TableA
Col1
----------
1
2
3
4....all the way to 27
I want to add a second column that assigns a number to groups of 5.
Results
Col1 Col2
----- ------
1 1
2 1
3 1
4 1
5 1
6 2
7 2
8 2...and so on
The 6th group should have 2 rows in it.
NTILE doesn't accomplish what I want because of the way NTILE handles the groups if they aren't divisible by the integer.
If the number of rows in a partition is not divisible by integer_expression, this will cause groups of two sizes that differ by one member. Larger groups come before smaller groups in the order specified by the OVER clause. For example if the total number of rows is 53 and the number of groups is five, the first three groups will have 11 rows and the two remaining groups will have 10 rows each. If on the other hand the total number of rows is divisible by the number of groups, the rows will be evenly distributed among the groups. For example, if the total number of rows is 50, and there are five groups, each bucket will contain 10 rows.
This is clearly demonstrated in this SQL Fiddle. Groups 4, 5, 6 each have 4 rows while the rest have 5. I have some started some solutions but they were getting lengthy and I feel like I'm missing something and that this could be done in a single line.
You can use this:
;WITH CTE AS
(
SELECT col1,
RN = ROW_NUMBER() OVER(ORDER BY col1)
FROM TableA
)
SELECT col1, (RN-1)/5+1 col2
FROM CTE;
In your sample data, col1 is a correlative without gaps, so you could use it directly (if it's an INT) without using ROW_NUMBER(). But in the case that it isn't, then this answer works too. Here is the modified sqlfiddle.
A bit of math can go a long way. subtracting 1 from all values puts the 5s (edge cases) into the previous group here, and 6's into the next. flooring the division by your group size and adding one give the result you're looking for. Also, the SQLFiddle example here fixes your iterative insert - the table only went up to 27.
SELECT col1,
floor((col1-1)/5)+1 as grpNum
FROM tableA
Related
I have a sample table named assets which looks like this:
id
name
block_no
1
asset1
2
2
asset2
2
3
asset3
3
There can be any number of assets in a specific block. I need a minimum of 100 rows from the table, and containing all the data from the block_no. Like, if there are 95 rows to block_no 2 and around 20 on block_no 3, I need all 20 of block_no 3 as if I am fetching data in packets based on block_no.
Is this possible and feasible?
Postgres 13 or later
There is a dead simple solution using WITH TIES in Postgres 13 or later:
SELECT *
FROM assets
WHERE block_no >= 2 -- your starting block
ORDER BY block_no
FETCH FIRST 100 ROWS WITH TIES;
This will return at least 100 rows (if enough qualify), plus all peers of the 100th row.
If your table isn't trivially small, an index on (block_no) is essential for performance.
See:
Get top row(s) with highest value, with ties
Older versions
Use the window function rank() in a subquery:
SELECT (a).*
FROM (
SELECT a, rank() OVER (ORDER BY block_no) AS rnk
FROM assets a
) sub
WHERE rnk <= 100;
Same result.
I use a little trick with the row type to strip the added rnk from the result. That's an optional addition.
See:
PostgreSQL equivalent for TOP n WITH TIES: LIMIT "with ties"?
I have a project where I am taking Documents from one system and importing them into another.
The first system has the documents and associated keywords stored. I have a query that will return the results which will then be used as the index file to import them into the new system. There are about 1.8 million documents involved so this means 1.8 million rows (One per document).
I need to divide the returned results into blocks of 40,000 to make importing them in batches of 40,000 at a time, rather than one long import.
I have the query to return the results I need. Just need to know how to take that and break it up for easier import. My apologies if I have included to little information. This is my first time here asking for help.
Use the built-in function ORA_HASH to divide the rows into 45 buckets of roughly the same number of rows. For example:
select * from some_table where ora_hash(id, 44) = 0;
select * from some_table where ora_hash(id, 44) = 1;
...
select * from some_table where ora_hash(id, 44) = 44;
The function is deterministic and will always return the same result for the same input. The resulting number starts with 0 - which is normal for a hash, but unusual for Oracle, so the query may look off-by-one at first. The hash works better with more distinct values, so pass in the primary key or another unique value if possible. Don't use a low-cardinality column, like a status column, or the buckets will be lopsided.
This process is in some ways inefficient, since you're re-reading the same table 45 times. But since you're dealing with documents, I assume the table scanning won't be the bottleneck here.
A prefered way to bucketing the ID is to use the NTILE analytic function.
I'll demonstrate this on a simplified example with a table with 18 rows that should be divided in four chunks.
select listagg(id,',') within group (order by id) from tab;
1,2,3,7,8,9,10,15,16,17,18,19,20,21,23,24,25,26
Note, that the IDs are not consecutive, so no arithmetic can be used - the NTILE gets the parameter of the requested number of buckets (4) and calculates the chunk_id
select id,
ntile(4) over (order by ID) as chunk_id
from tab
order by id;
ID CHUNK_ID
---------- ----------
1 1
2 1
3 1
7 1
8 1
9 2
10 2
15 2
16 2
17 2
18 3
19 3
20 3
21 3
23 4
24 4
25 4
26 4
18 rows selected.
All but the last bucket are of the same size, the last one can be smaller.
If you want to calculate the ranges - use simple aggregation
with chunk as (
select id,
ntile(4) over (order by ID) as chunk_id
from tab)
select chunk_id, min(id) ID_from, max(id) id_to
from chunk
group by chunk_id
order by 1;
CHUNK_ID ID_FROM ID_TO
---------- ---------- ----------
1 1 8
2 9 17
3 18 21
4 23 26
I am exploring windowing functions in Hive and I am able to understand the functionalities of all the UDFs. Although, I am not able to understand the partition by and order by that we use with the other functions. Following is the structure that is very similar to the query which I am planning to build.
