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  1. IMPALA
  2. IMPALA-8050

IS [NOT] NULL gives wrong selectivity when null count is missing

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Details

    • Bug
    • Status: Resolved
    • Minor
    • Resolution: Duplicate
    • Impala 3.1.0
    • Impala 4.0.0
    • Frontend
    • None
    • ghx-label-3

    Description

      Suppose we have the following query:

      select *
      from tpch.customer c
      where c.c_mktsegment is null
      

      If we have a null count, we can estimate selectivity based on that number. In the case of the TPC-H test data, after a recent fix to add null count back, null count is zero so the cardinality of the predicate c.c_mktsegment is null is 0 and no rows should be returned. Yet, the query plan shows:

      PLAN-ROOT SINK
      |
      00:SCAN HDFS [tpch.customer c]
         partitions=1/1 files=1 size=23.08MB row-size=218B cardinality=15.00K
         predicates: c.c_comment IS NULL
      

      So, the first bug is that the existing code which is supposed to consider null count (found in IsNullPredicate.analyzeImpl() does not work. Reason: the code in ColumnStats to check if we have nulls is wrong:

        public boolean hasNulls() { return numNulls_ > 0; }
      

      Zero is a perfectly valid null count: it means a NOT NULL column. The marker for a missing null count is -1 as shown in another method:

        public boolean hasStats() { return numNulls_ != -1 || numDistinctValues_ != -1; }
      

      This is probably an ambiguity in the name: does "has nulls" mean:

      • Do we have valid null count stats?
      • Do we have null count stats and we have at least some nulls?

      Fortunately, the only other use of this method is in (disabled) tests.

      Handle Missing Null Counts

      Second, if the null count is not available (for older stats), the next-best approximation is 1/NDV. The code currently guesses 0.1. The 0.1 estimate is fine if NDV is not available either.

      Note that to properly test some of these cases requires new tables in the test suite with no or partial stats.

      Special Consideration for Outer Joins

      When this predicate is applied to the result of an outer join, the estimation methods above will not work. Using the table null count to estimate an outer join null count is clearly wrong, as is using the table NDV value. The fall-back of .1 will tend to under-estimate an outer join.

      Instead, what is needed is a more complex estimate. Assume a left outer join (all rows from left, plus matching rows from right.)

      |join| = |left 𝜎 key is not null| * |right|/|key| + |left 𝜎 key is null|
      

      So, to estimate IS NULL or IS NOT NULL after an outer join must use a different algorithm then when estimating it in a scan.

      This suggests that expression selectivity is not an independent exercise as the code currently assumes it is. Instead, it must be aware of its context. In this case, the underlying null count for the column in the predicate must be adjusted when used in an outer join.

      The following TPC-H query gives a very clear example (see card-join.test):

      select c.c_custkey, o.o_orderkey
      from tpch.customer c
      left outer join tpch.orders o on c.c_custkey = o.o_custkey
      where o.o_clerk is null
      

      The plan, with the IS NULL filter applied twice (correct structure, wrong cardinality estimate):

      PLAN-ROOT SINK
      |
      02:HASH JOIN [RIGHT OUTER JOIN]
      |  hash predicates: o.o_custkey = c.c_custkey
      |  other predicates: o.o_clerk IS NULL
      |  runtime filters: RF000 <- c.c_custkey
      |  row-size=51B cardinality=0
      |
      |--00:SCAN HDFS [tpch.customer c]
      |     partitions=1/1 files=1 size=23.08MB row-size=8B cardinality=150.00K
      |
      01:SCAN HDFS [tpch.orders o]
         partitions=1/1 files=1 size=162.56MB row-size=43B cardinality=1.50M
         runtime filters: RF000 -> o.o_custkey
      

      The math:

      • The query obtains all customer rows, |customer| = 150K.
      • The query obtains all order rows where the clerk is null, which is none.
      • The query then left outer joins the customer table with orders. Since only 100K customers have orders, 50K do not. The result would be a join with 50K null clerks.
      • But, because the IS NULL calculations after the join consider only the orders null count, all the other rows are assumed discarded.

      So, it may be that the bug earlier, which accidentally turned off IS NULL handling, actually makes this particular use case work better...

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              Unassigned Unassigned
              Paul.Rogers Paul Rogers
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                Updated:
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