按组分区的单个列值的所有可能的唯一组合
All possible unique combination of a single column value partitioned by groups
我在 Google BigQuery 中有一个 table,如下所示。我正在尝试获取在 Group
.
上分区的 Item
列的所有可能的唯一组合(除空子集之外的所有子集)
Group Item
1 A
1 B
1 C
2 X
2 Y
2 Z
我正在寻找如下输出:
Group Item
1 A
1 B
1 C
1 A,B
1 B,C
1 A,C
1 A,B,C
2 X
2 Y
2 Z
2 X,Y
2 Y,Z
2 X,Z
2 X,Y,Z
我曾尝试使用这个公认的答案来合并 Group 但无济于事:
试试这个
with _data as
(
select 1 as _Group,'A' as Item union all
select 1 as _Group,'B' as Item union all
select 1 as _Group,'C' as Item union all
select 2 as _Group,'X' as Item union all
select 2 as _Group,'Y' as Item union all
select 2 as _Group,'Z' as Item
)
select distinct _Group ,Item from
(
select _Group,
Item
from _data
union all
select _Group,
string_agg(Item ,',') over(partition by _Group order by Item ) as item
from _data
union all
select a._Group ,
concat(a.item,',',b.item)
from _data a left join _data b on a._group = b._group and a.Item < b.Item
)
where item is not null
order by _group
考虑以下方法
CREATE TEMP FUNCTION generate_combinations(a ARRAY<STRING>)
RETURNS ARRAY<STRING>
LANGUAGE js AS '''
var combine = function(a) {
var fn = function(n, src, got, all) {
if (n == 0) {
if (got.length > 0) {
all[all.length] = got;
} return;
}
for (var j = 0; j < src.length; j++) {
fn(n - 1, src.slice(j + 1), got.concat([src[j]]), all);
} return;
}
var all = []; for (var i = 1; i < a.length; i++) {
fn(i, a, [], all);
}
all.push(a);
return all;
}
return combine(a)
''';
with your_table as (
select 1 as _Group,'A' as Item union all
select 1, 'B' union all
select 1, 'C' union all
select 2, 'X' union all
select 2, 'Y' union all
select 2, 'Z'
)
select _group, item
from (
select _group, generate_combinations(array_agg(item)) items
from your_table
group by _group
), unnest(items) item
有输出
我在 Google BigQuery 中有一个 table,如下所示。我正在尝试获取在 Group
.
Item
列的所有可能的唯一组合(除空子集之外的所有子集)
Group Item
1 A
1 B
1 C
2 X
2 Y
2 Z
我正在寻找如下输出:
Group Item
1 A
1 B
1 C
1 A,B
1 B,C
1 A,C
1 A,B,C
2 X
2 Y
2 Z
2 X,Y
2 Y,Z
2 X,Z
2 X,Y,Z
我曾尝试使用这个公认的答案来合并 Group 但无济于事:
试试这个
with _data as
(
select 1 as _Group,'A' as Item union all
select 1 as _Group,'B' as Item union all
select 1 as _Group,'C' as Item union all
select 2 as _Group,'X' as Item union all
select 2 as _Group,'Y' as Item union all
select 2 as _Group,'Z' as Item
)
select distinct _Group ,Item from
(
select _Group,
Item
from _data
union all
select _Group,
string_agg(Item ,',') over(partition by _Group order by Item ) as item
from _data
union all
select a._Group ,
concat(a.item,',',b.item)
from _data a left join _data b on a._group = b._group and a.Item < b.Item
)
where item is not null
order by _group
考虑以下方法
CREATE TEMP FUNCTION generate_combinations(a ARRAY<STRING>)
RETURNS ARRAY<STRING>
LANGUAGE js AS '''
var combine = function(a) {
var fn = function(n, src, got, all) {
if (n == 0) {
if (got.length > 0) {
all[all.length] = got;
} return;
}
for (var j = 0; j < src.length; j++) {
fn(n - 1, src.slice(j + 1), got.concat([src[j]]), all);
} return;
}
var all = []; for (var i = 1; i < a.length; i++) {
fn(i, a, [], all);
}
all.push(a);
return all;
}
return combine(a)
''';
with your_table as (
select 1 as _Group,'A' as Item union all
select 1, 'B' union all
select 1, 'C' union all
select 2, 'X' union all
select 2, 'Y' union all
select 2, 'Z'
)
select _group, item
from (
select _group, generate_combinations(array_agg(item)) items
from your_table
group by _group
), unnest(items) item
有输出