How can I list in one cell, unique values in a row
I have the following data in an Excel spreadsheet
Team1 Team2 Team3 Team4
C C All None
In the next cell in the row, I want to list values that are not duplicated (i.e. All and None) separated by a comma.
Team1 Team2 Team3 Team4
C C All None
In the next cell in the row, I want to list values that are not duplicated (i.e. All and None) separated by a comma.
SOLUTION
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ASKER CERTIFIED SOLUTION
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ASKER
Hello Syed,
Thank-you for your assistance. Although I was originally looking for a non-VBA solution, the UDF works beautifully. So I shall adopt it as my solution.
Elimishia
Thank-you for your assistance. Although I was originally looking for a non-VBA solution, the UDF works beautifully. So I shall adopt it as my solution.
Elimishia
ASKER
I've requested that this question be closed as follows:
Accepted answer: 0 points for elimishia's comment #a40695306
for the following reason:
It helped that I have some knowledge of VBA. However, the code was simple and it worked.
Accepted answer: 0 points for elimishia's comment #a40695306
for the following reason:
It helped that I have some knowledge of VBA. However, the code was simple and it worked.
ASKER
The solution is straight forward and would work well on spreadsheets with free columns available to hold the various formulae.
SOLUTION
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ASKER
Hello Syed,
Thank-you for this alternative suggestion. It too works very well.
I have tried to mark both your suggestions as ANSWERS, without success.
Thank-you for this alternative suggestion. It too works very well.
I have tried to mark both your suggestions as ANSWERS, without success.
ASKER
Thank-you for taking the time to create a sample spreadsheet. Unfortunately, my spreadsheet doesn't allow for the use of hidden columns to do the multiple IFs. Nesting the IFs would be cumbersome as there are likely to be up to 10 columns. I am sorry - I should have been more specific about needing the formula (as well as the result) to be in a single cell.
Using Match() in an array, I could see where the unique value were, but got stuck in not knowing how to extract the values.
Once again, thanks for your assistance. As your suggestion does provide a solution in different circumstances, I will make it as an answer.
Elimishia