set [i v] to [1]
repeat until <[i] = [654]>
delete (i v) of [list]
change [i v] by (1)
end

Evan2963 wrote:

Is there an easy way to delete say the first 654 items in a list? I tried

set [i v] to [1]
repeat until <[i] = [654]>
delete (i v) of [list]
change [i v] by (1)
end

TheCookieHero880 wrote:

Evan2963 wrote:

Is there an easy way to delete say the first 654 items in a list? I tried

set [i v] to [1]
repeat until <[i] = [654]>
delete (i v) of [list]
change [i v] by (1)
end

What's the problem with your current script?

set [i v] to [1]
repeat until <[i] = [654]>
delete (1 v) of [list]
change [i v] by (1)
end

repeat (654)
delete (1 v) of [list v]
end

TheCookieHero880 wrote:

set [i v] to [1]
repeat until <[i] = [654]>
delete (1 v) of [list]
change [i v] by (1)
end

Ayarkay wrote:

Let's say you run the above script on the following list:

{1, 2, 3, 4, 5, 6, 7, 8, 9, 10}

So we set i = 1, and we delete item number i
So we delete item number 1, and our list now becomes:
{2, 3, 4, 5, 6, 7, 8, 9, 10}

Note that in this new list, the first element is 2, the second element is 3, the third element is 4, etc…
Now we increase i by 1, so i = 2.
So again, we delete item number i from the list, so we delete the second item in the list, and our list now becomes:

{2, 4, 5, 6, 7, 8, 9, 10}

Now increase i by 1, so i=3
Delete the ith element from the list, so delete the third element from the list:

{2, 4, 6, 7, 8, 9, 10}

Now i=4, so we delete the 4th item in the list:

{2, 4, 6, 8, 9, 10}

Finally i=5, so we delete the 5th element from the list:

{2, 4, 6, 8, 10}



Every time you delete an element from a list, all elements after it get moved up by 1 position in the list.
If you wanted to delete the first 654 elements from the list, you'd just do:

repeat (654)
delete (1 v) of [list v]
end

That way you don't need to continuously add 1 to your counter and to check if your counter = 654

Evan2963 wrote:

repeat (654)
delete (1 v) of [list v]
end