Compare list content no matter on order

Hi Bernd,

It depends to some extent as to whether you have repeated entries in lists. Actually your code won't work anyway (assuming you fix the missing returnVal on line 15 because it will only actually tell you whether the last entry). but even if you used a forall instead of a foreach, it would not work if there was repetition in the first list but not in the second. In other words, if the code was:

forall(element listA
  member(element listB)
)

then it works if you have your listA/listB examples above, but if you have:

listA=list("a" "b" "b")
listB=list("a" "c" "b")

It would still return t. I guess you could use:

length(listA)==length(listB) && forall(element listA member(element listB)) && forall(element listB member(element listA))

Even that's not really correct though because you could have:

listA=list("a" "b" "b")
listB=list("b" "a" "a")

and the expression above would return t.

Also, the code wouldn't be terribly efficient for long lists. A couple of simple possibilities:

sort(copy(listA) 'alphalessp)==sort(copy(listB) 'alphalessp)

This stops it being destructive by using copy. Another (maybe less costly) implementation would be:

procedure(CCFcompareListContent(listA listB)
  let(((contentCount makeTable('counts 0)))
    foreach(elem listA contentCount[elem]=contentCount[elem]+1)
    foreach(elem listB contentCount[elem]=contentCount[elem]-1)
    forall(elem contentCount zerop(contentCount[elem]))
  )
)

I didn't bother checking if the lists where the same length first, because that adds two more list iterations to the code - a bit of profiling might show that to be beneficial for long lists.

I'm sure there are other ideas, but I'd thought I'd throw these out there.

Regards,

Andrew.

Hi Bernd,

It depends to some extent as to whether you have repeated entries in lists. Actually your code won't work anyway (assuming you fix the missing returnVal on line 15 because it will only actually tell you whether the last entry). but even if you used a forall instead of a foreach, it would not work if there was repetition in the first list but not in the second. In other words, if the code was:

forall(element listA
  member(element listB)
)

then it works if you have your listA/listB examples above, but if you have:

listA=list("a" "b" "b")
listB=list("a" "c" "b")

It would still return t. I guess you could use:

length(listA)==length(listB) && forall(element listA member(element listB)) && forall(element listB member(element listA))

Even that's not really correct though because you could have:

listA=list("a" "b" "b")
listB=list("b" "a" "a")

and the expression above would return t.

Also, the code wouldn't be terribly efficient for long lists. A couple of simple possibilities:

sort(copy(listA) 'alphalessp)==sort(copy(listB) 'alphalessp)

This stops it being destructive by using copy. Another (maybe less costly) implementation would be:

procedure(CCFcompareListContent(listA listB)
  let(((contentCount makeTable('counts 0)))
    foreach(elem listA contentCount[elem]=contentCount[elem]+1)
    foreach(elem listB contentCount[elem]=contentCount[elem]-1)
    forall(elem contentCount zerop(contentCount[elem]))
  )
)

I didn't bother checking if the lists where the same length first, because that adds two more list iterations to the code - a bit of profiling might show that to be beneficial for long lists.

I'm sure there are other ideas, but I'd thought I'd throw these out there.

Regards,

Andrew.