Block n of n?

[Simone]

Hi Jeremy,

I’m in the process of creating an experiment in which participants will see many trials divided into 9 blocks separated by a break. To keep them motivated, I’d therefore like to inform them how far along in the experiment they are, e.g., “This is block n of 9.”

Is there any way to do that? For reference, the relevant part of my code looks as follows:

PennController.ResetPrefix(null); // Shorten command names (keep this line here))

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); }

// Start with welcome screen, then present test trials in a random order,
// and show the final screen after sending the results
Sequence( "welcome" , "instructions", "practice1" , "practice2", "practice3", "practice4", "begin", sepWithN("break", randomize("test"), 108) , "send" , "final" )

[Jeremy]

Hi Simone,

Your code seems perfectly fine to me (assuming you have 9 blocks of 108 test trials—if you have 9 blocks of 12 trials, you need to write sepWithN("break", randomize("test"), 12)). You could write your “break” trial like this, for example:

newTrial("break", 
    newVar("block_n", 0).global().set(v=>v+1)
    ,
    newVar("text").set(getVar("block_n")).set(v=>"You've just finished block "+v+" of 9")
    ,
    newText("prompt", "").text(getVar("text")).print()
    ,
    newButton("Next").print().wait()
)

Jeremy

[Author]

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