Say I have the following data:
+--------+-------+
| Group | Data |
+--------+-------+
| 1 | row 1 |
| 1 | row 2 |
| 1 | row 3 |
| 20 | row 1 |
| 20 | row 3 |
| 10 | row 1 |
| 10 | row A |
| 10 | row 2 |
| 10 | row 3 |
+--------+-------+
Is it possible to draw a map that shows which groups have which rows? Groups may not be contagious, so they can be placed into a separate table and use the row index for the string index instead. Something like this:
+-------+
| Group |
+-------+
| 1 |
| 20 |
| 10 |
+-------+
+-------+----------------+
| Data | Found in group |
+-------+----------------+
| row 1 | 111 |
| row A | 1 |
| row 2 | 1 1 |
| row 3 | 111 |
+-------+----------------+
Where the first character represents Group 1, the 2nd is Group 20 and the 3rd is Group 10.
Ordering of the Group rows isn't critical so long as I can reference which row goes with which character.
I only ask this because I saw this crazy example in the documentation generating a fractal, but I can't quite get my head around it.
Is this doable?
To find the missing values, first thing is to prepare a dataset which have all possible combination. You can achieve that using CROSS JOIN.
Once you have that DataSet, compare it with the actual DataSet.
Considering the Order by is done in the Grp column, you can achieve it using below.
SELECT
a.Data,group_concat(case when base.Grp is null then "." else "1" end,'') as Found_In_Group
,group_concat(b.Grp) as Group_Order
FROM
(SELECT Data FROM yourtable Group By Data)a
CROSS JOIN
(SELECT Grp FROM yourtable Group By Grp Order by Grp)b
LEFT JOIN yourtable base
ON b.Grp=base.Grp
AND a.Data=base.Data
GROUP BY a.Data
Note: Considered . instead of blank for better visibility to represent missing Group.
Data
Found_In_Group
Group_Order
row 1
111
1,10,20
row 2
11.
1,10,20
row 3
111
1,10,20
row A
.1.
1,10,20
Demo: Try here
SELECT Data, group_concat("Group") AS "Found in group"
FROM yourtable
GROUP BY Data
will give you a CSV list of groups.
Related
I have a table that has a number column and an attribute column like this:
1.
+-----+-----+
| num | att |
-------------
| 1 | a |
| 1 | b |
| 1 | a |
| 2 | a |
| 2 | b |
| 2 | b |
+------------
I want to make the number unique, and the attribute to be whichever attribute occured most often for that number, like this (This is the end-product im interrested in) :
2.
+-----+-----+
| num | att |
-------------
| 1 | a |
| 2 | b |
+------------
I have been working on this for a while and managed to write myself a query that looks up how many times an attribute occurs for a given number like this:
3.
+-----+-----+-----+
| num | att |count|
------------------+
| 1 | a | 1 |
| 1 | b | 2 |
| 2 | a | 1 |
| 2 | b | 2 |
+-----------------+
But I can't think of a way to only select those rows from the above table where the count is the highest (for each number of course).
So basically what I am asking is given table 3, how do I select only the rows with the highest count for each number (Of course an answer describing providing a way to get from table 1 to table 2 directly also works as an answer :) )
You can use aggregation and window functions:
select num, att
from (
select num, att, row_number() over(partition by num order by count(*) desc, att) rn
from mytable
group by num, att
) t
where rn = 1
For each num, this brings the most frequent att; if there are ties, the smaller att is retained.
Oracle has an aggregation function that does this, stats_mode().:
select num, stats_mode(att)
from t
group by num;
In statistics, the most common value is called the mode -- hence the name of the function.
Here is a db<>fiddle.
You can use group by and count as below
select id, col, count(col) as count
from
df_b_sql
group by id, col
The following query gives me the information that I need but I want it to take it just a step further. In the table at the bottom (only showing a subset of the fields), I want to group by cust_line in an unusual way (at least to me it's unusual).
Let's look at the items with a cust_line of 2 as an example. I would like these to be represented by one line not 5. For this line, I would like to select all the fields except for the price field where the cust_part = "GROUPINVC". For the total field I would like it to be 'sum(total) as new_total' and for the price, I would like it to be new_total / qty_invoiced, where qty_invoiced is the value on the line where cust_part = "GROUPINV".
