Redefine value in each trial

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Redefine value in each trial

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skotturi	skotturi posted 10 Years Ago Topic Details Share Topic
Group: Forum Members Posts: 2, Visits: 4	Dear Community, I am trying to use the Prospective Time Estimation task provided by Inquisit. Instead of just doing one trial at a fixed time interval of 53 seconds, I am trying to do multiple trials of random length. Instead of using: <values> ... /timeinterval = 5300 </values> I'm using: <values> ... /timeinterval = ceil(rand(45,75))1000 </values> To achieve a random time interval in between 45 & 75 seconds. I'm currently just running two experiment blocks: <expt > /blocks = [1 = timeestimationstart, 2=timeestimationstart] </expt> This works except that the time interval variable isn't reassigned the second time, so my interval is the same for both trials. I have tried to force the reevaluation of the timeinterval variable: <trial start> /stimulustimes = [0=start, circle] /timeout = [values.timeinterval = ceil(rand(45, 75))1000] /branch = [trial.end] /recorddata = false </trial> But this fails with "Expression '[values.timeinterval' is invalid. Expression contains an unknown element or property name." I could just hardcode a second timeinterval, a second trial and second variable to be exported but I'd prefer a more elegant solution. It's been a while since I've coded much in Inquisit, been spending most of my time in python, js & R. Thus I'm undoubtedly over thinking this, I think... The code I'm working with can be found here (the prospective task): http://www.millisecond.com/download/library/TimeEstimation/ Cheers, Sante Tags change reevaluate trial value variable
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Dave	Dave posted 10 Years Ago ANSWER Post Details Share Post
Group: Administrators Posts: 12K, Visits: 98K	#1: Values are -- by design -- static. I.e. the expression in <values> ... /timeinterval = ceil(rand(45,75))1000 </values> will get evaluated once and then values.timeinterval will remain unchanged. It is supposed to. #2: This <trial start> ... /timeout = [values.timeinterval = ceil(rand(45, 75))1000] ... </trial> is simply invalid syntax. The correct way to express it would be simply <trial start> ... /timeout = ceil(rand(45, 75))1000 ... </trial> #3: The proper way to do this is <values> ... /timeinterval = 5300 </values> <trial start> ... /ontrialbegin = [values.timeinterval = ceil(rand(45, 75))1000] /timeout = values.timeinterval ... </trial>
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skotturi	skotturi posted 10 Years Ago ANSWER Post Details Share Post
Group: Forum Members Posts: 2, Visits: 4	Great, that works and makes a lot more sense. The /ontrialbegin method was just what I was looking for! Cheers, Sante
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