I have computational task which can be reduced to the follow problem:
I have a large set of pairs of integers (key, val) which I want to group into windows. The first window starts with the first pair p ordered by key attribute and spans all the pairs where p[i].key belongs to [p[0].key; p[0].key + N), with some arbitrary integer N, positive and common to all windows.
The next window starts with the first pair ordered by key not included in the previous windows and again spans all the pairs from its key to key + N, and so on for the following windows.
The last step is to sum second attribute for each window and display it together with the first key of the window.
For example, given list of records with values:
key
val
1
3
2
7
5
1
6
4
7
1
10
3
13
5
and N=3, the windows would be:
{(1,3),(2,7)},
{(5,1),(6,4),(7,1)},
{(10,3)}
{(13,5)}
The final result:
key
sum_of_values
1
10
5
6
10
3
13
5
This is easy to program with a standard programming language but I have no clue how to solve this with SQL.
Note: If clickhouse doesn't support the RECURSIVE keyword, just remove that keyword from the expression.
Clickhouse seems to use non-standard syntax for the WITH clause. The below uses standard SQL. Adjust as needed.
Sorry. clickhouse may not support this approach. If not, we would need to find another method of walking through the data.
Standard SQL:
There are a few ways. Here's one approach. First assign row numbers to allow recursively stepping through the rows. We could use LEAD as well.
Assign a group (key value) to each row based on the current key and the last group/key value and whether they are within some distance (N = 3, in this case).
The last step is to just SUM these values per group start_key and to use the start_key value as the starting key in each group.
WITH RECURSIVE nrows (xkey, val, n) AS (
SELECT xkey, val, ROW_NUMBER() OVER (ORDER BY xkey) FROM test
)
, cte (xkey, val, n, start_key) AS (
SELECT xkey, val, n, xkey FROM nrows WHERE n = 1
UNION ALL
SELECT t1.xkey, t1.val, t1.n
, CASE WHEN t1.xkey <= t2.start_key + (3-1) THEN t2.start_key ELSE t1.xkey END
FROM nrows AS t1
JOIN cte AS t2
ON t2.n = t1.n-1
)
SELECT start_key
, SUM(val) AS sum_values
FROM cte
GROUP BY start_key
ORDER BY start_key
;
Result:
+-----------+------------+
| start_key | sum_values |
+-----------+------------+
| 1 | 10 |
| 5 | 6 |
| 10 | 3 |
| 13 | 5 |
+-----------+------------+
I'd like to create a query that returns a column with a repeating number sequence in it.
For example:
row_num | repeat
----------------
1 | 1
2 | 2
3 | 3
4 | 1
5 | 2
6 | 3
I'm struggling to understand how I could achieve this with BigQuery Standard SQL.
So far i've generated the row number (ROW_NUMBER() OVER()) as row_num in my select, and then I was thinking I could use a modulus function to determine the repeat number, but this would split it into several separate columns, so I'd need additional steps to merge them into the one column. I wondered if there was a more elegant way of achieving this.
Many Thanks!
In fact, the modulus should work here. Assuming your table already has a row_num column, and you want to generate the repeat column, you may try:
SELECT
row_num,
MOD(row_num - 1, 3) + 1 AS repeat
FROM yourTable
ORDER BY
row_num;
i have a table name conversion and i have these below mentioned columns in it i want to multiply Length\width row elements l*w of 'dimension' values and display them in another new table
Please let me know if anything changes for the same logic in ms access
probably it is simple but i dont know exact query to solve the problem waiting for your solutions
ID area length/width dimensions **new column(L*W) here**
1 1 l 3 3*5=15
2 1 w 5
3 2 l 4
4 2 w 8
5 3 l 6
6 3 w 10
7 4 l 12
8 4 w 13
9 4 W 10
waiting for your reply
You could query the table twice: once for lengths and once for widths and then join by area and multiply the values:
select length.area, length.dimension * width.dimension
from
(select area, dimension from conversion where lenwidth = 'l') length
inner join
(select area, dimension from conversion where lenwidth = 'w') width
on length.area = width.area;
Two remarks:
I suppose that it is a typo that you have two width entries for area 4? Otherwise you would have to decide which value to take in above select statement.
It would not be a good idea to keep the old table and have a new table holding the results. What if you change a value? You would have to remember to change the result accordingly every time. So either ditch the old table or use a view instead of a new table.
Try this
select *,
dimensions*(lead(dimensions) over(order by id)) product
from table1;
Or if you want for the set of area then
select *,
case when length_width='l' and (lead(length_width) over(order by id))='w'
then dimensions*(lead(dimensions) over(order by id))
else 0
end as product
from table1;
fiddle
EDIT after #NealB solution: the #NealB's solution is very very fast comparated with any another one, and dispenses this new question about "add a constraint to improve performance". The #NealB's not need any improve, have O(n) time and is very simple.
The problem of "label transitive groups with SQL" have an elegant solution using recursion and CTE... But this solution consumes an exponential time (!). I need to work with 10000 itens: with 1000 itens need 1 second, with 2000 need 1 day...
