Is there a way to iterate over multiple lists to delete a duplicate element?

[Gayngel]

Is there a UDF somewhere that can iterate over multiple lists and delete duplicate elements and keep the element in just one list? (sort of like the .zip()function in Python) 

e.g:

list a = ["Apples"];

list b = ["Apples"];

list c = ["Apples"];

 

if the element "Apples" in list a is also in list b and list c delete it from both list b and c.

[Wulfie Reanimator]

The problem is that lists cannot contain lists in LSL, unlike Python, so you can't "iterate over multiple lists" unless you write a function that takes a specific amount of lists that never changes.

[Rolig Loon]

Take a look at the set of handy user-defined functions at http://wiki.secondlife.com/wiki/Category:LSL_List#Extended_List_Operations ..  I think there's one in there that will do what you want.  Or you can combine functions to do it.

Edited September 17, 2020 by Rolig Loon

[VirtualKitten]

You can add JSON constructs to lists to give you the same as a multi array. You said you want to compare lists against other lists is this all you are trying to do llListFindList will find a list item in a list if its an exact match you will have to parse one list with two others and then second with third. If its a partial match then you need a script:

//************************************ CORE ROUTINES BELOW WERE LSL LACKS FUNCTIONS **Virtual Kitten
integer _contains(string haystack, string needle) 
{
    return ~llSubStringIndex(haystack, needle);
}
integer _llStringPartialFindList(list list_1, string string_1) {
    integer s1 = llGetListLength(list_1);
    integer s2 = llGetListLength(list_1);
    integer  n;
    for(n=0;n<s2;n++) { 
     if (_contains(llList2String(list_1,n) ,string_1)!=0) return n;
    }
    return -1;
}

[animats]

llListSort might be helpful. Concatenate the lists, sort, remove sequential duplicates. SL's sort is O(N²) speed, which is embarrassingly slow and might be a problem for large lists.

[Mollymews]

the method depends on what the desired outcome is:

for example, given the following:

list a = ["apple","banana","carrot"];
list b = ["apple","banana"];
list c = ["carrot"];

is 7 possible outcomes

a = ["apple","banana","carrot"]; b = []; c = [];
a = ["apple","banana"]; b = []; c = ["carrot"];
a = ["apple"]; b = ["banana"]; c = ["carrot"];
a = ["banana","carrot"]; b = ["apple"]; c = [];
a = ["banana"]; b = ["apple"]; c = [carrot];
a = ["carrot"]; b = ["apple","banana"]; c = [];
a = []; b = ["apple","banana"]; c = ["carrot"];

each outcome depends on the order in which the lists are presented, and any other rules of precedence we might determine/code for our particular app use case

 

[DoteDote Edison]

A user function probably exists somewhere, but the gist of the process is... Take each item from List A one by one and llListFIndList() for that item in List B. If found, remove and if it’s possible to have duplicates in the same list, scan List B again. Else, scan List C. Repeat with item two.

After running through all items in List A, do the same with List B, but only llListFindList() in List C, since you already know there are no copies in List A.

No need to process items in List C, or whatever is the last list, since its dupes would’ve been found in scans from previous lists.

Probably best to start with the smallest list first, working towards the largest which, hopefully, won’t need to be processed at all.

Edited September 21, 2020 by DoteDote Edison

[VirtualKitten]

@DoteDote EdisonAlthough this is true of llListFindList()  it is not true of as a partial find as of the code I supplied as two of the lists would need to be checked as partial matches may not be found. Hope this clarifies?