Listing Unique Combinations

[hutcheedingo]

I've been trying to write a Custom Function that will give me a list of unique combinations between two occurrences of the same list.

 

The nearest I've come is with this Function, using the following list as an example 1,2,3,4

 

ListCombinations ( listofValues ; separator )

 

Let([

listLength = ValueCount(listOfValues);

lastValue = GetValue(listOfValues; listLength);

lesserList = Case(

listLength > 1; ListCombinations(LeftValues(listOfValues; listLength -1 ); separator);

"")

];

Case(

listLength = 0; "";

listLength = 1; RightValues(listOfValues; 1); /* Ensures ¶ at end */

lastValue & "¶" &

lesserList & /* already terminated by ¶ */

Substitute(lesserList; "¶"; separator & lastValue & "¶")

)

)

 

The output comes out looking like this:

4

3

2

1

1 | 2

2 | 3

1 | 3

1 | 2 | 3

3 | 4

2 | 4

1 | 4

1 | 2 | 4

2 | 3 | 4

1 | 3 | 4

1 | 2 | 3 | 4

 

What I want is for the output to be limited to only the 2 number values, shown below.  I've had a play with the Function, but I think I've been looking at it too long and can't make hide nor hair of what I need to change to produce the below list   

 

1 | 2

2 | 3

1 | 3

3 | 4

2 | 4

1 | 4

 

Thank You for An Help in Advance  

 

 

 

[comment]

See if this helps:

http://www.briandunning.com/cf/1638

[LaRetta]

That link opens for me but it is blank and I restarted also.  Can you check it again for us?  :-)

[hutcheedingo]

I went back and had a look at the calculations, and went through a bundle of CF's on Brian Dunnings Site...

 

This is what I am working with now, since it meets the requirement for two variables.  I added in the If Function highlighted below

 

Combine Values ( list1; list2; divider )

 

Let([$ƒCounter[1] = Case($ƒCounter[1]; $ƒCounter[1]; 1);

       $ƒCounter[2] = Case($ƒCounter[2]; $ƒCounter[2]; 1)

]; //end define Let

 

   Substitute( MiddleValues( list1; $ƒCounter[1]; 1 ); ¶; "" ) & divider &

   If ( MiddleValues ( list1; $ƒCounter[1]; 1) ≠ MiddleValues ( list2; $ƒCounter[2]; 1 ) ;

   Substitute( MiddleValues( list2; $ƒCounter[2]; 1 ); ¶; "" ) ; "0" ) &

 

   Case(

 

      $ƒCounter[2] < ValueCount(list2);

         Let( $ƒCounter[2] = $ƒCounter[2] + 1; ¶ & CombineValues(list1; list2; divider) );

 

      $ƒCounter[1] < ValueCount(list1);

         Let([$ƒCounter[1] = $ƒCounter[1] + 1; $ƒCounter[2] = 1]; ¶ & CombineValues(list1; list2; divider) );

 

      Let([$ƒCounter[1] = ""; $ƒCounter[2] = ""]; "") //clear variables when done

 

   ) //end Case

 

) //end Let

 

This gives me an output shown below

 

1 | 0

1 | 2

1 | 3

1 | 4

2 | 1

2 | 0

2 | 3

2 | 4

3 | 1

3 | 2

3 | 0

3 | 4

4 | 1

4 | 2

4 | 3

4 | 0

 

Now, I want to make it recognise that 1 | 2 is the same as 2 | 1. So basically, all the Italicised outputs just don't appear.  I'd also like to remove the 1 | 0 outputs from the list in the event that the GetValue of List is Even, but keep it for when GetValue is Odd.

 

Any Ideas? since I'm afraid to say I'm stumped again... 

 

Thanking  You in Advance 

[comment]

You are experiencing a temporary rift in the universe's time-space continuum, which may lead to the occurrence of local asynchronicity incidents.

[comment]

Try it now.

http://www.briandunning.com/cf/1638

[hutcheedingo]

Brilliant... Works an absolute treat

 

Many Thanks, from both myself, and the hole I was banging into the wall with my head

   

[LaRetta]

And well worth waiting for the rift to close.  

[hutcheedingo]

Brilliant... Works an absolute treat

 

Many Thanks, from both myself, and the hole I was banging into the wall with my head

   

 

Sorry to say, I spoke a little too soon... When I realised, that given the usage for this CF, I will need those pesky 1 | 0 outputs to be included if the list of Values is Odd... But if it's even, then the 1 | 0 outputs need not be there...

 

Any thoughts Comment?

[comment]

I will need those pesky 1 | 0 outputs to be included if the list of Values is Odd... But if it's even, then the 1 | 0 outputs need not be there...

 

It seems to me you're attacking this at the wrong end. You should count the values before calling the function - and if the count comes out odd, add a value to the call.