Constraint: in my case is possible to break the problem into pieces of ~100 itens or less, but only to select one group of ~10 itens, and discard all the other ~90 labeled itens...
There are a generic algotithm to add and use this kind of "pre-selection", to reduce the quadratic, O(N^2), time? Perhaps, as showed by comments and #wildplasser, a O(N log(N)) time; but I expect, with "pre-selection" to reduce to O(N) time.
(EDIT)
I try to use alternative algorithm, but it need some improvement to use as solution here; or, to really increase performance (to O(N) time), need to use "pre-selection".
The "pre-selection" (constraint) is based on a "super-set grouping"... Stating by the original "How to label 'transitive groups' with SQL?" question t1 table,
table T1
(original T1 augmented by "super-set grouping label" ssg, and more one row)
ID1 | ID2 | ssg
1 | 2 | 1
1 | 5 | 1
4 | 7 | 1
7 | 8 | 1
9 | 1 | 1
10 | 11 | 2
So there are three groups,
g1: {1,2,5,9} because "1 t 2", "1 t 5" and "9 t 1"
g2: {4,7,8} because "4 t 7" and "7 t 8"
g3: {10,11} because "10 t 11"
The super-group is only a auxiliary grouping,
ssg1: {g1,g2}
ssg2: {g3}
If we have M super-group-items and N total T1 items, the average group length will be less tham N/M. We can suppose (for my typical problem) also that ssg maximum length is ~N/M.
So, the "label algorithm" need to run only M times with ~N/M items if it use the ssg constraint.
An SQL only soulution appears to be a bit of a problem here. With the help of some procedural
programming on top of SQL the solution appears to be failry simple and efficient. Here is a brief outline
of a solution as could be implemented using any procedural language invoking SQL.
Declare table R with primary key ID where ID corresponds the same domain as ID1 and ID2 of table T1.
Table R contains one other non-key column, a Label number
Populate table R with the range of values found in T1. Set Label to zero (no label).
Using your example data, the initial setup for R would look like:
Table R
ID Label
== =====
1 0
2 0
4 0
5 0
7 0
8 0
9 0
Using a host language cursor plus an auxiliary counter, read each row from T1. Lookup ID1 and ID2 in R. You will find one of
four cases:
Case 1: ID1.Label == 0 and ID2.Label == 0
In this case neither one of these IDs have been "seen" before: Add 1 to the counter and then update both
rows of R to the value of the counter: update R set R.Label = :counter where R.ID in (:ID1, :ID2)
Case 2: ID1.Label == 0 and ID2.Label <> 0
In this case, ID1 is new but ID2 has already been assigned a label. ID1 needs to be assigned to the
same label as ID2: update R set R.Lablel = :ID2.Label where R.ID = :ID1
Case 3: ID1.Label <> 0 and ID2.Label == 0
In this case, ID2 is new but ID1 has already been assigned a label. ID2 needs to be assigned to the
same label as ID1: update R set R.Lablel = :ID1.Label where R.ID = :ID2
Case 4: ID1.Label <> 0 and ID2.Label <> 0
In this case, the row contains redundant information. Both rows of R should contain the same Label value. If not,
there is some sort of data integrity problem. Ahhhh... not quite see edit...
EDIT I just realized that there are situations where both Label values here could be non-zero and different. If both are non-zero and different then two Label groups need to be merged at this point. All you need to do is choose one Label and update the others to match with something like: update R set R.Label to ID1.Label where R.Label = ID2.Label. Now both groups have been merged with the same Label value.
Upon completion of the cursor, table R will contain Label values needed to update T2.
Table R
ID Label
== =====
1 1
2 1
4 2
5 1
7 2
8 2
9 1
Process table T2
using something along the lines of: set T2.Label to R.Label where T2.ID1 = R.ID. The end result should be:
table T2
ID1 | ID2 | LABEL
1 | 2 | 1
1 | 5 | 1
4 | 7 | 2
7 | 8 | 2
9 | 1 | 1
This process is puerly iterative and should scale to fairly large tables without difficulty.
I suggest you check this and use some
general-purpose language for solving it.
http://en.wikipedia.org/wiki/Disjoint-set_data_structure
Traverse the graph, maybe run DFS or BFS from each node,
then use this disjoint set hint. I think this should work.
The #NealB solution is the faster(!) See an example of PostgreSQL implementation here.
Below an example of another "brute force algorithm", only for curiosity!
As #peter.petrov and #RBarryYoung suggested, some performance problems can be avoided abandoning the CTE recursion... I do some issues at the basic labeler, and, abover I add the constraint for grouping by a super-set label. This new transgroup1_loop() function is working!
PS: this solution still have performance limitations, please post your answer with better, or with some adaptation of this one.