Is what I am asking for completely ridiculous? Is it even possible? I'm not advanced at SQL so it may also be easy and I just don't know how to approach it. I thought of using 'partition by' but I couldn't imagine how I would get it to work as I figured it would still return 5 rows where I only want 1.
I've also looked at these questions with similar titles but not really what I am looking for:
SQL query that returns aggregate AND non aggregate results
Combined aggregated and non-aggregate query in SQL
SELECT L.CUST_LINE, I.LINE_NO, I.ORDER_NO, I.STAGE, I.ORDER_LINE_POS, I.CUST_PART,
I.LINE_ITEM_NO, I.QTY_INVOICED, I.CUST_DESC, I.DESCRIPTION, I.SALE_UNIT_PRICE, I.PRICE_TOTAL,
I.INVOICE_NO, I.CUSTOMER_PO_NO, I.ORDER_NO, I.CUSTOMER_NO, I.CATALOG_DESC, I.ORDER_LINE_NOTES
FROM
(SELECT CUST_LINE, ORDER_NO, LINE_NO
FROM CUSTOMER_ORDER_LINE
GROUP BY CUST_LINE, ORDER_NO, LINE_NO
) L
INNER JOIN CUSTOMER_ORDER_IVC_REP I
ON I.ORDER_NO = L.ORDER_NO
WHERE RESULT_KEY = 999999
AND I.LINE_NO = L.LINE_NO
ORDER BY L.CUST_LINE;
| cust_line | line_no | cust_part | qty_invoiced | cust_desc | price | total |
| 1 | 4 | ... | 1 | ... | 55 | 55 |
| 2 | 1 | GROUPINV | 1 | some part | 0 | 0 |
| 2 | 6 | ... | 3 | ... | 0 | 0 |
| 2 | 2 | ... | 1 | ... | 0 | 0 |
| 2 | 3 | ... | 1 | ... | 0 | 0 |
| 2 | 7 | ... | 2 | ... | 10 | 20 |
| 3 | 7 | ... | 1 | ... | 67 | 67 |
You can use an analytic function to calculate a total over multiple rows of a result set, then filter out the rows you don't want.
Leaving out all the extra columns for sake of brevity:
SELECT cust_line, qty_invoiced, order_total/qty_invoiced AS price
FROM (
SELECT l.cust_line, qty_invoiced,
SUM(total) OVER (PARTITION BY l.cust_line) AS order_total,
COUNT(cust_line) OVER (PARTITION BY l.cust_line) AS group_count
FROM
(SELECT CUST_LINE, ORDER_NO, LINE_NO
FROM CUSTOMER_ORDER_LINE
GROUP BY CUST_LINE, ORDER_NO, LINE_NO
) L
INNER JOIN CUSTOMER_ORDER_IVC_REP I
ON I.ORDER_NO = L.ORDER_NO
WHERE RESULT_KEY = 999999
AND I.LINE_NO = L.LINE_NO
)
WHERE ( cust_part = 'GROUPINV' OR group_count = 1 )
ORDER BY cust_line
I am guessing on what you want in the PARTITION BY clause; this is essentially a GROUP BY that applies only to the SUM function. Not sure if you might also want order_no in the partition.
The trick is to select all the rows in the inner query, applying SUM across them all; then filter out the rows you are not interested in in the outermost query.
t1
id | name | include
-------------------
1 | foo | true
2 | bar | true
3 | bum | false
t2
id | some | table_1_id
-------------------------
1 | 42 | 1
2 | 43 | 1
3 | 42 | 2
4 | 44 | 1
5 | 44 | 3
Desired output:
name | count(some)
------------------
foo | 3
bar | 1
What I have currently from looking through other solutions here:
SELECT a.name,
COUNT(r.some)
FROM t1 a
JOIN t2 r on a.id=r.table_1_id
WHERE a.include = 'true'
GROUP BY a.id,
r.some;
but that seems to get me
name | count(r.some)
--------------------
foo | 1
foo | 1
bar | 1
foo | 1
I'm no sql expert (I can do simple queries) so I'm googling around as well but finding most of the solutions I find give me this result. I'm probably missing something really easy.
Just remove the second column from the group by clause
SELECT a.name,
COUNT(r.some)
FROM t1 a
JOIN t2 r on a.id=r.table_1_id
WHERE a.include = 'true'
GROUP BY a.name
Columns you want to use in an aggregate function like sum() or count() must be left out of the group by clause. Only put the columns in there you want to be unique outputted.