SELECT a, RANK() OVER(partition by b order by c) as d from xyz;
Just trying to understand the background process involved for both keywords.
Appreciate the help :)
RANK() analytic function assigns a rank to each row in each partition in the dataset.
PARTITION BY clause determines how the rows to be distributed (between reducers if it is hive).
ORDER BY determines how the rows are being sorted in the partition.
First phase is distribute by, all rows in a dataset are distributed into partitions. In map-reduce each mapper groups rows according to the partition by and produces files for each partition. Mapper does initial sorting of partition parts according to the order by.
Second phase, all rows are sorted inside each partition.
In map-reduce, each reducer gets partitions files (parts of partitions) produced by mappers and sorts rows in the whole partition (sort of partial results) according to the order by.
Third, rank function assigns rank to each row in a partition. Rank function is being initialized for each partition.
For the first row in the partition rank starts with 1. For each next row Rank=previous row rank+1. Rows with equal values (specified in the order by) given the same rank, if the two rows share the same rank, next row rank is not consecutive.
Different partitions can be processed in parallel on different reducers. Small partitions can be processed on the same reducer. Rank function re-initializes when it crossing the partition boundary and starts with rank=1 for each partition.
Example (rows are already partitioned and sorted inside partitions):
SELECT a, RANK() OVER(partition by b order by c) as d from xyz;
a, b, c, d(rank)
----------------
1 1 1 1 --starts with 1
2 1 1 1 --the same c value, the same rank=1
3 1 2 3 --rank 2 is skipped because second row shares the same rank as first
4 2 3 1 --New partition starts with 1
5 2 4 2
6 2 5 3
If you need consecutive ranks, use dense_rank function. dense_rank will produce rank=2 for the third row in the above dataset.
row_number function will assign a position number to each row in the partition starting with 1. Rows with equal values will receive different consecutive numbers.
SELECT a, ROW_NUMBER() OVER(partition by b order by c) as d from xyz;
a, b, c, d(row_number)
----------------
1 1 1 1 --starts with 1
2 1 1 2 --the same c value, row number=2
3 1 2 3 --row position=3
4 2 3 1 --New partition starts with 1
5 2 4 2
6 2 5 3
Important note: For rows with the same values row_number or other such analytic function may have non-deterministic behavior and produce different numbers from run to run. First row in the above dataset may receive number 2 and second row may receive number 1 and vice-versa, because their order is not determined unless you will add one more column a to the order by clause. In this case all rows will always have the same row_number from run to run, their order values are different.
Consider the following table:
id gap groupID
0 0 1
2 3 1
3 7 2
4 1 2
5 5 2
6 7 3
7 3 3
8 8 4
9 2 4
Where groupID is the desired, computed column, such as its value is incremented whenever the gap column is greater than a threshold (in this case 6). The id column defines the sequential order of appearance of the rows (and it's already given).
Can you please help me figure out how to dynamically fill out the appropriate values for groupID?
I have looked in several other entries here in StackOverflow, and I've seen the usage of sum as an aggregate for a window function. I can't use sum because it's not supported in MonetDB window functions (only rank, dense_rank, and row_num). I can't use triggers (to modify the record insertion before it takes place) either because I need to keep the data mentioned above within a stored function in a local temporary table -- and trigger declarations are not supported in MonetDB function definitions.
I have also tried filling out the groupID column value by reading the previous table (id and gap) into another temporary table (id, gap, groupID), with the hope that this would force a row-by-row operation. But this has failed as well because it gives the groupID 0 to all records:
declare threshold int;
set threshold = 6;
insert into newTable( id, gap, groupID )
select A.id, A.gap,
case when A.gap > threshold then
(select case when max(groupID) is null then 0 else max(groupID)+1 end from newTable)
else
(select case when max(groupID) is null then 0 else max(groupID) end from newTable)
end
from A
order by A.id asc;
Any help, tip, or reference is greatly appreciated. It's been a long time already trying to figure this out.
BTW: Cursors are not supported in MonetDB either --
You can assign the group using a correlated subquery. Simply count the number of previous values that exceed 6:
select id, gap,
(select 1 + count(*)
from t as t2
where t2.id <= t.id and t2.gap > 6
) as Groupid
from t;
I have some data like this as shown below:
Acc_Id || Row_No
1 1
2 1
2 2
2 3
3 1
3 2
3 3
3 4
and I need a query to get the results as shown below:
Acc_Id || Row_No
1 1
2 3
3 4
Please consider that I'm a beginner in SQL.
I assume you want the Count of the row
SELECT Acc_Id, COUNT(*)
FROM Table
GROUP BY Acc_Id
Try this:
select Acc_Id, MAX(Row_No)
from table
group by Acc_Id
As a beginner then this is your first exposure to aggregation and grouping. You may want to look at the documentation on group by now that this problem has motivated your interest in a solutions. Grouping operates by looking at rows with common column values, that you specify, and collapsing them into a single row which represents the group. In your case values in Acc_Id are the names for your groups.
The other answers are both correct in the the final two columns are going to be equivalent with your data.
select Acc_Id, count(*), max(Row_No)
from T
group by Acc_Id;
If you have gaps in the numbering then they won't be the same. You'll have to decide whether you're actually looking for a count of rows of a maximum of a value within a column. At this point you can also consider a number of other aggregate functions that will be useful to you in the future. (Note that the actual values here are pretty much meaningless in this context.)
select Acc_Id, min(Row_No), sum(Row_No), avg(Row_No)
from T
group by Acc_Id;