How to retrieve python list of SQLAlchemy result set?
Question:
I have the following query to retrieve a single column of data:
routes_query = select(
[schema.stop_times.c.route_number],
schema.stop_times.c.stop_id == stop_id
).distinct(schema.stop_times.c.route_number)
result = conn.execute(routes_query)
return [r['route_number'] for r in result]
I am wondering if there is a cleaner way to retrieve a native list of the data rows returned.
Answers:
the most succinct way to pull out a list of 1-element tuples into a list is:
result = [r[0] for r in result]
or:
result = [r for r, in result]
This is what I would use:
return zip(*result)[0]
It’s more succinct that the the list comprehension methods in zzzeek’s answer (22 characters rather than 29 or 31 characters), and for larger result sets the timings in this answer to a similar question show that it’s faster too.
I have the following query to retrieve a single column of data:
routes_query = select(
[schema.stop_times.c.route_number],
schema.stop_times.c.stop_id == stop_id
).distinct(schema.stop_times.c.route_number)
result = conn.execute(routes_query)
return [r['route_number'] for r in result]
I am wondering if there is a cleaner way to retrieve a native list of the data rows returned.
the most succinct way to pull out a list of 1-element tuples into a list is:
result = [r[0] for r in result]
or:
result = [r for r, in result]
This is what I would use:
return zip(*result)[0]
It’s more succinct that the the list comprehension methods in zzzeek’s answer (22 characters rather than 29 or 31 characters), and for larger result sets the timings in this answer to a similar question show that it’s faster too.