How can I sum values per record from a table?
If I have
row
key
value
expected
row 1
A
1
1
row 2
A
5
6
row 3
B
10
10
row 4
B
1
11
row 5
B
1
12
can you point me some directions?
thanks
You can use Windowing Functions
Something like
select t.*,
sum(value) over (partition by key order by "row" rows unbounded preceding) expected
from tab t;
Note that ROW is a reserved word. You can create a table with a column row, but you better use another name.
Related
Say I have a table with two columns: the time and the value. I want to be able to get a table with :
for each time get the max values of every next n seconds.
If I want the max value of every next 3 seconds, the following table:
time
value
1
6
2
1
3
4
4
2
5
5
6
1
7
1
8
3
9
7
Should return:
time
value
max
1
6
6
2
1
4
3
4
5
4
2
5
5
5
5
6
1
3
7
1
7
8
3
NULL
9
7
NULL
Is there a way to do this directly with an sql query?
You can use the max window function:
select *,
case
when row_number() over(order by time desc) > 2 then
max(value) over(order by time rows between current row and 2 following)
end as max
from table_name;
Fiddle
The case expression checks that there are more than 2 rows after the current row to calculate the max, otherwise null is returned (for the last 2 rows ordered by time).
Similar Version to Zakaria, but this solution uses about 40% less CPU resources (scaled to 3M rows for benchmark) as the window functions both use the same exact OVER clause so SQL can better optimize the query.
Optimized Max Value of Rolling Window of 3 Rows
SELECT *,
MaxValueIn3SecondWindow = CASE
/*Check 3 rows exists to compare. If 3 rows exists, then calculate max value*/
WHEN 3 = COUNT(*) OVER (ORDER BY [Time] ROWS BETWEEN CURRENT ROW AND 2 FOLLOWING)
/*Returns max [Value] between the current row and the next 2 rows*/
THEN MAX(A.[Value]) OVER (ORDER BY [Time] ROWS BETWEEN CURRENT ROW AND 2 FOLLOWING)
END
FROM #YourTable AS A
Say I have the following table:
ID
T
R
1
2
1
3
Y
1
4
1
5
1
6
Y
1
7
I would like to add a column which equals the value from column T based on the last non-null value from column R. This means the following:
ID
T
R
GOAL
1
2
1
3
Y
1
4
Y
3
1
5
4
1
6
Y
4
1
7
6
I do have many ID's so I need to make use of the OVER (PARTITION BY ...) clause. Also, if possible, I would like to use a single statement, like
SELECT *
, GOAL
FROM TABLE
So without any extra select statement.
T is in ascending order so just null it out according to R and take the maximum looking backward.
select *,
max(case when R is not null then T end)
over (
partition by id
order by T
rows between unbounded preceding and 1 preceding
) as GOAL
from TBL
http://sqlfiddle.com/#!18/c927a5/5
I have table something like:
GROUP
NAME
Value_1
Value_2
1
ABC
0
0
1
DEF
4
4
50
XYZ
6
6
50
QWE
6
7
100
XYZ
26
2
100
QWE
26
2
What I would like to do is to groupby group and select the name with highest value_1. If their value_1 are the same, compare and select the max with value_2. If they're still the same, select the first one.
The output will be something like:
GROUP
NAME
Value_1
Value_2
1
DEF
4
4
50
QWE
6
7
100
XYZ
26
2
The challenge for me here is I don't know how many categories in NAME so a simple case when is not working. Thanks for help
You can use window functions to solve the bulk of your problem:
select t.*
from (select t.*,
row_number() over (partition by group order by value1 desc, value2 desc) as seqnum
from t
) t
where seqnum = 1;
The one caveat is the condition:
If they're still the same, select the first one.
SQL tables represent unordered (multi-) sets. There is no "first" one unless a column specifies the ordering. The best you can do is choose an arbitrary value when all the other values are the same.
That said, you might have another column that has an ordering. If so, add that as a third key to the order by.
i have a table like this
code Quantity
1 5
1 6
2 2
2 1-
3 4
.
.
how can made it like this
code Quantity remain
1 5 5
1 6 11
2 2 2
2 1- 1
3 4 4
.
.
Your query presumes an ordering of the rows. I will assume you have such a column.
Assuming the values are numbers (1- ???), then you can simply use a cumulative sum:
select t.*,
sum(quantity) over (partition by code order by ?) as remaining
from t;
The ? is for the column that specifies the ordering.
You can do a window sum, but you need a column to unambiguously order the records within groups sharing the same code. I assumed that this column is called id.
select t.*, sum(quantity) over(partition by code order by id) remain from mytable t
I have a table of consecutive ids (integers, 1 ... n), and values (integers), like this:
Input Table:
id value
-- -----
1 1
2 1
3 2
4 3
5 1
6 1
7 1
Going down the table i.e. in order of increasing id, I want to count how many times in a row the same value has been seen consecutively, i.e. the position in a run:
Output Table:
id value position in run
-- ----- ---------------
1 1 1
2 1 2
3 2 1
4 3 1
5 1 1
6 1 2
7 1 3
Any ideas? I've searched for a combination of windowing functions including lead and lag, but can't come up with it. Note that the same value can appear in the value column as part of different runs, so partitioning by value may not help solve this. I'm on Hive 1.2.
One way is to use a difference of row numbers approach to classify consecutive same values into one group. Then a row number function to get the desired positions in each group.
Query to assign groups (Running this will help you understand how the groups are assigned.)
select t.*
,row_number() over(order by id) - row_number() over(partition by value order by id) as rnum_diff
from tbl t
Final Query using row_number to get positions in each group assigned with the above query.
select id,value,row_number() over(partition by value,rnum_diff order by id) as pos_in_grp
from (select t.*
,row_number() over(order by id) - row_number() over(partition by value order by id) as rnum_diff
from tbl t
) t