Millisecond Forums

Aha, that makes a lot of sence, thank you so much,

<trial five>
/ ontrialbegin = [values.listnumber = 1; ==> instead of listnumber = 2

In this case, the three positions in the 3-position list would be sampled without replacement, then the list would reset, i.e. all  3 positions become available again, and then two positions are sampled again (without replacement). At the end of the trial, one unsampled position would remain in the list. The position would be the 1st position sampled the next time a trial samples from that list.

Aha, right that makes, sense, however, if I set trial5 to the "wrong list", namely the list for trial3, the experiment still runs, even though the list only has 3 values, and picks without replacemant for 5 items. Does that mean that 2 of the positions will simply not be changed, and remain the default value 0? A bit of a nonesense question, but it would help woth understanding how the programs handles these situations since it does not throw an error.

<trial five>
/ ontrialbegin = [values.listnumber = 1; ==> instead of listnumber = 2

> - selectionmode in the list hpos differentiates between a setsize of either 3 or 5?

Yes, that is correct.

> - how are positions chosen from one trial to the next?

Positions are selected randomly without replacement, which is the <list> element's default behavior.

> if I try to 'break' the experiment by setting the listnumber for 5hpos to 1 (list only has 3 positions);
> the experiment does not crash, how are the positions now determined?

I don't understand this question. values.listnumber in

<list hpos>
/ items = (list.3hpos.nextvalue, list.5hpos.nextvalue)
/ selectionmode = values.listnumber
/ selectionrate = always
</list>

determines from which "sub-list" positions are sampled. If values.listnumnber is set to 1, sampling occurs from the 1st <list> listed in /items, here: <list 3hpos>. If values.listnumber is set to 2, sampling occurs from the 2nd <list> listed in /items, here: < list 5hpos>. You can change the order in /items as needed, and replace or add further sub-lists.

Hi Dave,
That looks very useful, thanks. Just to check my understanding:
- selectionmode in the list hpos differentiates between a setsize of either 3 or 5?
- how are positions chosen from one trial to the next? selection rate is set to always: does this mean one random choice per trial, after which the other items are chosen in sequential order?

if I try to 'break' the experiment by setting the listnumber for 5hpos to 1 (list only has 3 positions); the experiment does not crash, how are the positions now determined?

Lastly, just a general question, I was wondering if there are any more tutorials online: I know the tutorial by Ron Dotsch, but this does not address any of the somewhat more advanced operations, of course the forum is a great resource already, but it would be handy to go over some general principles, like the one you outline here.


Roelof

You could use <list>s to assign positions, with one pair of <list>s for the 9-position case, one pair for the 7-position case, and finally one pair of <list>s for the 5-position case.
Note that you can "pipe through" or "nest" lists like so: https://www.millisecond.com/forums/FindPost20679.aspx

While the linked example concerns item selection, this would also work for positions:

<values>
/ listnumber = 1
/ t_h = 0
/ d1_h = 0
/ d2_h = 0
/ d3_h = 0
/ d4_h = 0
</values>

<block example>
/ trials = [1-10 = noreplace(three, five)]
</block>


<trial three>
/ ontrialbegin = [values.listnumber = 1;
    values.t_h = list.hpos.nextvalue;
    values.d1_h = list.hpos.nextvalue;
    values.d2_h = list.hpos.nextvalue;
]
/ stimulusframes = [1=t,d1,d2]
/ validresponse = (57)
</trial>


<trial five>
/ ontrialbegin = [values.listnumber = 2;
    values.t_h = list.hpos.nextvalue;
    values.d1_h = list.hpos.nextvalue;
    values.d2_h = list.hpos.nextvalue;
    values.d3_h = list.hpos.nextvalue;
    values.d4_h = list.hpos.nextvalue;
]
/ stimulusframes = [1=t,d1,d2,d3,d4]
/ validresponse = (57)
</trial>


<list hpos>
/ items = (list.3hpos.nextvalue, list.5hpos.nextvalue)
/ selectionmode = values.listnumber
/ selectionrate = always
</list>

//list #1: 3 horizontal positions
<list 3hpos>
/ items = (25%, 50%, 75%)
/ selectionrate = always
</list>

//list #2: 5 horizontal positions
<list 5hpos>
/ items = (30%, 40%, 50%, 60%, 70%)
/ selectionrate = always
</list>

//the target
<text t>
/ items = ("T")
/ hposition = values.t_h
</text>

//the distractors
<text d1>
/ items = ("D1")
/ hposition = values.d1_h
</text>

<text d2>
/ items = ("D2")
/ hposition = values.d2_h
</text>

<text d3>
/ items = ("D3")
/ hposition = values.d3_h
</text>

<text d4>
/ items = ("D4")
/ hposition = values.d4_h
</text>

As for linking x/y positions, the previous approach applies: https://www.millisecond.com/forums/FindPost22664.aspx

In other words: There should be no need for elaborate if-else logic, simple paired <list>s of x/y positions should suffice.

Hope this helps.