I need to get the max values from a list of values obtained from a query.
Basically, the problem is this:
I have 2 tables:
Lawyer
id (PK)
surname
name
Case
id (PK)
id_Client
date
id_Lawyer (FK)
And I need to get the Lawyer with the largest number of cases...(There is not problem with that) but, if exist more than one lawyer with the largest number of cases, I should list them.
Any help on this would be appreciated.
SELECT l.*, cases
FROM (
SELECT "id_Lawyer", count(*) AS cases, rank() OVER (ORDER BY count(*) DESC) AS rnk
FROM "Case"
GROUP BY 1
) c
JOIN "Lawyer" l ON l.id = c."id_Lawyer"
WHERE c.rnk = 1;
Basics for the technique (like #FuzzyTree provided):
PostgreSQL equivalent for TOP n WITH TIES: LIMIT "with ties"?
You only need a single subquery level since you can run window functions over aggregate functions:
Get the distinct sum of a joined table column
Best way to get result count before LIMIT was applied
Aside: It's better to use legal, lower case, unquoted identifiers in Postgres. Never use a reserved word like Case, that can lead to very confusing errors.
Related
Let's say I have an Oracle table with measurements in different categories:
CREATE TABLE measurements (
category CHAR(8),
value NUMBER,
error NUMBER,
created DATE
)
Now I want to find the "best" row in each category, where "best" is defined like this:
It has the lowest errror.
If there are multiple measurements with the same error, the one that was created most recently is the considered to be the best.
This is a variation of the greatest N per group problem, but including two columns instead of one. How can I express this in SQL?
Use ROW_NUMBER:
WITH cte AS (
SELECT m.*, ROW_NUMBER() OVER (PARTITION BY category ORDER BY error, created DESC) rn
FROM measurements m
)
SELECT category, value, error, created
FROM cte
WHERE rn = 1;
For a brief explanation, the PARTITION BY clause instructs the DB to generate a separate row number for each group of records in the same category. The ORDER BY clause places those records with the smallest error first. Should two or more records in the same category be tied with the lowest error, then the next sorting level would place the record with the most recent creation date first.
I have a table of existing customers and another table of potential customers. I want to return a list of potential customers rank ordered by the number of radii of existing purchasers that they appear in.
There are many rows in the potential customers table per each existing customer, and the radius around a given existing customer could encompass multiple potential customers. I want to return a list of potential customers ordered by the count of the existing customer radii that they fall within.
SELECT pur.contact_id AS purchaser, count(pot.*) AS nearby_potential_customers
FROM purchasers_geocoded pur, potential_customers_geocoded pot
WHERE ST_DWithin(pur.geom,pot.geom,1000)
GROUP BY purchaser;
Does anyone have advice on how to proceed?
EDIT:
With some help, I wrote this query, which seems to do the job, but I'm verifying now.
WITH prequalified_leads_table AS (
SELECT *
FROM nearby_potential_customers
WHERE market_val > 80000
AND market_val < 120000
)
, proximate_to_existing AS (
SELECT pot.prop_id AS prequalified_leads
FROM purchasers_geocoded pur, prequalified_leads_table pot
WHERE ST_DWithin(pot.geom,pur.geom,100)
)
SELECT prequalified_leads, count(prequalified_leads)
FROM proximate_to_existing
GROUP BY prequalified_leads
ORDER BY count(*) DESC;
I want to return a list of potential customers ordered by the count of the existing customer radii that they fall within.
Your query tried the opposite of your statement, counting potential customers around existing ones.
Inverting that, and after adding some tweaks:
SELECT pot.contact_id AS potential_customer
, rank() OVER (ORDER BY pur.nearby_customers DESC
, pot.contact_id) AS rnk
, pur.nearby_customers
FROM potential_customers_geocoded pot
LEFT JOIN LATERAL (
SELECT count(*) AS nearby_customers
FROM purchasers_geocoded pur
WHERE ST_DWithin(pur.geom, pot.geom, 1000)
) pur ON true
ORDER BY 2;
I suggest a subquery with LEFT JOIN LATERAL ... ON true to get counts. Should make use of the spatial index that you undoubtedly have:
CREATE INDEX ON purchasers_geocoded USING gist (geom);
Thereby retaining rows with 0 nearby customers in the result - your original join style would exclude those. Related:
What is the difference between LATERAL and a subquery in PostgreSQL?
