I don't understand how sum works.
For a PostgreSQL table in dbeaver:
a
b
c
d
1
2
3
2
1
2
4
3
2
1
3
2
2
1
4
2
3
2
4
2
the query
select a, b, c, d, sum(c) as sum_c, sum(d) as sum_d from abc a group by a, b, c, d
produces
a
b
c
d
sum_c
sum_d
1
2
3
2
3
2
1
2
4
3
4
3
2
1
3
2
3
2
2
1
4
2
4
2
3
2
4
2
4
2
and I don't understand why: I expected sum_c would be 18 in each row, which is the sum of values in c, and sum_d would be 11 for the same reason.
Why do sum_c and sum_d just copy the values from c and d in each row?
You can't get the result that you want with group by.
When you aggregate with group by you create groups for all the columns that are after group by and for each of these groups you get the aggregated results.
For your sample data, one group is 1,2,3,2 and for this combination of values you get the sum of c which is 3 since there is only 1 row with c=3 in that group.
Use SUM() window function:
SELECT a, b, c, d,
SUM(c) OVER () sum_c,
SUM(d) OVER () sum_d
FROM abc
Related
There is a table with a column that I would like to break into multiple records. For example
q)tab:([]a:1 2 3;b:(`a;`$"b c";`d);c:2 3 4)
q)tab
a b c
-------
1 a 2
2 b c 3
3 d 4
There is a space between b and c in the second entry of column b, I would like the table to become
a b c
-----
1 a 2
2 b 3
2 c 3
3 d 4
I tried
" " string vs exec b from tab
but didn't work.
Any idea?
Since b is the column with multiple entries per row, you can count each value and expand the corresponding row entries accordingly. Then ungroup like Terry mentioned should work.
q)t:([]a:1 2 3;b:(`a;`b`c;`d);c:2 3 4)
q)![t;();0b;{x!(enlist({(count each x)#'y};`b)),/:x}cols t]
a b c
------------
,1 ,`a ,2
2 2 `b`c 3 3
,3 ,`d ,4
q)ungroup ![t;();0b;{x!(enlist({(count each x)#'y};`b)),/:x}cols t]
a b c
-----
1 a 2
2 b 3
2 c 3
3 d 4
EDIT: Realised after your comment that the input is different. I think this is what you want.
q)t:([]a:1 2 3;b:(`a;`$"b c";`d);c:2 3 4)
q)ungroup update`$" "vs'string b from t
a b c
-----
1 a 2
2 b 3
2 c 3
3 d 4
You would normally do this using ungroup:
q)ungroup([]a:1 2 3;b:((),`a;`b`c;(),`d);c:2 3 4)
a b c
-----
1 a 2
2 b 3
2 c 3
3 d 4
I would like to know what is the proper way in SPSS to group data by specydic column and then find top n max values.
For example I have below columns:
x<-c(3,2,1,8,7,11,10,9,7,5,4)
y<-c("a","a","a", "b","b","c","c","c","c","c","c")
z<-c(2,2,2,1,1,3,3,3,3,3,3)
I want to select top max n values from column X for each group by column y
x y
1 3 a
2 2 a
3 1 a
4 8 b
5 7 b
6 11 c
7 10 c 3
8 9 c 3
9 7 c 3
10 5 c 3
11 4 c 3
I have data which includes 2 columns, ages, and groups similar to
B 1 B 1 B 1 B 4 B 5 B 8 D 2 D 2 D 3 D 3 D 3 D 4 D 6 D 7 D 9 D 9
In Tableau, I wish to plot a line for each group B and D, % number of records(observations) (of group in group), against the age range, 1 to 9.
So B 1 - 3/6*100, B 5 1/6*100, D 3 - 3/10*100.
Any help or pointers would be really appreciated.
Enda
Drag 'Age' measure in columns.
Drag 'Group' dimension in 'Color'
Drag the tableau default measure of 'Number of Records' in rows. Make it's aggregation as 'sum', add quick table calculation of 'Percent of Total'. Change it's 'Compute Using' to 'Age'.
That's it! Hopefully this is what you were trying to do.
x a b c d
----------
A 1 2 3 4
B 5 6 7 8
C 6 7 8 9
I want my sum of A = 1 + 2 + 3 + 4 and so for B and C, Is there any command that can sum row of data in PostgreSQL?
There is no such built-in function, but you can simply do the following:
select x, a+b+c+d as column_sum from mytable
Assuming, of course, that the data type of a, b, c and d are numeric.
I have a table rCom which has various columns. I would like to sum across each row..
for example:
Date TypeA TypeB TypeC TypeD
date1 40.5 23.1 45.1 65.2
date2 23.3 32.2 56.1 30.1
How can I write a q query to add a fourth column 'Total' that sums across each row?
why not just:
update Total: TypeA+TypeB+TypeC+TypeD from rCom
?
Sum will work just fine:
q)flip`a`b`c!3 3#til 9
a b c
-----
0 3 6
1 4 7
2 5 8
q)update d:sum(a;b;c) from flip`a`b`c!3 3#til 9
a b c d
--------
0 3 6 9
1 4 7 12
2 5 8 15
Sum has map reduce which will be better for a huge table.
One quick point regarding summing across rows. You should be careful about nulls in 1 column resulting in a null result for the sum. Borrowing #WooiKent Lee's example.
We put a null into the first position of the a column. Notice how our sum now becomes null
q)wn:.[flip`a`b`c!3 3#til 9;(0;`a);first 0#] //with null
q)update d:sum (a;b;c) from wn
a b c d
--------
3 6
1 4 7 12
2 5 8 15
This is a direct effect of the way nulls in q are treated. If you sum across a simple list, the nulls are ignored
q)sum 1 2 3 0N
6
However, a sum across a general list will not display this behavior
q)sum (),/:1 2 3 0N
,0N
So, for your table situation, you might want to fill in with a zero beforehand
q)update d:sum 0^(a;b;c) from wn
a b c d
--------
3 6 9
1 4 7 12
2 5 8 15
Or alternatively, make it s.t. you are actually summing across simple lists rather than general lists.
q)update d:sum each flip (a;b;c) from wn
a b c d
--------
3 6 9
1 4 7 12
2 5 8 15
For a more complete reference on null treatment please see the reference website
This is what worked:
select Answer:{[x;y;z;a] x+y+z+a }'[TypeA;TypeB;TypeC;TypeD] from
([] dt:2014.01.01 2014.01.02 2014.01.03; TypeA:4 5 6; TypeB:1 2 3; TypeC:8 9 10; TypeD:3 4 5)