Forum Discussion
Grouping data together
I have a dataset which includes the person, and their average score. I need to break those scores out into buckets of >90, >80, etc. with the count of people who averaged that score. However, I am having a hard time grouping them in this fashion: I am able to set up the people, and see their average scores, but grouping them is where I get lost. Please help!!
What I need to show is two more columns showing a score range and then the count of people who's average score falls into that range. so it should look like this:
| Person | Average Score | Score Range | Number in column |
| A | 90 | 90-100% | 1 |
| B | 80 | 80-89% | 1 |
C | 70 | 70-79% | 1 |
| D | 60 | 60-69% | 1 |
| e | 58 | <69% | 3 |
| f | 58 |
|
|
| g | 58 |
|
|
Hi HunterAlpert ，
Based on your description, you can do some steps as follows.
- Create a calculated column to mark the range.
Score Range = SWITCH(
TRUE(),
[Average Scores]>=90&&[Average Scores]<100,"90-100%",
[Average Scores]>=80&&[Average Scores]<90,"80-89%",
[Average Scores]>=70&&[Average Scores]<800,"70-79%",
[Average Scores]>=60&&[Average Scores]<70,"60-69%",
"<69%")2. Create a measure to count filtered rows.
Number in column = COUNTROWS(FILTER(ALL('Table 3'),'Table 3'[Score Range]=SELECTEDVALUE('Table 3'[Score Range])))Result:
Hope that's what you were looking for.
Best Regards,
Yuna
If this post helps, then please consider Accept it as the solution to help the other members find it more quickly.
5 Replies
Hi HunterAlpert ，
Based on your description, you can do some steps as follows.
- Create a calculated column to mark the range.
Score Range = SWITCH(
TRUE(),
[Average Scores]>=90&&[Average Scores]<100,"90-100%",
[Average Scores]>=80&&[Average Scores]<90,"80-89%",
[Average Scores]>=70&&[Average Scores]<800,"70-79%",
[Average Scores]>=60&&[Average Scores]<70,"60-69%",
"<69%")2. Create a measure to count filtered rows.
Number in column = COUNTROWS(FILTER(ALL('Table 3'),'Table 3'[Score Range]=SELECTEDVALUE('Table 3'[Score Range])))Result:
Hope that's what you were looking for.
Best Regards,
Yuna
If this post helps, then please consider Accept it as the solution to help the other members find it more quickly.
This is exactly what I needed! Thank you!
Hi,
Is it me that you replied to? If my reply helped, please mark it as Answer.