Then ORDER BY the resulting nearby_customers in the outer query (not: nearby_potential_customers).
It's not clear whether you want to add an actual rank. Use the window function rank() if so. I made the rank deterministic while being at it, breaking ties with an additional ORDER BY expression: pot.contact_id. Else, peers are returned in arbitrary order which can change for every execution.
ORDER BY 2 is short syntax for "order by the 2nd out column". See:
Select first row in each GROUP BY group?
Related:
How do I query all rows within a 5-mile radius of my coordinates?
I have a table Orders(id_trip, id_order), table Trip(id_hotel, id_bus, id_type_of_trip) and table Hotel(id_hotel, name).
I would like to get name of the most frequent hotel in table Orders.
SELECT hotel.name from Orders
JOIN Trip
on Orders.id_trip = Trip.id_hotel
JOIN hotel
on trip.id_hotel = hotel.id_hotel
FROM (SELECT hotel.name, rank() over (order by cnt desc) rnk
FROM (SELECT hotel.name, count(*) cnt
FROM Orders
GROUP BY hotel.name))
WHERE rnk = 1;
The "most frequently occurring value" in a distribution is a distinct concept in statistics, with a technical name. It's called the MODE of the distribution. And Oracle has the STATS_MODE() function for it. https://docs.oracle.com/cd/B19306_01/server.102/b14200/functions154.htm
For example, using the EMP table in the standard SCOTT schema, select stats_mode(deptno) from scott.emp will return 30 - the number of the department with the most employees. (30 is the department "name" or number, it is NOT the number of employees in that department!)
In your case:
select stats_mode(h.name) from (the rest of your query)
Note: if two or more hotels are tied for "most frequent", then STATS_MODE() will return one of them (non-deterministic). If you need all the tied values, you will need a different solution - a good example is in the documentation (linked above). This is a documented flaw in Oracle's understanding and implementation of the statistical concept.
Use FIRST for a single result:
SELECT MAX(hotel.name) KEEP (DENSE_RANK FIRST ORDER BY cnt DESC)
FROM (
SELECT hotel.name, COUNT(*) cnt
FROM orders
JOIN trip USING (id_trip)
JOIN hotel USING (id_hotel)
GROUP BY hotel.name
) t
Here is one method:
select name
from (select h.name,
row_number() over (order by count(*) desc) as seqnum -- use `rank()` if you want duplicates
from orders o join
trip t
on o.id_trip = t.id_trip join -- this seems like the right join condition
hotels h
on t.id_hotel = h.id_hotel
) oth
where seqnum = 1;
** Getting the most recent statistical mode out of a data sample **
I know it's more than a year, but here's my answer. I came across this question hoping to find a simpler solution than what I know, but alas, nope.
I had a similar situation where I needed to get the mode from a data sample, with the requirement to get the mode of the most recently inserted value if there were multiple modes.
In such a case neither the STATS_MODE nor the LAST aggregate functions would do (as they would tend to return the first mode found, not necessarily the mode with the most recent entries.)
In my case it was easy to use the ROWNUM pseudo-column because the tables in question were performance metric tables that only experienced inserts (not updates)
In this oversimplified example, I'm using ROWNUM - it could easily be changed to a timestamp or sequence field if you have one.
SELECT VALUE
FROM
(SELECT VALUE ,
COUNT( * ) CNT,
MAX( R ) R
FROM
( SELECT ID, ROWNUM R FROM FOO
)
GROUP BY ID
ORDER BY CNT DESC,
R DESC
)
WHERE
(
ROWNUM < 2
);
That is, get the total count and max ROWNUM for each value (I'm assuming the values are discrete. If they aren't, this ain't gonna work.)
Then sort so that the ones with largest counts come first, and for those with the same count, the one with the largest ROWNUM (indicating most recent insertion in my case).
Then skim off the top row.
Your specific data model should have a way to discern the most recent (or the oldest or whatever) rows inserted in your table, and if there are collisions, then there's not much of a way other than using ROWNUM or getting a random sample of size 1.
If this doesn't work for your specific case, you'll have to create your own custom aggregator.
Now, if you don't care which mode Oracle is going to pick (your bizness case just requires a mode and that's it, then STATS_MODE will do fine.
