SQL count (*) function query for a baseball data table explanation - sql

This line of code is simple and I understand what the query will return. I am trying to understand the third line of this code.
select First
, sum(rbi) as 'rbis'
**, count(*) as 'num_players'**
from #batter_stats
group by First
having count(*) >= 5
order by [rbis] desc
The count(*) as 'num_players' is confusing me as to what this does in this query.
Thanks you. I provided both the line of code and the data table.

In your code the aggregation functions
, sum(rbi) as rbis
, count(*) as num_players
agregated by column First
count(*) retunr the number of rows for the corresponding groped value
so in your case retunr the number of rows in your table for the corresponding value of olumn First

Related

Can't understand a deduplication code in BigQuery

I'm new to SQL, and i got this piece of code, inspired by this post Delete duplicate rows from a BigQuery table
I ran the code, and understood that it deduplicates my table, but i'm not sure how.
The code is the one that follows
WITH product_query AS(
SELECT
DISTINCT
v2ProductName,
productSKU
FROM `data-to-insights.ecommerce.all_sessions_raw`
WHERE v2ProductName IS NOT NULL
)
SELECT k.* FROM (
#aggregate the products into an array and only take 1 result
SELECT ARRAY_AGG (x LIMIT 1)[OFFSET(0)] k
FROM product_query x
GROUP BY productSKU
)
I think i got the first part, but, starting from "SELECT k.*"... i'm not sure what am i looking at. What the "k" and the "x" are supposed to mean? And what is the OFFSET function?

How to compare ordered datasets with the dataset before?

I have the following query:
select * from events order by Source, DateReceived
This gives me something like this:
I would like to get the results which i marked blue -> When there are two or more equal ErrorNr-Entries behind each other FROM THE SAME SOURCE.
So I have to compare every row with the row before. How can I achieve that?
This is what I want to get:
Apply the row number over partition by option on your table:
SELECT
ROW_NUMBER() OVER(PARTITION BY Source ORDER BY datereceived)
AS Row,
* FROM events
Either you can run a (max) having > 1 option on the result set's row number. Or if you need the details, apply the same query deducting the row nuber with 1.
Then you can make a join on the source and the row numbers and if the error nr is the same then you have a hit.
You can use the partition by as below.
select * from(select
*,row_number()over(partition by source,errornr order by Source, DateReceived) r
from
[yourtable])t
where r>1
You can specify your column names in the outer select.

Transposing and summing the top 5 results in Teradata SQL Assistant

I have a query that I converted from Access and is currently working correctly in Teradata SQL Assistant. The data pulled is just a standard table full of all of the data I need.
What I am wondering is: Can something be added to this query that will essentially sum up all of the Exposure values and then only show the top 5 Divisions by greatest to smallest sum (of those Top 5). Also, transposing the data so that my Topics are the left most column.
Here is the working code, details omitted.
SELECT
A.AS_OF_DT
, B.DIVISION
, B.CLASS
, Sum(A.BALANCE/1000000) AS "Bal in MMs"
, Sum(A.EXPOSURE/1000000) AS "Exp in MMs"
, Sum(CASE WHEN A.STATUS = 'NACC' THEN (B.BALANCE/1000000) ELSE 0 END) AS "NPL Bal as MMs"
FROM DB.TABLE1 A LEFT JOIN DB.TABLE2 B ON A.NAICS = B.NAICS_CD
WHERE A.AS_OF_DT= '2017-03-31'
GROUP BY
A.AS_OF_DT,
B.DIVISION,
B.CLASS
ORDER BY SUM (A.EXPOSURE/1000000) DESC
Essentially I want the columns to be the following:
DIVISION|DATE|
Below DIVISION would only be the Top 5 DIVISIONS summarized by EXPOSURE (under DATE)
I can try and clarify if needed. Just let me know.
Thanks!
End result is to have a datapaste I can throw into Excel without the manual work of transposing the data in Excel along with writing formulas to rummage through the 1000's of results of the base query to find summarize the individual Divisions and then picking the top 5 each month.
Thanks!
Shill
To get the 5 top for each division, you can use QUALIFY.
Add this to the end of you query:
QUALIFY ROW_NUMBER() over (PARTITION BY AS_OF_DATE,DIVISION order by (SUM (A.EXPOSURE/1000000))
For your other questions, SQL Assistant isn't much of a presentation tool, it won't do what you are asking for.
If your query already work,
try replacing:
SELECT
By:
SELECT top 10
(line 1)

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.

Oracle Group by issue

I have the below query. The problem is the last column productdesc is returning two records and the query fails because of distinct. Now i need to add one more column in where clause of the select query so that it returns one record. The issue is that the column i need
to add should not be a part of group by clause.
SELECT product_billing_id,
billing_ele,
SUM(round(summary_net_amt_excl_gst/100)) gross,
(SELECT DISTINCT description
FROM RES.tariff_nt
WHERE product_billing_id = aa.product_billing_id
AND billing_ele = aa.billing_ele) productdescr
FROM bil.bill_sum aa
WHERE file_id = 38613 --1=1
AND line_type = 'D'
AND (product_billing_id, billing_ele) IN (SELECT DISTINCT
product_billing_id,
billing_ele
FROM bil.bill_l2 )
AND trans_type_desc <> 'Change'
GROUP BY product_billing_id, billing_ele
I want to modify the select statement to the below way by adding a new filter to the where clause so that it returns one record .
(SELECT DISTINCT description
FROM RRES.tariff_nt
WHERE product_billing_id = aa.product_billing_id
AND billing_ele = aa.billing_ele
AND (rate_structure_start_date <= TO_DATE(aa.p_effective_date,'yyyymmdd')
AND rate_structure_end_date > TO_DATE(aa.p_effective_date,'yyyymmdd'))
) productdescr
The aa.p_effective_date should not be a part of GROUP BY clause. How can I do it? Oracle is the Database.
So there are multiple RES.tariff records for a given product_billing_id/billing_ele, differentiated by the start/end dates
You want the description for the record that encompasses the 'p_effective_date' from bil.bill_sum. The kicker is that you can't (or don't want to) include that in the group by. That suggests you've got multiple rows in bil.bill_sum with different effective dates.
The issue is what do you want to happen if you are summarising up those multiple rows with different dates. Which of those dates do you want to use as the one to get the description.
If it doesn't matter, simply use MIN(aa.p_effective_date), or MAX.
Have you looked into the Oracle analytical functions. This is good link Analytical Functions by Example