Linked/ conditioned consecutive randomization


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Dave
Dave
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Rebsu - 5/27/2019

At some point in the <trial Suche>, so if ZR1 or ZR1k is chosen in <trial Suche>, the other one (ZR1 or ZR1k) must also be chosen in <trial Suche>

Do you mean the same instance of <trial Suche> or some other, later instance of <trial Suche>?

Rebsu
Rebsu
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Dave - 5/27/2019
Rebsu - 5/27/2019

At some point in the <trial Suche>, so if ZR1 or ZR1k is chosen in <trial Suche>, the other one (ZR1 or ZR1k) must also be chosen in <trial Suche>

Do you mean the same instance of <trial Suche> or some other, later instance of <trial Suche>?
It should be within the same person, in one of the overall 20 <trial Suche>.
Dave
Dave
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Rebsu - 5/27/2019
It should be within the same person, in one of the overall 20 <trial Suche>.

Wait -- how can there be 20 instances of <trial Suche> if you only have 20 items (ZR1 to ZR10 plus the copies ZR1k to ZR10k) AND the selection of items in <trial Suche> is supposed to eliminate items that can be displayed by <trial AltPresentation>? That would mean that there can be NO items left for <trial AltPresentation>.

Rebsu
Rebsu
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Dave - 5/27/2019
Rebsu - 5/27/2019
It should be within the same person, in one of the overall 20 <trial Suche>.

Wait -- how can there be 20 instances of <trial Suche> if you only have 20 items (ZR1 to ZR10 plus the copies ZR1k to ZR10k) AND the selection of items in <trial Suche> is supposed to eliminate items that can be displayed by <trial AltPresentation>? That would mean that there can be NO items left for <trial AltPresentation>.

There are 20 ZR items and 20 NA items to choose from. (That wasn't in our old design so it's not in the code yet).
As I said, I'm not sure the set-up of item lists as it is now is the best way to do it.


Dave
Dave
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Rebsu - 5/27/2019
Dave - 5/27/2019
Rebsu - 5/27/2019
It should be within the same person, in one of the overall 20 <trial Suche>.

Wait -- how can there be 20 instances of <trial Suche> if you only have 20 items (ZR1 to ZR10 plus the copies ZR1k to ZR10k) AND the selection of items in <trial Suche> is supposed to eliminate items that can be displayed by <trial AltPresentation>? That would mean that there can be NO items left for <trial AltPresentation>.

There are 20 ZR items and 20 NA items to choose from. (That wasn't in our old design so it's not in the code yet).
As I said, I'm not sure the set-up of item lists as it is now is the best way to do it.


> There are 20 ZR items and 20 NA items to choose from.

Well, no. It doesn't work that way. <trial Suche> can sample from one of those (let's say ZR), but you need a separate <trial> to sample from the other (NA).




Rebsu
Rebsu
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Dave - 5/27/2019
Rebsu - 5/27/2019
Dave - 5/27/2019
Rebsu - 5/27/2019
It should be within the same person, in one of the overall 20 <trial Suche>.

Wait -- how can there be 20 instances of <trial Suche> if you only have 20 items (ZR1 to ZR10 plus the copies ZR1k to ZR10k) AND the selection of items in <trial Suche> is supposed to eliminate items that can be displayed by <trial AltPresentation>? That would mean that there can be NO items left for <trial AltPresentation>.

There are 20 ZR items and 20 NA items to choose from. (That wasn't in our old design so it's not in the code yet).
As I said, I'm not sure the set-up of item lists as it is now is the best way to do it.


> There are 20 ZR items and 20 NA items to choose from.

Well, no. It doesn't work that way. <trial Suche> can sample from one of those (let's say ZR), but you need a separate <trial> to sample from the other (NA).




Okay. I guess the best workaround then really are pseudorandomizations i.e. 6 groups where in each group it is clear which of the ZR items and which of the NA items are chosen for <trial Suche>.
Do I have to do that by creating 6 different <item ZR> lists --> 6 <trial Suche> -->6 different blocks for exp1 to exp6 or is there some better way?

