Swapping two items in a list reliably?

[helicoptur]

I need a block that can do that. Could someone help?

[slate_technologies]

You mean swap two items with different items in a list? Or do you mean reordering them?

I could help you make a block for that.

[bh]

[ego-lay_atman-bay]

Here's how I made it

https://snap.berkeley.edu/snap/snap.html#present:Username=ego-lay_atman-bay&ProjectName=swap%20items

[kinestheticlearning]

Can you explain why yours is more complex than the Brian's one; is it because of allowing for randomness or something else makes it more complex?

[ego-lay_atman-bay]

Ok, so here's what the script does

[scratchblocks]
set [list v] to ([id v] of (list) :: operators)
[/scratchblocks]
This is to create a copy of the list so I'm not modifying the original one. If the block was a command block, then I wouldn't do this.

[scratchblocks]
if <(1::variables) =[random]> ::control
set [1 v] to (pick random (1) to ([length v] of (list) :: list)
[/scratchblocks]
This is to make the random option working. Can be removed if you don't want this. If I didn't do this, all the values would be mixed up, and then you might lose an item instead of swapping it. I also did this because all other list blocks with an item number value has options for 1, random, and last.

[scratchblocks]
set [1 v] to (list (1::variables) (item (1::variables) of (list)) @delInput @addInput::list)
[/scratchblocks]
This is essentially just setting a temporary variable to the item value. I just didn't want to bother with script vars.

[scratchblocks]
replace item (item (1 v) of (1::variables)) of (list) with (item (2 v) of (2::variables)
[/scratchblocks]
This is where the actual swapping part is. It just replaces the first item with the sencond item value, and vice versa.

Does this help you understand how mine works?

[kinestheticlearning]

@ego-lay_atman-bay

This is essentially just setting a temporary variable to the item value.

It was not just setting it to the item value, but setting it to a new list, which has confused me, I guess, where the new list is made of the index value and the item at that index value.

As a non-math person, I had to think through your clever trick to understand it.

I like it now.