-- DROP table transgroup1;
CREATE TABLE transgroup1 (
id serial NOT NULL PRIMARY KEY,
items integer[], -- two or more items in the transitive relationship
ssg_label varchar(12), -- the super-set gropuping label
dels integer[] DEFAULT array[]::integer[]
);
INSERT INTO transgroup1(items,ssg_label) values
(array[1, 2],'1'),
(array[1, 5],'1'),
(array[4, 7],'1'),
(array[7, 8],'1'),
(array[9, 1],'1'),
(array[10, 11],'2');
-- or SELECT array[id1, id2],ssg_label FROM t1, with 10000 items
them, with these two functions we can solve the problem,
CREATE FUNCTION transgroup1_loop(p_ssg varchar, p_max_i integer DEFAULT 100)
RETURNS integer AS $funcBody$
DECLARE
cp_dels integer[];
i integer;
BEGIN
i:=1;
LOOP
UPDATE transgroup1
SET items = array_uunion(transgroup1.items,t2.items),
dels = transgroup1.dels || t2.id
FROM transgroup1 AS t1, transgroup1 AS t2
WHERE transgroup1.id=t1.id AND t1.ssg_label=$1 AND
t1.id>t2.id AND t1.items && t2.items;
cp_dels := array(
SELECT DISTINCT unnest(dels) FROM transgroup1
); -- ensures all itens to del
RAISE NOTICE '-- bug, repeting dels, item-%; % dels! %', i, array_length(cp_dels,1), array_to_string(cp_dels,';','*');
EXIT WHEN i>p_max_i OR array_length(cp_dels,1)=0;
DELETE FROM transgroup1
WHERE ssg_label=$1 AND id IN (SELECT unnest(cp_dels));
UPDATE transgroup1 SET dels=array[]::integer[];
i:=i+1;
END LOOP;
UPDATE transgroup1 -- only to beautify
SET items = ARRAY(SELECT unnest(items) ORDER BY 1 desc);
RETURN i;
END;
$funcBody$ LANGUAGE plpgsql VOLATILE;
to run and see results, you can use
SELECT transgroup1_loop('1'); -- run with ssg-1 items only
SELECT transgroup1_loop('2'); -- run with ssg-2 items only
-- show all with a sequential group label:
SELECT *, dense_rank() over (ORDER BY id) AS group_label from transgroup1;
results:
id | items | ssg_label | dels | group_label
----+-----------+-----------+------+-------------
4 | {8,7,4} | 1 | {} | 1
5 | {9,5,2,1} | 1 | {} | 2
6 | {11,10} | 2 | {} | 3
PS: the function array_uunion() is the same as original,
CREATE FUNCTION array_uunion(anyarray,anyarray) RETURNS anyarray AS $$
-- ensures distinct items of a concatemation
SELECT ARRAY(SELECT unnest($1) UNION SELECT unnest($2))
$$ LANGUAGE sql immutable;
I am a long time fan of Stack Overflow but I've come across a problem that I haven't found addressed yet and need some expert help.
I have a query that is sorted chronologically with a date-time compound key (unique, never deleted) and several pieces of data. What I want to know is if there is a way to find the start (or end) of a region where a value changes? I.E.
DateTime someVal1 someVal2 someVal3 target
1 3 4 A
1 2 4 A
1 3 4 A
1 2 4 B
1 2 5 B
1 2 5 A
and my query returns rows 1, 4 and 6. It finds the change in col 5 from A to B and then from B back to A? I have tried the find duplicates method and using min and max in the totals property however it gives me the first and last overall instead of the local max and min? Any similar problems?
I didn't see any purpose for the someVal1, someVal2, and someVal3 fields, so I left them out. I used an autonumber as the primary key instead of your date/time field; but this approach should also work with your date/time primary key. This is the data in my version of your table.
pkey_field target
1 A
2 A
3 A
4 B
5 B
6 A
I used a correlated subquery to find the previous pkey_field value for each row.
SELECT
m.pkey_field,
m.target,
(SELECT Max(pkey_field)
FROM YourTable
WHERE pkey_field < m.pkey_field)
AS prev_pkey_field
FROM YourTable AS m;
Then put that in a subquery which I joined to another copy of the base table.
SELECT
sub.pkey_field,
sub.target,
sub.prev_pkey_field,
prev.target AS prev_target
FROM
(SELECT
m.pkey_field,
m.target,
(SELECT Max(pkey_field)
FROM YourTable
WHERE pkey_field < m.pkey_field)
AS prev_pkey_field
FROM YourTable AS m) AS sub
LEFT JOIN YourTable AS prev
ON sub.prev_pkey_field = prev.pkey_field
WHERE
sub.prev_pkey_field Is Null
OR prev.target <> sub.target;
This is the output from that final query.
pkey_field target prev_pkey_field prev_target
1 A
4 B 3 A
6 A 5 B
Here is a first attempt,
SELECT t1.Row, t1.target
FROM t1 WHERE (((t1.target)<>NZ((SELECT TOP 1 t2.target FROM t1 AS t2 WHERE t2.DateTimeId<t1.DateTimeId ORDER BY t2.DateTimeId DESC),"X")));