analyze_column(): handle null values in the common-value analysis

Three ways a column containing nulls produced a wrong answer, all from
nulls not being accounted for.

1. The "single distinct value" shortcut. count(distinct col) skips
   nulls in SQL, so a column holding nulls plus one other value reports
   num_distinct == 1 and took the shortcut, which reports that one value
   with a count of total_rows:

       insert [{'c': None}, {'c': None}, {'c': 'a'}, {'c': 'a'}]
       analyze_column('c').most_common   # was [(None, 4)]
       select c, count(*) group by c     # is [(None, 2), ('a', 2)]

   The shortcut now requires that the column has no nulls, so those
   columns fall through to the group by like any other.

2. The tie-break sort compared the values directly, so a null and a
   string with the same count raised TypeError:

       analyze_column('c', common_limit=5)
       # TypeError: '<' not supported between instances of 'NoneType' and 'str'

   sort_key() keeps nulls out of the comparison and sorts them last,
   which keeps the existing ordering of non-null ties - the test for
   'Terryterryterry' ahead of 'Kumar' at equal counts still passes.

3. truncate() turned a null into the string 'None', so a null was
   indistinguishable from a column really holding 'None'. That is the
   path analyze-tables takes, since it passes value_truncate=80.

num_distinct itself is unchanged: it still follows SQL and reports 0 for
an all-null column, which test_analyze_table_column_all_nulls pins. But
num_distinct can therefore be lower than len(most_common) when a column
holds nulls alongside all-distinct values - {'c': None, 'a', 'b'} gives
num_distinct 2 and three most_common entries. Whether null should count
as a distinct value is a question about what the number means, so I have
left it alone rather than change an output the tests already pin.

Tests: 6 new cases in tests/test_analyze.py, all failing on 6bc1d33.
Suite 1506 passed, 16 skipped. mypy, pyright, flake8, ty, black and
cog --check clean.
This commit is contained in:
feiiiiii5 2026-09-28 18:55:53 +08:00
commit 8e5dfdd409
No known key found for this signature in database
2 changed files with 77 additions and 3 deletions

View file

@ -5323,6 +5323,8 @@ class Table(Queryable):
total_rows = db[table].count
def truncate(value):
if value is None:
return None
if value_truncate is None or isinstance(value, (float, int)):
return value
value = str(value)
@ -5330,6 +5332,13 @@ class Table(Queryable):
value = value[:value_truncate] + "..."
return value
def sort_key(pair):
# Counts are the primary key. Ties are broken on the value
# itself, so keep nulls out of the comparison - None cannot be
# ordered against a string or a number - and sort them last.
value = pair[0]
return (pair[1], value is not None, "" if value is None else value)
table_quoted = quote_identifier(table)
column_quoted = quote_identifier(column)
num_null = db.execute(
@ -5343,7 +5352,8 @@ class Table(Queryable):
).fetchone()[0]
most_common_results = None
least_common_results = None
if num_distinct == 1:
if num_distinct == 1 and not num_null:
# Every row holds that one value, so skip the group by
value = db.execute(
f"select {column_quoted} from {table_quoted} limit 1"
).fetchone()[0]
@ -5363,7 +5373,7 @@ class Table(Queryable):
f"limit {common_limit}"
).fetchall()
]
most_common_results.sort(key=lambda p: (p[1], p[0]), reverse=True)
most_common_results.sort(key=sort_key, reverse=True)
if least_common:
if num_distinct <= common_limit:
# No need to run the query if it will just return the results in reverse order
@ -5378,7 +5388,7 @@ class Table(Queryable):
f"limit {common_limit}"
).fetchall()
]
least_common_results.sort(key=lambda p: (p[1], p[0]))
least_common_results.sort(key=sort_key)
return ColumnDetails(
self.name,
column,

View file

@ -51,3 +51,67 @@ def test_analyze_index_by_name(db):
assert list(db.table("sqlite_stat1").rows) == [
{"tbl": "two_indexes", "idx": "idx_two_indexes_species", "stat": "1 1"},
]
@pytest.mark.parametrize(
"values,expected_num_distinct",
(
# count(distinct col) skips nulls, which is why a column of nulls
# reports no distinct values at all
([None, None, None], 0),
([None, None, "a", "a"], 1),
(["a", "a", "b", "b"], 2),
),
)
def test_analyze_column_num_distinct_ignores_nulls(
fresh_db, values, expected_num_distinct
):
fresh_db["t"].insert_all([{"c": value} for value in values])
assert fresh_db["t"].analyze_column("c").num_distinct == expected_num_distinct
@pytest.mark.parametrize("value_truncate", (None, 80))
def test_analyze_column_most_common_includes_nulls(fresh_db, value_truncate):
# One distinct non-null value plus nulls used to take the
# "single distinct value" shortcut, which reported that one value
# with a count of total_rows and left the nulls out entirely
fresh_db["t"].insert_all([{"c": None}, {"c": None}, {"c": "a"}, {"c": "a"}])
details = fresh_db["t"].analyze_column(
"c", total_rows=4, value_truncate=value_truncate
)
assert dict(details.most_common) == {None: 2, "a": 2}
@pytest.mark.parametrize("value_truncate", (None, 80))
def test_analyze_column_most_common_ties_with_nulls(fresh_db, value_truncate):
# A null and a string with the same count used to be compared
# directly by the tie-break sort, which raised TypeError
fresh_db["t"].insert_all(
[{"c": None}, {"c": None}, {"c": "a"}, {"c": "a"}, {"c": "b"}]
)
details = fresh_db["t"].analyze_column("c", value_truncate=value_truncate)
assert dict(details.most_common) == {None: 2, "a": 2, "b": 1}
def test_analyze_column_least_common_ties_with_nulls(fresh_db):
# A null and a string share the lowest count, so the tie-break has
# to order them without comparing None against a string
fresh_db["t"].insert_all(
[{"c": None}, {"c": "a"}] + [{"c": c} for c in "bcdefg" for _ in range(2)]
)
details = fresh_db["t"].analyze_column("c", common_limit=5)
assert dict(details.least_common) == {
None: 1,
"a": 1,
"e": 2,
"f": 2,
"g": 2,
}
def test_analyze_column_does_not_render_null_as_the_string_none(fresh_db):
# analyze-tables passes value_truncate=80, and the str() of a null
# was indistinguishable from a column that really holds "None"
fresh_db["t"].insert_all([{"c": None}, {"c": None}, {"c": "None"}])
details = fresh_db["t"].analyze_column("c", total_rows=3, value_truncate=80)
assert dict(details.most_common) == {None: 2, "None": 1}