Hello

I'm trying to produce a list of all possible unique combinations from another list.

The order does not matter.

For example:

combinaisons_uniques (a, b, c, d) will return :

a, b, c, d

a, b, c

a, b, d

a, b

a, c

a, d

a, c, d

b, c, d

b, c

b d

b

c, d

c

d

any Idea?

Edited April 9, 201015 yr by Guest

There is no apparent need to go recursive here since the number of combinations is down to 16 (15) - making it recursive involves an overhead and safeguarding not needed here.

The function can then by and large be writen like you have put it, including a pilcrow at the end of each line...

Where is the real problem here?

--sd

the real problem is that i don't know how to do... :-(

Your example list has 4 items.

Do you know in advance if all of the lists for your custom function will have 4 items?

If it can be different from 4, is there a maximum number of items it will have?

Edited April 9, 201015 yr by Guest

no, it could be from 2 to 10 items in the list.

I found a java source :

http://stackoverflow.com/questions/403865/algorithm-to-sum-up-a-list-of-numbers-for-all-combinations

http://www.briandunning.com/cf/1023

unique_combinations_counter (number_values)

Edited April 9, 201015 yr by Guest

Although I spotted you missed out a combination, could it be witten this way:

--sd

Benka.zip

Thanks.

But there is a problem.

With 4 items, there are 15 combinations, but

with 7 items, there are 127 combinations...

I'm trying to make a custom function to do this.

Ah yes then it begins to make sense to make it recursive instead.

But it was a little unclear in the beginning of the thread!

--sd

Edited April 9, 201015 yr by Guest

Usage:

ListCombinations("a¶b¶c¶d", ", ")

Custom Function: (uses recursion)

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 & "¶")

)

)

Edited April 9, 201015 yr by Guest

:goodpost:

It works! Thanks a lot! :thankyou:

Edited April 9, 201015 yr by Guest

:clap:

--sd

Hello

I'm trying to produce a list of all possible unique consecutive combinations from another list.

The order does matter.

For example:

combinaisons_uniques (a, b, c, d) will return

a,b,c,d

a,b,c

a,b

a

b,c,d

b,c

b

c,d

c

d

any Idea?

Whats wrong with the last reply you got in this direction?

http://fmforums.com/forum/showtopic.php?tid/214260/post/354242/hl//fromsearch/1/#354242

...is it just the sortorder that makes it wrong?

--sd

This situation is different from the older one in that each result in the list must contain sequential letters from the original sequence without skipping a letter.

Notice that "b, d", "a, d", and so on are not in the list as they were in the older post.

From what I can tell, this is essentially the result of Middle("abcd", n, m) where n and m are appropriately chosen and commas are added.

In designing the function I would use MiddleValues("a¶b¶c¶d"; n;m) and substitute "¶" for "," at the end.

Ah yes I now as well spotted the difference -

--sd

I think the custom function described here provides most of this functionality.

That function has duplicate entries and doesn't separate by commas, but it does provide a starting point.

This works with the caveat that it produces some duplicates in the results. (Maybe someone can suggest a tweak that prevents the duplicates.)

ListSequences("a¶b¶c¶d"; ",") yields

a,b,c,d

a,b,c

a,b

a

b

b,c

b

c

b,c,d

b,c

b

c

c,d

c

d

// ListSequences(listOfValues; separator)

Let(

[

listLength = ValueCount(listOfValues);

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

];

Case(

listLength = 0; "";

listLength = 1; properList;

Substitute(

Left(

properList;

Length(properList) - 1); "¶"; separator) & "¶" &

ListSequences(LeftValues(properList; listLength-1); separator) &

ListSequences(RightValues(properList; listLength-1); separator)

)

)

Edited April 29, 201015 yr by Guest

Hello

Thank you all for your answers!

Sorry for the response time I'm in France.

I'll try to implement a duplicate management in this function.