Output of CSUM() in teradata - sql

Can anyone please help me in unstanding below csum function.
What will be the output in each case.
csum(1,1),
csum(1,1) + emp_no
csum(1,emp_no)+emp_no

CSUM is an old deprecated function from V2R3, over 15 years ago. It can always be rewritten using newer Standard SQL compliant syntax.
CSUM(1,1) returns the same as ROW_NUMBER() OVER (ORDER BY 1), a sequence starting with 1.
But you should never use it like that as ORDER BY 1 within a Windowed Aggregate Function is not the same as the final ORDER BY 1 of a SELECT, it's ordering all rows by the same value 1. Teradata calculates those functions in parallel based on the values in PARTITION BY and ORDER BY, this means all rows with the same PARTITION/ORDER data are processed on a single AMP, if there's only a single value one AMP will process all rows, resulting in a totally skewed distribution.
Instead of ORDER BY 1 you should use a column which is more or less unique in best case.
csum(1,emp_no)+emp_no is probably used with another SELECT to get the current maximum value of a column and add the new sequential values to it, i.e. creating your own gap-less sequence numbers.
This is the best way to do it:
SELECT ROW_NUMBER() OVER (ORDER BY column(s)_with_a_low_number_of_rows_per_value)
+ COALESCE((SELECT MAX(seqnum) FROM table),0)
,....
FROM table

Related

Finding the last 4, 3, 2, 1 months consecutive order drops among clients based on drop variance

Here I have this query that finds out the drop percentage of a bunch of clients based on the orders they have received(i.e. It finds the percentage difference in orders by comparing the current month with the previous month). What I want to achieve here is to have a field where I can see the clients who had 4 months continuous drop, 3 months drop, 2 months drop, and 1 month drop.
I know, it can only be achieved by comparing the last 4 months using the lag function or sub queries. can you guys pls help me out on this one, would appreciate it very much
select
fd.customers2, fd.Month1, fd.year1, fd.variance, case when
(fd.variance < -0.00001 and fd.year1 = '2022.0' and fd.Month1 = '1')
then '1month drop' else fd.customers2 end as 1_most_host_drop
from 
(SELECT
c.*,
sa.customers as customers2,
sum(sa.order) as orders,
date_part(mon, sa.date) as Month1,
date_part(year, sa.date) as year1,
(cast(orders - LAG(orders) OVER(Partition by customers2 ORDER BY
 year1, Month1) as NUMERIC(10,2))/NULLIF(LAG(orders) 
OVER(partition by customers2 ORDER BY year1, Month1) * 1, 0)) AS variance
FROM stats sa join (select distinct
    d.id, d.customers 
     from configer d 
    ) c on sa.customers=c.customers
WHERE sa.date >= '2021-04-1' 
GROUP BY Month1, sa.customers, c.id,  year1, 
     c.customers)fd
In a spirit of friendliness: I think you are a little premature in posting this here as there are several issues with the syntax before even reaching the point where you can solve the problem:
You have at least two places with a comma immediately preceding the word FROM:
...AS variance, FROM stats_archive sa ...
...d.customers, FROM config d...
Recommend you don't use VARIANCE as an alias (it is a system function in PostgreSQL and so is likely also a system function name in Redshift)
Not super important, but there's no need for c.* - just select the columns you will use
DATE_PART requires a string as the first parameter DATE_PART('mon',current_date)
I might be wrong about this, but I suspect you cannot use column aliases in the partition by or order by of a window function. Put the originating expressions there instead:
... OVER (PARTITION BY customers2 ORDER BY DATE_PART('year',sa.date),DATE_PART('mon',sa.date))
LAG has three parameters. (1) The column you want to retrieve the value from, (2) the row offset, where a positive integer indicates how many rows prior to the current row you should retrieve a value from according to the partition and order context and (3) the value the function should return as a default (in case of the first row in the partition). As such, you don't need NULLIF. So, to get the row immediately prior to the current row, or return 0 in case the current row is the first row in the partition:
LAG(orders,1,0) OVER (PARTITION BY customers2 ORDER BY DATE_PART('year',sa.date),DATE_PART('mon',sa.date))
If you use 0 as a default in the calculation of what is currently aliased variance, you will almost certainly run into a div/0 error either now, or worse, when you least expect it in the future. You should protect against that with some CASE logic or better, provide a more appropriate default value or even better, calculate the LAG with the default 0, then filter out the 0 rows before doing the calculation.
You can't use column aliases in the GROUP BY. You must reference each field that is not participating in an aggregate in the group by, whether through direct mention (sa.date) or indirectly in an expression (DATE_PART('mon',sa.date))
Your date should be '2021-04-01'
All in all, without sample data, expected results using the posted sample data and without first removing syntax errors, it is a tall order to attempt to offer advice on the problem which is any more specific than:
Build the source of the calculation as a completely separate query first. Calculate the LAG in that source query. Only when you've run that source query and verified that the LAG is producing the correct result should you then wrap it as a sub-query or CTE (not sure if Redshift supports these, but presumably) at which point you can filter out the rows with a zero as the denominator (the first month of orders for each customer).
Good luck!

