removing repetitions from list with same elements but in reverse order

Hi Raghu,

Assuming that 'a' is the original list of four sub-lists you can return the 'odd' or 'even' elements like this:

    idx=0 setof(item a idx++ && oddp(idx) && a)
=> ((1 2) 
    (1 3)
    )
idx=0 setof(item a idx++ && evenp(idx) && a)
=> ((2 1) 
    (3 1)
    )

Best regards,
Lawrence.

Hi lawrence,

That is just an example i gave . I am not dealing with numbers in real. Idea is to filter list with same elements i.e; removing repetition

Thanks,

Raghu

Hi Raghu,

Your description in the original post was rather unclear - I assumed the same as Lawrence.

Something like this is probably what you want?

procedure(CCFfilterReversedLists(l)
    let(((revListTable makeTable('reversedLists nil)))
        setof(sublist l
            unless(revListTable[sublist]
                when(listp(sublist)
                    revListTable[reverse(sublist)]=t
                )
                t
            )
        )
    )
)

Regards,

Andrew.

Hi Raghu,

Your description in the original post was rather unclear - I assumed the same as Lawrence.

Something like this is probably what you want?

procedure(CCFfilterReversedLists(l)
    let(((revListTable makeTable('reversedLists nil)))
        setof(sublist l
            unless(revListTable[sublist]
                when(listp(sublist)
                    revListTable[reverse(sublist)]=t
                )
                t
            )
        )
    )
)

Regards,

Andrew.

Hi Andrew,

Thank you so much Andrew.This is what i exactly wanted.

Regards,

Raghu