And thank you for your help up until now already!!
Dave
Dave
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Rebsu - 5/27/2019
Dave - 5/27/2019
Rebsu - 5/27/2019
Dave - 5/27/2019
Rebsu - 5/27/2019
It should be within the same person, in one of the overall 20 <trial Suche>.

Wait -- how can there be 20 instances of <trial Suche> if you only have 20 items (ZR1 to ZR10 plus the copies ZR1k to ZR10k) AND the selection of items in <trial Suche> is supposed to eliminate items that can be displayed by <trial AltPresentation>? That would mean that there can be NO items left for <trial AltPresentation>.

There are 20 ZR items and 20 NA items to choose from. (That wasn't in our old design so it's not in the code yet).
As I said, I'm not sure the set-up of item lists as it is now is the best way to do it.


> There are 20 ZR items and 20 NA items to choose from.

Well, no. It doesn't work that way. <trial Suche> can sample from one of those (let's say ZR), but you need a separate <trial> to sample from the other (NA).




Okay. I guess the best workaround then really are pseudorandomizations i.e. 6 groups where in each group it is clear which of the ZR items and which of the NA items are chosen for <trial Suche>.
Do I have to do that by creating 6 different <item ZR> lists --> 6 <trial Suche> -->6 different blocks for exp1 to exp6 or is there some better way?

And thank you for your help up until now already!!

I don't think you need to do that. If you set up two <trial> elements -- <trial sucheZR> and <trial sucheNA> -- you can prevent one selecting an item # that has already been selected by the other. Maybe let's start with this and perhaps you can build from there (since I'm still not entirely clear what you want the end result to be):

<values>
/ itemnumber = 1
</values>

// will run 5 sucheZR trials and 5 sucheNA trials
<block exampleblock>
/ trials = [1-10 = noreplace(sucheZR, sucheNA)]
</block>

<trial sucheZR>
/ ontrialbegin = [values.itemnumber = list.itemnumbers.nextindex;]
/ stimulusframes = [1=ZR]
/ validresponse = (57)
</trial>

<trial sucheNA>
/ ontrialbegin = [values.itemnumber = list.itemnumbers.nextindex;]
/ stimulusframes = [1=NA]
/ validresponse = (57)
</trial>

// 10 items in this example
// item # selected for ZR cannot be used for NA
// and vice versa
<list itemnumbers>
/ poolsize = 10
</list>


<text ZR>
/ items = ZRitems
/ select = values.itemnumber
</text>

<item ZRitems>
/1 = "ZR1.jpg"
/2 = "ZR2.jpg"
/3 = "ZR3.jpg"
/4 = "ZR4.jpg"
/5 = "ZR5.jpg"
/6 = "ZR6.jpg"
/7 = "ZR7.jpg"
/8 = "ZR8.jpg"
/9 = "ZR9.jpg"
/10 = "ZR10.jpg"
</item>

<text NA>
/ items = NAitems
/ select = values.itemnumber
</text>

<item NAitems>
/1 = "NA1.jpg"
/2 = "NA2.jpg"
/3 = "NA3.jpg"
/4 = "NA4.jpg"
/5 = "NA5.jpg"
/6 = "NA6.jpg"
/7 = "NA7.jpg"
/8 = "NA8.jpg"
/9 = "NA9.jpg"
/10 = "NA10.jpg"
</item>




Rebsu
Rebsu
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Dave - 5/27/2019
Rebsu - 5/27/2019
Dave - 5/27/2019
Rebsu - 5/27/2019
Dave - 5/27/2019
Rebsu - 5/27/2019
It should be within the same person, in one of the overall 20 <trial Suche>.

Wait -- how can there be 20 instances of <trial Suche> if you only have 20 items (ZR1 to ZR10 plus the copies ZR1k to ZR10k) AND the selection of items in <trial Suche> is supposed to eliminate items that can be displayed by <trial AltPresentation>? That would mean that there can be NO items left for <trial AltPresentation>.