Splitting table PK values into roughly same-size ranges

I have a table in Postgres with about half a million rows and an integer primary key.
I'd like to split its entire PK space into N ranges of approximately same size for independent processing. How do I best do it?
I apparently can do it by fetching all PK values to a client and remember every N-th value. This does a full scan and a fetch of all the values, while I only want no more than N+1 of them.
I can select min and max values and cut the range, but if the PKs are not distributed quite evenly, it may give me some ranges of seriously different sizes.
I want ranges for index-based access later on, so any modulo-based tricks do mot apply.
Is there any nice SQL-based solution that does not involve fetching all the keys to a client? Writing an N-specific query, e.g. with N clauses, if fine.
An example:
IDs in a range, say, from 1234 to 567890, N = 4.
I'd like to get 4 numbers, say 127123, 254789, 379860, so than there are approximately 125k records in each of the ranges of IDs [1234, 127123], [127123, 254789], [254789, 379860], [379860, 567890].
Update:
I've come up with a solution like this:
select
percentile_disc(0.25) within group (order by c.id) over() as pct_25
,percentile_disc(0.50) within group (order by c.id) over() as pct_50
,percentile_disc(0.75) within group (order by c.id) over() as pct_75
from customer c
limit 1
;
It does a decent job of giving me the exact range boundaries, and runs only a few seconds, which is fine for my purposes.
What bothers me is that I have to add the limit 1 clause to get just one row. Without it, I receive identical rows, one per record in the table. Is there a better way to get just a one row of the percentiles?
I think you can use row_number() for this purpose. Something like this:
select t.*,
floor((seqnum * N) / cnt) as range
from (select t.*,
row_number() over (order by pk) - 1 as seqnum,
count(*) over () as cnt
from t
) t;
This assumes by range that you mean ranges on pk values. You can also move the range expression to a where clause to just select one particular range.

Postgresql Writing max() Window function with multiple partition expressions?

I am trying to get the max value of column A ("original_list_price") over windows defined by 2 columns (namely - a unique identifier, called "address_token", and a date field, called "list_date"). I.e. I would like to know the max "original_list_price" of rows with both the same address_token AND list_date.
E.g.:
SELECT
address_token, list_date, original_list_price,
max(original_list_price) OVER (PARTITION BY address_token, list_date) as max_list_price
FROM table1
The query already takes >10 minutes when I use just 1 expression in the PARTITION (e.g. using address_token only, nothing after that). Sometimes the query times out. (I use Mode Analytics and get this error: An I/O error occurred while sending to the backend) So my questions are:
1) Will the Window function with multiple PARTITION BY expressions work?
2) Any other way to achieve my desired result?
3) Any way to make Windows functions, especially the Partition part run faster? e.g. use certain data types over others, try to avoid long alphanumeric string identifiers?
Thank you!
The complexity of the window functions partitioning clause should not have a big impact on performance. Do realize that your query is returning all the rows in the table, so there might be a very large result set.
Window functions should be able to take advantage of indexes. For this query:
SELECT address_token, list_date, original_list_price,
max(original_list_price) OVER (PARTITION BY address_token, list_date) as max_list_price
FROM table1;
You want an index on table1(address_token, list_date, original_list_price).
You could try writing the query as:
select t1.*,
(select max(t2.original_list_price)
from table1 t2
where t2.address_token = t1.address_token and t2.list_date = t1.list_date
) as max_list_price
from table1 t1;
This should return results more quickly, because it doesn't have to calculate the window function value first (for all rows) before returning values.

