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Anonymous
Not applicable

Count with condition and not include duplicate

Hi guys!

 

How can we Count with some conditions but exclude doublicates?

For example: a Table like this:

No.   Name   Status

1       AAA      YES

2       BBB       NO

3       CCC      YES

4       AAA      NO

5       AAA      YES

6       AAA      NO

7       AAA      YES

8       DDD      NO

 

How can I have a formula to:

       return 2 with status = YES (for AAA and CCC);

       return 3 with status = NO (for BBB, AAA, DDD);

Thanks in advanced!

1 ACCEPTED SOLUTION

Accepted Solutions

Hi,

 

You don't have to push the list to collection, You can directly use the collection in the formulae and add as many conditions as you would like.

 

Mark answer as verified if it answers your question.

Regards,

Pavan Kumar Garlapati

View solution in original post

4 REPLIES 4
Responsive Resident
Responsive Resident

Hi Trung,

 

You can use below formulae to get values avoiding duplicates. Lets say your actual list (which you have mentioned below) is in List Status Collection.

 

ClearCollect(StatusCollection,
{ Name: "AAA", Status: "YES"},
{ Name: "BBB", Status: "NO"},
{ Name: "CCC", Status: "YES"},
{ Name: "AAA", Status: "NO"},
{ Name: "AAA", Status: "YES"},
{ Name: "AAA", Status: "NO"},
{ Name: "AAA", Status: "YES"},
{ Name: "DDD", Status: "NO"}
);

 

//List which gets Unique Yes Names

ClearCollect(DistinctYesStatus, Distinct(Filter(StatusCollection, Status ="YES"),Name));

 

//List which gets Unique No Names.

ClearCollect(DistinctNOStatus, Distinct(Filter(StatusCollection, Status ="NO"),Name));

 

Mark it as verified if it answers your problem.

Regards,

Pavan Kumar Garlapati

Anonymous
Not applicable

Thank @PavanKumar;

My list is in Excel and the conditions are complicated (many conditions).

I am looking for a solution without push the list to a Collection.

Hi,

 

You don't have to push the list to collection, You can directly use the collection in the formulae and add as many conditions as you would like.

 

Mark answer as verified if it answers your question.

Regards,

Pavan Kumar Garlapati

View solution in original post

Anonymous
Not applicable

Thanks! It worked!

I used 

CountRows(Distinct(Filter(Table;Conditions1...));Name))

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