i have a table like this:
ID var_1 var_2 Var_3 .....Var_n
1 1 1 2 3
2 4 5 6 8
3 2 5 5 5
i tried
Select *, sum(week_1_week_2+…+week_n)/n.0 from table group by ID;
is there any fast way to do it ??
The sum() and group by do not seem appropriate:
Select t.*, (week_1_week_2+…+week_n)/n.0
from table t;
Related
I'd appreciate some help on the following SQL problem:
I have a table of 3 columns:
ID Group Value
1 1 5
1 1 5
1 2 10
1 2 10
1 3 20
2 1 5
2 1 5
2 1 5
2 2 10
2 2 10
3 1 5
3 2 10
3 2 10
3 2 10
3 4 50
I need to group by ID, and I would like to SUM the values based on DISTINCT values in Group. So the value for a group is only accounted for once even though it may appear multiple for times for a particular ID.
So for IDs 1, 2 and 3, it should return 35, 15 and 65, respectively.
ID SUM
1 35
2 15
3 65
Note that each Group doesn't necessarily have a unique value
Thanks
the CTE will remove all duplicates, so if there a sdiffrenet values for ID and Group, it will be counted.
The next SELECT wil "GROUP By" ID
For Pstgres you would get
WITH CTE as
(SELECT DISTINCT "ID", "Group", "Value" FROM tablA
)
SELECT "ID", SUM("Value") FROM CTE GROUP BY "ID"
ORDER BY "ID"
ID | sum
-: | --:
1 | 35
2 | 15
3 | 65
db<>fiddle here
Given what we know at the moment this is what I'm thinking...
The CTE/Inline view eliminate duplicates before the sum occurs.
WITH CTE AS (SELECT DISTINCT ID, Group, Value FROM TableName)
SELECT ID, Sum(Value)
FROM CTE
GROUP BY ID
or
SELECT ID, Sum(Value)
FROM (SELECT DISTINCT * FROM TableName) CTE
GROUP BY ID
I have a table like the one below:
ID
RID
Count
1
1
1
2
1
3
3
1
5
4
1
1
5
2
1
6
2
6
7
2
3
8
2
2
9
2
4
I am trying to retrieve the rows of each RID until the rolling sum of Count is ≤ 10.
In this example I need all rows of RID = 1 and only rows 1, 2 and 3 of RID = 2.
Expected answer:
ID
RID
Count
Sum_Count
1
1
1
NULL
2
1
3
4
3
1
5
9
4
1
1
10
5
2
1
NULL
6
2
6
7
7
2
3
10
I tried with ROWNUM, inner query, etc. but nothing worked out.
Can someone please point me in the right direction?
You need to use a cumulative sum of Count by RID and then select all rows where the cumulative count is less than or equal to 10.
Try this:
select
*
from
(
select
ID,
RID,
COUNT,
sum(COUNT) over (partition by RID order by ID) as cum_count
from
my_table
)
where
cum_count <= 10
I want to group records by row numbers.
Like from row 1-3 in group 1 , 4-6 in group 2 , 7-9 in group 3 and so on.
Suppose below is the table structure:
Row NumberDataValue
1 A 10
2 A 5
3 A 1
4 A 33
5 A 2
6 A 127
1 B 1
2 B 0
3 B 7
4 B 7
5 B 5
6 B 8
7 B 1
8 B 0
I want a output like this:
GroupValue
1 10
1 5
1 1
2 33
2 2
2 127
1 1
1 0
1 7
2 7
2 5
2 8
3 1
3 0
I am using Oracle 11G.
I can achieve this using PL/SQL. But I have to use SQL only. As I have to use this query in a reporting tool.
If this is a duplicate question please provide the link of the answered question.
Subtract 1 from the column "RowNumber" and divide by 3.
Then use TRUNC() to get the integer part:
SELECT TRUNC(("RowNumber" - 1) / 3) + 1 "Group",
"Value"
FROM tablename
See the demo.
I would assume the name of the first column is ordering.
You can do:
select
1 + trunc(row_number() over(partition by data order by ordering) - 1) / 3,
value
from t
What you show looks like the output from something like this:
select ceil(rn/3) as grp, value
from your_table
order by rn;
Note that "row number" and "group" are reserved words/phrases which should not be used as column names. I used rn and grp instead.
I think the ceiling function is the simplest way to arrive at what you want. If you want to base it on the RowNumber column:
select ceil( RowNumber / 3.0) as grouping
If you want to calculate it yourself using row_number():
select ceil( row_number() over (order by RowNumber) / 3.0 ) as grouping
I have data that looks like this:
ID num_of_days
1 0
2 0
2 8
2 9
2 10
2 15
3 10
3 20
I want to add another column that increments in value only if the num_of_days column is divisible by 5 or the ID number increases so my end result would look like this:
ID num_of_days row_num
1 0 1
2 0 2
2 8 2
2 9 2
2 10 3
2 15 4
3 10 5
3 20 6
Any suggestions?
Edit #1:
num_of_days represents the number of days since the customer last saw a doctor between 1 visit and the next.
A customer can see a doctor 1 time or they can see a doctor multiple times.
If it's the first time visiting, the num_of_days = 0.
SQL tables represent unordered sets. Based on your question, I'll assume that the combination of id/num_of_days provides the ordering.
You can use a cumulative sum . . . with lag():
select t.*,
sum(case when prev_id = id and num_of_days % 5 <> 0
then 0 else 1
end) over (order by id, num_of_days)
from (select t.*,
lag(id) over (order by id, num_of_days) as prev_id
from t
) t;
Here is a db<>fiddle.
If you have a different ordering column, then just use that in the order by clauses.
The following is an example of what I have to work with.
Sample data :
ID RANK
---------
1 2
1 3
2 4
2 1
3 2
2 3
4 2
SQLFiddle
I am trying to combine the rows with like IDs and sum the RANKs for these IDs into a single row:
ID SUM(rank)
1 5
2 8
3 2
4 2
You can use sum aggregate function together with the group by clause:
select [ID]
, sum([RANK])
from [STUFF]
group by [ID]
SQLFiddle