I have a scenario where there are multiple IDs in a column (uniqueid) of data in a Google Spreadsheet, some IDs are also duplicated in different rows.
Some of these ids are labeled as left', 'right' or 'both' in another column (obs).
uniqueid | obs |
m1 | right |
m2 | left |
m1 | both |
m3 | right |
m4 | left |
I want to be able to create a counting field that indicates the result of the sum that totals the highest number of distinct ids between only 'left' + has 'both' (m2+m4+m1) or only 'right' + has 'both' (m3+m1), in this case it should appear = 3. In case of equal sums between these two conditions, show the sum of only 'right' + has 'both'
I tried looking for help on the forum but I can't adapt the conditional counting formulas using two columns.
Solved! Go to Solution.
I think I got this to work in Looker Studio! I used several fields, but you could do it with as little as 1 if you like.
The basic logic can be summed up in this IF statement:
`if([count distinct ids with right or both] > [count distinct ids with left or both] , [count distinct ids with right or both] , [count distinct ids with left or both] )`
In other words, return the larger of the 2 count distincts.
To compute a count distinct that only looks at right or both, use an IF inside of a count distinct:
`COUNT_DISTINCT(IF([obs]="right" OR [obs]="both",[uniqueid],null))`
I hope this helps! Happy to clarify any part of this if you'd like.
I think I got this to work in Looker Studio! I used several fields, but you could do it with as little as 1 if you like.
The basic logic can be summed up in this IF statement:
`if([count distinct ids with right or both] > [count distinct ids with left or both] , [count distinct ids with right or both] , [count distinct ids with left or both] )`
In other words, return the larger of the 2 count distincts.
To compute a count distinct that only looks at right or both, use an IF inside of a count distinct:
`COUNT_DISTINCT(IF([obs]="right" OR [obs]="both",[uniqueid],null))`
I hope this helps! Happy to clarify any part of this if you'd like.
According to this logic it worked, thank you
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