How to: For each unique id, for each unique version, grab the best score and organize it into a table - sql

Question

Just wanted to preface this by saying while I do have a basic understanding, I am still fairly new to using Bigquery tables and sql statements in general.
I am trying to make a new view out of a query that grabs all of the best test scores for each version by each employee:
select emp_id,version,max(score) as score from `project.dataset.table` where type = 'assessment_test' group by version,emp_id order by emp_id
I'd like to take the results of that query, and make a new table comprised of employee id's with a column for each versions best score for that rows emp_id. I know that I can manually make a table for each version by including a "where version = a", "where version = b", etc.... and then joining all of the tables at the end but that doesn't seem like the most elegant solution plus there is about 20 different versions in total.
Is there a way to programmatically create a column for each unique version or at the very least use my initial query as maybe a subquery and just reference it, something like this:
with a as (
select id,version,max(score) as score
from `project.dataset.table`
where type = 'assessment_test' and version is not null and score is not null and id is not null
group by version,id
order by id),
version_a as (select score from a where version = 'version_a')
version_b as (select score from a where version = 'version_b')
version_c as (select score from a where version = 'version_c')
select
a.id as id,
version_a.score as version_a,
version_b.score as version_b,
version_c.score as version_c
from
a,
version_a,
version_b,
version_c
Example Picture: left table is example data, right table is expected output
Example Data:
id
version
score
1
a
88
1
b
93
1
c
92
2
a
89
2
b
99
2
c
78
3
a
95
3
b
83
3
c
89
4
a
90
4
b
90
4
c
86
5
a
82
5
b
78
5
c
98
1
a
79
1
b
97
1
c
77
2
a
100
2
b
96
2
c
85
3
a
83
3
b
87
3
c
96
4
a
84
4
b
80
4
c
77
5
a
95
5
b
77
Expected Output:
id
a score
b score
c score
1
88
97
92
2
100
99
85
3
95
87
96
4
90
90
86
5
95
78
98
Thanks in advance and feel free to ask any clarifying questions

Answer 1

Use below approach
select * from your_table
pivot (max(score) score for version in ('a', 'b', 'c'))
if applied to sample data in your question - output is
In case if versions is not known in advance - use below
execute immediate (select '''
select * from your_table
pivot (max(score) score for version in (''' || string_agg(distinct "'" || version || "'") || "))"
from your_table
)