'Count flags for a variable (big) number of colums
I have a table which looks like this: http://i.stack.imgur.com/EyKt3.png
And I want a result like this:
Conditon COL
ted1 4
ted2 1
ted3 2
I.e., the count of the number of '1' only in this case.
I want to know the total no. of 1's only (check the table), neglecting the 0's. It's like if the condition is true (1) then count +1.
Also consider: what if there are many columns? I want to avoid typing expressions for every single one, like in this case ted1 to ted80.
Solution 1:[1]
Using proc means is the most efficient method:
proc means data=have noprint;
var ted:; *captures anything that starts with Ted;
output out=want sum =;
run;
proc print data=want;
run;
Solution 2:[2]
Try this
select
sum(case when ted1=1 then 1 else 0 end) as ted1,
sum(case when ted2=1 then 1 else 0 end) as ted2,
sum(case when ted3=1 then 1 else 0 end) as ted3
from table
Solution 3:[3]
In PostgreSQL (tested with version 9.4) you could unpivot with a VALUES expression in a LATERAL subquery. You'll need dynamic SQL.
This works for any table with any number of columns matching any pattern as long as selected columns are all numeric or all boolean. Only the value 1 (true) is counted.
Create this function once:
CREATE OR REPLACE FUNCTION f_tagcount(_tbl regclass, col_pattern text)
RETURNS TABLE (tag text, tag_ct bigint)
LANGUAGE plpgsql AS
$func$
BEGIN
RETURN QUERY EXECUTE (
SELECT
'SELECT l.tag, count(l.val::int = 1 OR NULL)
FROM ' || _tbl || ', LATERAL (VALUES '
|| string_agg( format('(%1$L, %1$I)', attname), ', ')
|| ') l(tag, val)
GROUP BY 1
ORDER BY 1'
FROM pg_catalog.pg_attribute
WHERE attrelid = _tbl
AND attname LIKE col_pattern
AND attnum > 0
AND NOT attisdropped
);
END
$func$;
Call:
SELECT * FROM f_tagcount('tbl', 'ted%');
Result:
tag | tag_ct
-----+-------
ted1 | 4
ted2 | 1
ted3 | 2
The 1st argument is a valid table name, possibly schema-qualified. Defense against SQL-injection is built into the data type regclass.
The 2nd argument is a LIKE pattern for the column names. Hence the wildcard %.
Related:
Sources
This article follows the attribution requirements of Stack Overflow and is licensed under CC BY-SA 3.0.
Source: Stack Overflow
| Solution | Source |
|---|---|
| Solution 1 | Reeza |
| Solution 2 | Madhivanan |
| Solution 3 |