comparing current row and previous in oracle sql

I need to compare current row and previous row and based on some comparision need to derive a column value. Currently apporach I m following is making two differnt record sets and then use rank function and then by joining rank functions I m able to achiieve this. However, this seems to be tedious apporach, is there a better way to achieve this. I m currently writing query something like below :-
select
< comparing columns from two record sets and deriving column value>
(
select(<some complex logic>, rank from a) rcdset,
(select <some complex logic>, rank +1 from a) rcdset2 where rcdset.rnk = rcdset1.rnk (+)
Database - Oracle 10g
Use LAG(value_expr) OVER (ORDER BY rank_col) to retrieve the value (value_expr) from previous row (order defined by rank_col), see http://oracle-base.com/articles/misc/lag-lead-analytic-functions.php

Select finishes where athlete didn't finish first for the past 3 events

Suppose I have a database of athletic meeting results with a schema as follows
DATE,NAME,FINISH_POS
I wish to do a query to select all rows where an athlete has competed in at least three events without winning. For example with the following sample data
2013-06-22,Johnson,2
2013-06-21,Johnson,1
2013-06-20,Johnson,4
2013-06-19,Johnson,2
2013-06-18,Johnson,3
2013-06-17,Johnson,4
2013-06-16,Johnson,3
2013-06-15,Johnson,1
The following rows:
2013-06-20,Johnson,4
2013-06-19,Johnson,2
Would be matched. I have only managed to get started at the following stub:
select date,name FROM table WHERE ...;
I've been trying to wrap my head around the where clause but I can't even get a start
I think this can be even simpler / faster:
SELECT day, place, athlete
FROM (
SELECT *, min(place) OVER (PARTITION BY athlete
ORDER BY day
ROWS 3 PRECEDING) AS best
FROM t
) sub
WHERE best > 1
->SQLfiddle
Uses the aggregate function min() as window function to get the minimum place of the last three rows plus the current one.
The then trivial check for "no win" (best > 1) has to be done on the next query level since window functions are applied after the WHERE clause. So you need at least one CTE of sub-select for a condition on the result of a window function.
Details about window function calls in the manual here. In particular:
If frame_end is omitted it defaults to CURRENT ROW.
If place (finishing_pos) can be NULL, use this instead:
WHERE best IS DISTINCT FROM 1
min() ignores NULL values, but if all rows in the frame are NULL, the result is NULL.
Don't use type names and reserved words as identifiers, I substituted day for your date.
This assumes at most 1 competition per day, else you have to define how to deal with peers in the time line or use timestamp instead of date.
#Craig already mentioned the index to make this fast.
Here's an alternative formulation that does the work in two scans without subqueries:
SELECT
"date", athlete, place
FROM (
SELECT
"date",
place,
athlete,
1 <> ALL (array_agg(place) OVER w) AS include_row
FROM Table1
WINDOW w AS (PARTITION BY athlete ORDER BY "date" ASC ROWS BETWEEN 3 PRECEDING AND CURRENT ROW)
) AS history
WHERE include_row;
See: http://sqlfiddle.com/#!1/fa3a4/34
The logic here is pretty much a literal translation of the question. Get the last four placements - current and the previous 3 - and return any rows in which the athlete didn't finish first in any of them.
Because the window frame is the only place where the number of rows of history to consider is defined, you can parameterise this variant unlike my previous effort (obsolete, http://sqlfiddle.com/#!1/fa3a4/31), so it works for the last n for any n. It's also a lot more efficient than the last try.
I'd be really interested in the relative efficiency of this vs #Andomar's query when executed on a dataset of non-trivial size. They're pretty much exactly the same on this tiny dataset. An index on Table1(athlete, "date") would be required for this to perform optimally on a large data set.
; with CTE as
(
select row_number() over (partition by athlete order by date) rn
, *
from Table1
)
select *
from CTE cur
where not exists
(
select *
from CTE prev
where prev.place = 1
and prev.athlete = cur.athlete
and prev.rn between cur.rn - 3 and cur.rn
)
Live example at SQL Fiddle.