eq => {
+ match (left.as_ref(), right.as_ref()) {
+ (Expr::Column(_), Expr::Literal(_)) => Some(Cow::Owned(InList {
+ expr: left.clone(),
Review Comment:
Of course, removing cloned is my main goal
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