Metadata wide_table_type_distribution

Metadata benchmark · all Metadata queries

Representative SQL

Rendered in the benchmark’s native SQL with default parameters (no dialect translation available).

SELECT
  table_name,
  data_type,
  COUNT(*) AS type_count,
  COUNT(*) * 100.0 / SUM(COUNT(*)) OVER (PARTITION BY table_name) AS percentage
FROM information_schema.columns
WHERE
  table_name LIKE 'benchbox_wide_%'
GROUP BY
  table_name,
  data_type
ORDER BY
  table_name,
  type_count DESC

Representative DataFrame

No DataFrame implementation is registered for this query.

Get the exact query for your run

The blocks above are representative. To get the precise statement a specific platform runs at a specific scale, substitute your own values:

from benchbox import MetadataPrimitives

bench = MetadataPrimitives(scale_factor=1.0)            # your scale factor
print(bench.get_queries()['wide_table_type_distribution'])

Source

Query text: benchbox/core/metadata_primitives/queries.py