I have this query which finds the number of properties handled by each staff member along with their branch number:
SELECT s.branchNo, s.staffNo, COUNT(*) AS myCount
FROM Staff s, PropertyForRent p
WHERE s.staffNo=p.staffNo
GROUP BY s.branchNo, s.staffNo
The two relations are:
Staff{staffNo, fName, lName, position, sex, DOB, salary, branchNO}
PropertyToRent{propertyNo, street, city, postcode, type, rooms, rent, ownerNo, staffNo, branchNo}
How does SQL know what COUNT(*) is referring to? Why does it count the number of properties and not (say for example), the number of staff per branch?
This is a bit long for a comment.
COUNT(*) is counting the number of rows in each group. It is not specifically counting any particular column. Instead, what is happening is that the join is producing multiple properties, because the properties are what cause multiple rows for given values of s.branchNo and s.staffNo.
It gets even a little more "confusing" if you include a column name. The following would all typically return the same value:
COUNT(*)
COUNT(s.branchNo)
COUNT(s.staffNo)
COUNT(p.propertyNo)
With a column name, COUNT() determines the number of rows that do not have a NULL value in the column.
And finally, you should learn to use proper, explicit join syntax in your queries. Put join conditions in the on clause, not the where clause:
SELECT s.branchNo, s.staffNo, COUNT(*) AS myCount
FROM Staff s JOIN
PropertyForRent p
ON s.staffNo = p.staffNO
GROUP BY s.branchNo, s.staffNo;
GROUP BY clauses partition your result set. These partitions are all the sql engine needs to know - it simply counts their sizes.
Try your query with only count(*) in the select part.
In particular, COUNT(*) does not produce the number of distinct rows/columns in your result set!
Some people might think that count(*) really count all the columns, however the sql optimizer is smarter than that.
COUNT(*) returns the number of rows in a specified table without getting rid of duplicates. Which mean that you can't use Distinct with count(*)
Count(*) will return the cardinality (elements in table) of the specified mapping.
What you have to remember is that when using count over a specific column, null won't be allowed while count(*) will allow null in the rows as it could be any field.
How does SQL know what COUNT(*) is referring to?
I'm pretty sure, however not 100% sure as I can't find in doc, that the sql optimizer simply do a count on the primary key (not null) instead of trying to handle null in rows.
Say that I have some SELECT statement:
SELECT id, name FROM people
ORDER BY name ASC;
I have a few million rows in the people table and the ORDER BY clause can be much more complex than what I have shown here (possibly operating on a dozen columns).
I retrieve only a small subset of the rows (say rows 1..11) in order to display them in the UI. Now, I would like to solve following problems:
Find the number of a row with a given id.
Display the 5 items before and the 5 items after a row with a given id.
Problem 2 is easy to solve once I have solved problem 1, as I can then use something like this if I know that the item I was looking for has row number 1000 in the sorted result set (this is the Firebird SQL dialect):
SELECT id, name FROM people
ORDER BY name ASC
ROWS 995 TO 1005;
I also know that I can find the rank of a row by counting all of the rows which come before the one I am looking for, but this can lead to very long WHERE clauses with tons of OR and AND in the condition. And I have to do this repeatedly. With my test data, this takes hundreds of milliseconds, even when using properly indexed columns, which is way too slow.
Is there some means of achieving this by using some SQL:2003 features (such as row_number supported in Firebird 3.0)? I am by no way an SQL guru and I need some pointers here. Could I create a cached view where the result would include a rank/dense rank/row index?
Firebird appears to support window functions (called analytic functions in Oracle). So you can do the following:
To find the "row" number of a a row with a given id:
select id, row_number() over (partition by NULL order by name, id)
from t
where id = <id>
This assumes the id's are unique.
To solve the second problem:
select t.*
from (select id, row_number() over (partition by NULL order by name, id) as rownum
from t
) t join
(select id, row_number() over (partition by NULL order by name, id) as rownum
from t
where id = <id>
) tid
on t.rownum between tid.rownum - 5 and tid.rownum + 5
I might suggest something else, though, if you can modify the table structure. Most databases offer the ability to add an auto-increment column when a row is inserted. If your records are never deleted, this can server as your counter, simplifying your queries.