How do I randomize the order of may blocks except the first one ?


How do I randomize the order of may blocks except the first one ?

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labdandeneau@gmail.com
labdandeneau@gmail.com
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Hey, I tried to  make it so all participants in my task would first all get the same practice block but then have a different order for the other blocks depending. Is this the accurate syntax for it ?
<expt>
/ subjects = (1 of 6)
/ groupassignment = groupnumber
/ postinstructions = (end)
/ onexptend = [values.completed = 1]
/ blocks = [1 = wordmemory0; 2 = wordmemory; 3 = wordmemory2; 4 = wordmemory3;]
</expt>

<expt>
/ subjects = (2 of 6)
/ groupassignment = groupnumber
/ postinstructions = (end)
/ onexptend = [values.completed = 1]
/ blocks = [1 = wordmemory0; 2 = wordmemory; 3 = wordmemory3; 4 = wordmemory2;]
</expt>

<expt>
/ subjects = (3 of 6)
/ groupassignment = groupnumber
/ postinstructions = (end)
/ onexptend = [values.completed = 1]
/ blocks = [1 = wordmemory0; 2 = wordmemory2; 3 = wordmemory3; 4 = wordmemory;]
</expt>

<expt>
/ subjects = (4 of 6)
/ groupassignment = groupnumber
/ postinstructions = (end)
/ onexptend = [values.completed = 1]
/ blocks = [1 = wordmemory0; 2 = wordmemory2; 3 = wordmemory; 4 = wordmemory3;]
</expt>

<expt>
/ subjects = (5 of 6)
/ groupassignment = groupnumber
/ postinstructions = (end)
/ onexptend = [values.completed = 1]
/ blocks = [1 = wordmemory0; 2 = wordmemory3; 3 = wordmemory; 4 = wordmemory2;]
</expt>

<expt>
/ subjects = (6 of 6)
/ groupassignment = groupnumber
/ postinstructions = (end)
/ onexptend = [values.completed = 1]
/ blocks = [1 = wordmemory0; 2 = wordmemory3; 3 = wordmemory2; 4 = wordmemory;]



Attachments
TheWordMemoryTaskModified2DATA.iqx (147 views, 65.00 KB)
Dave
Dave
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labdandeneau@gmail.com - Monday, March 4, 2019
Hey, I tried to  make it so all participants in my task would first all get the same practice block but then have a different order for the other blocks depending. Is this the accurate syntax for it ?
<expt>
/ subjects = (1 of 6)
/ groupassignment = groupnumber
/ postinstructions = (end)
/ onexptend = [values.completed = 1]
/ blocks = [1 = wordmemory0; 2 = wordmemory; 3 = wordmemory2; 4 = wordmemory3;]
</expt>

<expt>
/ subjects = (2 of 6)
/ groupassignment = groupnumber
/ postinstructions = (end)
/ onexptend = [values.completed = 1]
/ blocks = [1 = wordmemory0; 2 = wordmemory; 3 = wordmemory3; 4 = wordmemory2;]
</expt>

<expt>
/ subjects = (3 of 6)
/ groupassignment = groupnumber
/ postinstructions = (end)
/ onexptend = [values.completed = 1]
/ blocks = [1 = wordmemory0; 2 = wordmemory2; 3 = wordmemory3; 4 = wordmemory;]
</expt>

<expt>
/ subjects = (4 of 6)
/ groupassignment = groupnumber
/ postinstructions = (end)
/ onexptend = [values.completed = 1]
/ blocks = [1 = wordmemory0; 2 = wordmemory2; 3 = wordmemory; 4 = wordmemory3;]
</expt>

<expt>
/ subjects = (5 of 6)
/ groupassignment = groupnumber
/ postinstructions = (end)
/ onexptend = [values.completed = 1]
/ blocks = [1 = wordmemory0; 2 = wordmemory3; 3 = wordmemory; 4 = wordmemory2;]
</expt>

<expt>
/ subjects = (6 of 6)
/ groupassignment = groupnumber
/ postinstructions = (end)
/ onexptend = [values.completed = 1]
/ blocks = [1 = wordmemory0; 2 = wordmemory3; 3 = wordmemory2; 4 = wordmemory;]



If you want to treat the order of non-practice blocks (wordmemory to wordmemory3) as a between-subjects factor, then yes, that would be the way to do it.

