## MySQL: Efficient way computing set powers of Venn-Diagram

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Given the 4 tables, each containing items and representing one set, how to get the count of the items in each compartment required to draw a Venn diagram as shown below. The calculation should take place in the MySQL server avoiding transmission of items to the application server.

Example tables:

s1: s2: s3: s4: +------+ +------+ +------+ +------+ | item | | item | | item | | item | +------+ +------+ +------+ +------+ | a | | a | | a | | a | +------+ +------+ +------+ +------+ | b | | b | | b | | c | +------+ +------+ +------+ +------+ | c | | c | | d | | d | +------+ +------+ +------+ +------+ | d | | e | | e | | e | +------+ +------+ +------+ +------+ | ... | | ... | | ... | | ... |

Now, I think I would calculate some set powers. Some examples with `I`

corresponding to `s1`

, `II`

to `s2`

, `III`

to `s3`

and `IV`

to `s4`

:

If I reinterpret `sx`

as being a set, I would write:

`|s1 ∩ s2 ∩ s3 ∩ s4|`

- the white 25 in the center`|(s1 ∩ s2 ∩ s4) \ s3|`

- the white 15 below right in relation to the center`|(s1 ∩ s4) \ (s2 ∪ s3)|`

- the white 5 on the bottom`|s1 \ (s2 ∪ s3 ∪ s4)|`

- the dark blue 60 on the blue ground- ... till 15.

How to calculate those powers efficiently on the MySQL server? Does MySQL provide a function aiding in the calculation?

A naive approach would be running a query for 1.

SELECT count(*) FROM( SELECT item FROM s1 INTERSECT SELECT item FROM s2 INTERSECT SELECT item FROM s3 INTERSECT SELECT item FROM s4);

and another query for 2.

SELECT count(*) FROM( SELECT item FROM s1 INTERSECT SELECT item FROM s2 INTERSECT SELECT item FROM s4 EXCEPT SELECT item FROM s3);

and so on, resulting in 15 queries.

Try something like this:

with universe as ( select * from s1 union select * from s2 union select * from s3 union select * from s4 ), regions as ( select case when s1.item is null then '0' else '1' end || case when s2.item is null then '0' else '1' end || case when s3.item is null then '0' else '1' end || case when s4.item is null then '0' else '1' end as Region from universe u left join s1 on u.item = s1.item left join s2 on u.item = s2.item left join s3 on u.item = s3.item left join s4 on u.item = s4.item ) select Region, count(*) from regions group by Region

Disclaimer: I only tested this in SQLite. You might need to `SET sql_mode='PIPES_AS_CONCAT'`

for the ANSI string concatenation to work in MySQL, or use the `concat`

function instead. The `WITH`

syntax is only supported starting from version 8.0 of MySQL, but you can use temporary tables or nested queries appropriately instead.

If the sets are very large you might want to index the `item`

column before querying in case the SQL optimizer won't figure it out by itself.

**Newest 'venn-diagram' Questions - Page 2,** MySQL: Efficient way computing set powers of Venn-Diagram · mysql venn-diagram set-intersection · Nov 25 '18 at 20:36 Rainer Rillke. 0. 0 up vote 3 down vote favorite Given the 4 tables, each containing items and representing one set, how to get the count of the ite

Following procedure:

- Created a stored procedure, which creates temporary in-memory tables containing the sets.
- Mind that MySQL does not allow you refer to a temporary in-memory table more than one time in a query.
- As noted, MySQL does not have an
`INTERSECT`

or`EXCEPT`

. But you can emulate them. By removing duplicates from your raw data/ raw sets, emulation can be even more simplified. - Decided to store the computed value into a variable each and output a table consisting of all 15 of those values corresponding to components.

What I came up with is currently https://gist.github.com/Rillke/c2da0921f8f2a047615f41fab8781c11

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The question is a little complex so the answers are. Let me explain K.T.'s answer

with universe as ( select * from s1 union select * from s2 union select * from s3 union select * from s4 ), regions as ( select case when s1.item is null then '0' else '1' end || case when s2.item is null then '0' else '1' end || case when s3.item is null then '0' else '1' end || case when s4.item is null then '0' else '1' end as Region from universe u left join s1 on u.item = s1.item left join s2 on u.item = s2.item left join s3 on u.item = s3.item left join s4 on u.item = s4.item ) select Region, count(*) from regions group by Region

The `universe`

results in the UNION of all tables (duplicates eliminated), something like

+------+ | item | +------+ | a | +------+ | b | +------+ | c | +------+ | d | +------+ | e | +------+ | ... | +------+

Then, s1, s2, s3 and s4 are joined

+------+---------+---------+---------+---------+ | item | s1.item | s2.item | s3.item | s4.item | +------+---------+---------+---------+---------+ | a | a | a | a | a | +------+---------+---------+---------+---------+ | b | b | b | b | NULL | +------+---------+---------+---------+---------+ | c | c | c | NULL | c | +------+---------+---------+---------+---------+ | d | d | NULL | d | d | +------+---------+---------+---------+---------+ | e | NULL | e | e | e | +------+---------+---------+---------+---------+ | ... | ... | ... | ... | ... | +------+---------+---------+---------+---------+

and converted to a binary string (0: if cell is NULL; 1: else) called `Region`

where the first digit corresponds to s1, the second to s2 and so on

+------+--------+ | item | Region | +------+--------+ | a | 1111 | +------+--------+ | b | 1110 | +------+--------+ | c | 1101 | +------+--------+ | d | 1011 | +------+--------+ | e | 0111 | +------+--------+ | ... | ... | +------+--------+

and finally aggregated and grouped by Region

+--------+-------+ | Region | count | +--------+-------+ | 1111 | 1 | +--------+-------+ | 1110 | 1 | +--------+-------+ | 1101 | 1 | +--------+-------+ | 1011 | 1 | +--------+-------+ | 0111 | 1 | +--------+-------+ | ... | | +--------+-------+

Note that regions having 0 set elements in them do not show up in the results and `0000`

never will (=item not part of any set s1, s2, s3, s4) so there are 15 regions.

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##### Comments

- If someone tells me convincingly it would be a lot easier to do it with Postgres, I would change the question accordingly. It should probably read "Open Source DBMS: ..." but that's too broad for SO.
- There is no
`INTERSECT`

and`EXCEPT`

in MySQL. So, you could use other RDBMS, which provides these features. - @MadhurBhaiya Wasn't aware of that. MariaDB introduced set operations with 10.3.
- Current solution: gist.github.com/Rillke/c2da0921f8f2a047615f41fab8781c11