This is because multiple column group requires the all column values to be same.
See this link for more info., Using group by on multiple columns
Actually in you case., if some are equal, table_1_id is not equal (And Vice versa). so grouping cannot occur. So all are displayed individually.
If the entries are like,
id | some | table_1_id
-------------------------
1 | 42 | 1
2 | 43 | 1
3 | 42 | 2
4 | 42 | 1
Then the output would have been.,
name | count
------------------
foo | 2 (for 42)
foo | 1 (for 43)
bar | 1 (for 42)
Actually, if you want to group on 1 column as Juergen said, you could remove r.some; from groupby clause.
I have the following table:
+----+-------+
| id | value |
+----+-------+
| 1 | 10 |
| 2 | 11 |
| 3 | 12 |
+----+-------+
I want to calculate a column on the fly to sum value of all the previous rows, to come up with something like this:
+----+-------+--------+
| id | value | offset |
+----+-------+--------+
| 1 | 10 | 0 |
| 2 | 11 | 10 |
| 3 | 12 | 21 |
+----+-------+--------+
What is an efficient way to do this?
Credit goes to Egor Skriptunoff.
select
id,
value,
nvl(
sum(value) over (
order by id rows between unbounded preceding and 1 preceding
), 0) as offset
from table
The great thing about analytic function sum is it's progressive, in the sense that in each iteration the engine remembers the value that was calculated for the previous row and only adds value of the previous row to the total. In other words, for each offset to be calculated, it is summing the previous row offset with value. This is very efficient and scales up nicely.
If your id values will be in sequence like 1,2,3 etc.. then
select a.*,(select sum(decode(a.id,1,0,b.value)) off_set from table b where b.id<=a.id-1)
from table a;
If your id's are not in sequence then try below code
select a.*,(select sum(decode(a.rn,1,0,b.value)) off_set from (select table.*,rownum rn from table) b
where b.rn<=a.rn-1)
from (select table.*,rownum rn from table) a;
I want to create a window function that will count how many times the value of the field in the current row appears in the part of the ordered partition coming before the current row. To make this more concrete, suppose we have a table like so:
| id| fruit | date |
+---+--------+------+
| 1 | apple | 1 |
| 1 | cherry | 2 |
| 1 | apple | 3 |
| 1 | cherry | 4 |
| 2 | orange | 1 |
| 2 | grape | 2 |
| 2 | grape | 3 |
And we want to create a table like so (omitting the date column for clarity):
| id| fruit | prior |
+---+--------+-------+
| 1 | apple | 0 |
| 1 | cherry | 0 |
| 1 | apple | 1 |
| 1 | cherry | 1 |
| 2 | orange | 0 |
| 2 | grape | 0 |
| 2 | grape | 1 |
Note that for id = 1, moving along the ordered partition, the first entry 'apple' doesn't match anything (since the implied set is empty), the next fruit, 'cherry' also doesn't match. Then we get to 'apple' again, which is a match and so on. I'm imagining the SQL looks something like this:
SELECT
id, fruit,
<some kind of INTERSECT?> OVER (PARTITION BY id ORDER by date) AS prior
FROM fruit_table;
But I cannot find anything that looks right. FWIW, I'm using PostgreSQL 8.4.
You could solve that without a window function rather elegantly with a self-left join and a count():
SELECT t.id, t.fruit, t.day, count(t0.*) AS prior
FROM tbl t
LEFT JOIN tbl t0 ON (t0.id, t0.fruit) = (t.id, t.fruit) AND t0.day < t.day
GROUP BY t.id, t.day, t.fruit
ORDER BY t.id, t.day
I renamed the date column day because date is a reserved word in every SQL standard and in PostgreSQL.
I corrected a mistake in your sample data. They way you had it, it did not check out. Might confuse people.
If your point is to do it with a window function, this one should work:
SELECT id, fruit, day
,count(*) OVER (PARTITION BY id, fruit ORDER BY day) - 1 AS prior
FROM tbl
ORDER BY id, day
This works, because, I quote the manual:
If frame_end is omitted it defaults to CURRENT ROW.
You effectively count how many rows had the same (id, fruit) on prior days - including the current row. That's what the - 1 is for.