If you merely want to randomize the order (not treating it as a systematic between-subjects manipulation), then you would simply make do with a single <expt> element:

<expt>
/ postinstructions = (end)
/ onexptend = [values.completed = 1]
/ blocks = [1 = wordmemory0; 2-4 = noreplace(wordmemory, wordmemory2, wordmemory3)]
</expt>

labdandeneau@gmail.com
labdandeneau@gmail.com
Partner Member
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Dave - 3/4/2019
labdandeneau@gmail.com - Monday, March 4, 2019
Hey, I tried to  make it so all participants in my task would first all get the same practice block but then have a different order for the other blocks depending. Is this the accurate syntax for it ?
<expt>
/ subjects = (1 of 6)
/ groupassignment = groupnumber
/ postinstructions = (end)
/ onexptend = [values.completed = 1]
/ blocks = [1 = wordmemory0; 2 = wordmemory; 3 = wordmemory2; 4 = wordmemory3;]
</expt>

<expt>
/ subjects = (2 of 6)
/ groupassignment = groupnumber
/ postinstructions = (end)
/ onexptend = [values.completed = 1]
/ blocks = [1 = wordmemory0; 2 = wordmemory; 3 = wordmemory3; 4 = wordmemory2;]
</expt>

<expt>
/ subjects = (3 of 6)
/ groupassignment = groupnumber
/ postinstructions = (end)
/ onexptend = [values.completed = 1]
/ blocks = [1 = wordmemory0; 2 = wordmemory2; 3 = wordmemory3; 4 = wordmemory;]
</expt>

<expt>
/ subjects = (4 of 6)
/ groupassignment = groupnumber
/ postinstructions = (end)
/ onexptend = [values.completed = 1]
/ blocks = [1 = wordmemory0; 2 = wordmemory2; 3 = wordmemory; 4 = wordmemory3;]
</expt>

<expt>
/ subjects = (5 of 6)
/ groupassignment = groupnumber
/ postinstructions = (end)
/ onexptend = [values.completed = 1]
/ blocks = [1 = wordmemory0; 2 = wordmemory3; 3 = wordmemory; 4 = wordmemory2;]
</expt>

<expt>
/ subjects = (6 of 6)
/ groupassignment = groupnumber
/ postinstructions = (end)
/ onexptend = [values.completed = 1]
/ blocks = [1 = wordmemory0; 2 = wordmemory3; 3 = wordmemory2; 4 = wordmemory;]



If you want to treat the order of non-practice blocks (wordmemory to wordmemory3) as a between-subjects factor, then yes, that would be the way to do it.

If you merely want to randomize the order (not treating it as a systematic between-subjects manipulation), then you would simply make do with a single <expt> element:

<expt>
/ postinstructions = (end)
/ onexptend = [values.completed = 1]
/ blocks = [1 = wordmemory0; 2-4 = noreplace(wordmemory, wordmemory2, wordmemory3)]
</expt>

thank you

I wanted it to be a between-subject factor. I sent a link to my experiment to a few people via email but they all encountered the same block order, why is that?
Dave
Dave
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thank you

I wanted it to be a between-subject factor. I sent a link to my experiment to a few people via email but they all encountered the same block order, why is that?

I cannot answer that without (a) the relevant code from the script and (b) the link you sent out.

labdandeneau@gmail.com
labdandeneau@gmail.com
Partner Member
Partner Member (538 reputation)Partner Member (538 reputation)Partner Member (538 reputation)Partner Member (538 reputation)Partner Member (538 reputation)Partner Member (538 reputation)Partner Member (538 reputation)Partner Member (538 reputation)Partner Member (538 reputation)

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Posts: 10, Visits: 65
Dave - 7/22/2019

thank you

I wanted it to be a between-subject factor. I sent a link to my experiment to a few people via email but they all encountered the same block order, why is that?

I cannot answer that without (a) the relevant code from the script and (b) the link you sent out.

I have just sent you the link with the file so you can access the script. I could not copy paste the relevant syntax in the op however.
Dave
Dave
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Dave - 7/22/2019

thank you

I wanted it to be a between-subject factor. I sent a link to my experiment to a few people via email but they all encountered the same block order, why is that?

I cannot answer that without (a) the relevant code from the script and (b) the link you sent out.

I have just sent you the link with the file so you can access the script. I could not copy paste the relevant syntax in the op however.

Thank you -- I have responded via private message.

GO


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