There are 20 ZR items and 20 NA items to choose from. (That wasn't in our old design so it's not in the code yet).
As I said, I'm not sure the set-up of item lists as it is now is the best way to do it.


> There are 20 ZR items and 20 NA items to choose from.

Well, no. It doesn't work that way. <trial Suche> can sample from one of those (let's say ZR), but you need a separate <trial> to sample from the other (NA).




Okay. I guess the best workaround then really are pseudorandomizations i.e. 6 groups where in each group it is clear which of the ZR items and which of the NA items are chosen for <trial Suche>.
Do I have to do that by creating 6 different <item ZR> lists --> 6 <trial Suche> -->6 different blocks for exp1 to exp6 or is there some better way?

And thank you for your help up until now already!!

I don't think you need to do that. If you set up two <trial> elements -- <trial sucheZR> and <trial sucheNA> -- you can prevent one selecting an item # that has already been selected by the other. Maybe let's start with this and perhaps you can build from there (since I'm still not entirely clear what you want the end result to be):

<values>
/ itemnumber = 1
</values>

// will run 5 sucheZR trials and 5 sucheNA trials
<block exampleblock>
/ trials = [1-10 = noreplace(sucheZR, sucheNA)]
</block>

<trial sucheZR>
/ ontrialbegin = [values.itemnumber = list.itemnumbers.nextindex;]
/ stimulusframes = [1=ZR]
/ validresponse = (57)
</trial>

<trial sucheNA>
/ ontrialbegin = [values.itemnumber = list.itemnumbers.nextindex;]
/ stimulusframes = [1=NA]
/ validresponse = (57)
</trial>

// 10 items in this example
// item # selected for ZR cannot be used for NA
// and vice versa
<list itemnumbers>
/ poolsize = 10
</list>


<text ZR>
/ items = ZRitems
/ select = values.itemnumber
</text>

<item ZRitems>
/1 = "ZR1.jpg"
/2 = "ZR2.jpg"
/3 = "ZR3.jpg"
/4 = "ZR4.jpg"
/5 = "ZR5.jpg"
/6 = "ZR6.jpg"
/7 = "ZR7.jpg"
/8 = "ZR8.jpg"
/9 = "ZR9.jpg"
/10 = "ZR10.jpg"
</item>

<text NA>
/ items = NAitems
/ select = values.itemnumber
</text>

<item NAitems>
/1 = "NA1.jpg"
/2 = "NA2.jpg"
/3 = "NA3.jpg"
/4 = "NA4.jpg"
/5 = "NA5.jpg"
/6 = "NA6.jpg"
/7 = "NA7.jpg"
/8 = "NA8.jpg"
/9 = "NA9.jpg"
/10 = "NA10.jpg"
</item>




Ok, thanks! The only thing is, that I need the copies of ZR and NA in order for them to be selected again --> for both AZ1 ans BZ1/ AN1 and BN1 to be presented. Is there a way to make that work? If not, I think I'm really just gonna go with the 6 groups and get it over with haha :)
Dave
Dave
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Rebsu - 5/27/2019

Ok, thanks! The only thing is, that I need the copies of ZR and NA in order for them to be selected again --> for both AZ1 ans BZ1/ AN1 and BN1 to be presented. Is there a way to make that work? If not, I think I'm really just gonna go with the 6 groups and get it over with haha :)

I'm fairly sure there's a way to make that work, but I'm still not clear on that part, i.e. how you actually want this to behave.

Rebsu
Rebsu
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Posts: 11, Visits: 33
Dave - 5/27/2019
Rebsu - 5/27/2019

Ok, thanks! The only thing is, that I need the copies of ZR and NA in order for them to be selected again --> for both AZ1 ans BZ1/ AN1 and BN1 to be presented. Is there a way to make that work? If not, I think I'm really just gonna go with the 6 groups and get it over with haha :)

I'm fairly sure there's a way to make that work, but I'm still not clear on that part, i.e. how you actually want this to behave.

I'm just gonna do the typing and set up 6 experiment groups, that's probably faster and definitely easier for both :)

Thank you so so much for your help and patience!
GO

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