Adding break trials

PennController for IBEX Forums Support Adding break trials

Viewing 8 posts - 1 through 8 (of 8 total)
  • Author
    Posts
  • #5646
    angelica
    Participant

    Hi Jeremy,

    I’m working on a speeded grammaticality judgment task, and I’d like to insert a break trial every n experimental trials, so the participant has an opportunity to rest.

    I modified the running order manually :

    PennController.ResetPrefix(null);                       // Initiates PennController
    
    // source: https://github.com/addrummond/ibex/blob/master/docs/manual.md#modifying-the-running-order-manually
    function modifyRunningOrder(ro) {
    	var n = 5 ;
        for (var i = 0; i < ro.length; ++i) {
            if (i % n == (n-1)) {
                ro[i].push(new DynamicElement(
        			"PennController",
    			newTrial("break",
    			...
    			)
        		,
        		false
    		));
            }
        }
        return ro;
    }
    
    Sequence("practice", rshuffle("vpe", "fillers"), "exit", "send", "confirmation")
    
    ////////////////////////////////////////////////////////////////////////////////
    
    // Experiment
    
    // Trial template
    customTrial = label => variable => newTrial( label ,
        ...
        ,
        // RSVP sentence
        newController("dash", "DashedSentence", {s:variable.sentence, "mode":"speeded acceptability", "display":"in place", "wordTime":200})
            .cssContainer({"margin":"110px 0 0 0", "font-size": "150%",})
            .log()
            .print()
            .wait()
            .remove()
        ,
        ...
    )
    
    // Items
    Template("practice.csv",       	customTrial("practice"))
    Template("fillers.csv",    	customTrial("fillers"))
    Template("vpe.csv",            	customTrial("vpe"))
    
    // Post-experiment
    newTrial("exit",
    	...
    )
    
    // Send results
    PennController.SendResults("send");
    
    // End-of-experiment confirmation
    newTrial("confirmation",
    	...
    )

    Is it possible to restrict modifyRunningOrder() to certain parts of a Sequence()? I set n to 5 for testing, and so the “break” trial displays after the fifth “practice” trial. However, I’d like to only insert break trials in the rshuffle(“vpe”, “fillers”) section.

    Best,
    Angelica

    #5647
    Jeremy
    Keymaster

    Hi Angelica,

    Rather than using modifyRunningOrder, prefer using a custom function in Sequence, as described on this thread. You could do this to insert a break-labeled trial every 5 trials into your rshuffle subsequence:

    Sequence("practice", sepWithN( "break" , rshuffle("vpe", "fillers") , 5 ) , "exit", "send", "confirmation")

    Best,
    Jeremy

    #5648
    angelica
    Participant

    Hi Jeremy,

    I completely missed that thread, thank you for the help!

    #5659
    mrhelfrich
    Participant

    This was very helpful but I am encountering a minor problem with the the Break trial as it was described in this thread and the linked one. My experiment has 63 trials and the number of breaks I want to include doesn’t cleanly fit into that number (I also don’t need to give the participant a break after they have completed all 63 trials, they can just reach the end of the experiment at that point.
    I first tried to give breaks every 16 trials in the hopes that it would give a break at 16, 32, and 48, and since my .csv file only has 63 trials it would skip doing a fourth break since no 64th trial exists but I still get the break occurring. I even tried 17 just to see if it had something to do with 63 being so close to 64 and it still occurred.

    For reference, the code I am using is

    function SepWithN(sep, main, n) {
        this.args = [sep,main];
    
        this.run = function(arrays) {
            assert(arrays.length == 2, "Wrong number of arguments (or bad argument) to SepWithN");
            assert(parseInt(n) > 0, "N must be a positive number");
            let sep = arrays[0];
            let main = arrays[1];
    
            if (main.length <= 1)
                return main
            else {
                let newArray = [];
                while (main.length){
                    for (let i = 0; i < n && main.length>0; i++)
                        newArray.push(main.shift());
                    for (let j = 0; j < sep.length; ++j)
                        newArray.push(sep[j]);
                }
                return newArray;
            }
        }
    }
    function sepWithN(sep, main, n) { return new SepWithN(sep, main, n); }

    And my sequence looks like this:
    Sequence(sepWithN("break",randomize("exptrial"),17),"endexp")

    Thanks for the help,
    Max

    • This reply was modified 3 years, 2 months ago by Jeremy. Reason: replaced main.pop() with main.shift()
    #5660
    Jeremy
    Keymaster

    Hi Max,

    Replace this line:

                    for (let j = 0; j < sep.length; ++j)

    with this:

                    for (let j = 0; j < sep.length && main.length>0; ++j)

    Jeremy

    #6400
    aliona
    Participant

    Hi Jeremy,

    would it be possible to use the above mentioned SepWithN function while having different number of critical and filler items? In our experiment we have 20 criticals and 30 fillers devided into three blocks by 2 breaks (7/7/6 +10 fillers/block).
    My sequence looks like this:
    PennController.Sequence(…sepWithN(“break” , rshuffle(“critical_trials”, “fillers”) , 17 )…);

    How can I specify the number of the critical items and fillers/block? I tried adding the desired numbers after “critical_trials” and “fillers” in the sequence but that didn’t work, the proportion of criticls and fillers is still random.

    Thanks in advance for your responce!
    Aliona

    #6406
    Jeremy
    Keymaster

    Hi Aliona,

    One option would be to use the pick function defined in this thread:

    critical = randomize("critical_trials")
    fillers = randomize("fillers")
    
    Sequence( 
      rshuffle(pick(critical,7),pick(fillers,10)),"break",
      rshuffle(pick(critical,7),pick(fillers,10)),"break",
      rshuffle(pick(critical,6),pick(fillers,10))
    )

    Let me know if you have questions

    Jeremy

    #6415
    aliona
    Participant

    Hi Jeremy,

    it works perfectly! Thanks a lot for your help and all the great work on PC!

    Aliona

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