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# No more than two items per side (I can't use a constrain bc every item is randomly selected)

Hello,

So I have a paradigm where three items are presented one is presented to the left, one is presented to the right and one is presented on the top. If the top item is the same as the right item than the participant should click on the right item. If the top card is the same as the left item than the participant should click on the left item. So the correct answer is to click on the bottom right when the top card (target) is the same as the bottom right card and the correct answer is to click on the bottom left card when the top card (target) is the same as the bottom left card. I have the paradigm working nicely. Every item is shuffled so the item that is on the right and left is selected randomly and the item that is shown on the top is selected randomly. This is great and how I want it, except that I can't control for the fact that the correct answer could end up on the same side several times in a row. I would like to make it so the correct answer does not end up on the right side more than twice in a row and the same for the left side. I have attached the experiment. Any help would be greatly appreciated. Thank you so much.

Mollie

• Hi Mollie,
Do you still need help with this question, or has it been solved in the other discussion you had with Sebastiaan?

Eduard

• Hi Eduard,

I would still love help with this problem if you could help!

Thank you so much,

Mollie

• Hi Mollie,

Have you tried advances loop operations: https://osdoc.cogsci.nl/3.2/manual/structure/loop/#pseudorandomization

It sounds like you just need to constrain that maxrep value of the correct side to 1.

Is that possible?

(sorry for the tardy response, btw),

Eduard

• Hey Eduard,

No worries! Thanks for taking the time to respond. The advanced loop operations won't work because I do not have a column that accounts for correct side. The correct answer is defined later on with an in_line script. I have the rows shuffled horizontally such that a random item is selected at random so even I don't know which side the correct object will occur on (ie the match).

I have defined the correct asnwer as

if var.left==var.target and var.cursor_roi=='target_l'or var.right==var.target and var.cursor_roi=='target_r':

var.correct=1

else:

var.correct=0

So basically I do not want var.left==var.target more than twice in a row or var.right==var.target more than twice in a row.

Sorry it's a bit confusing. Thanks so much!

• Hi Mollie,

How about you add a variable that tracks the previous targets? So in the end of your trial loop add an `inline_script`, incl. something like this `var.prev_targets.append(var.target)` (also make sure you initialize the variable before the loop to make the script to crash on the first trial).

YOu can use this variable then, to select an appropriate target, like so:

```import random
# to make the operation easy, I just assume target 1 and target 2, are coded as 0 and 1
if len(var.prev_targets)>1:
# some the previous 2 entries
if sum(var.previous_targets[-2:]) == 2:
# if sum is 2, then there were 2 target 2 trials in a row
var.target = 0
elif sum(var.previous_targets[-2:]) == 0:
# if sum is 0, then there were 2 target 1 trials in a row
var.target = 1
else:
# if it is neither (1), then choose one target randomly
var.target = random.choice([0,1])﻿
```

Does that make sense?

Eduard

• Hey Eduard,

Thank you so so much for your help. Unfortunately, I keep getting a syntax error... I have tried to play around with it but I have not been successful.

Also, I just wanted to clarify. I put this script in an inline script in the run tab before the loop. and at the end of my loop I add var.prev_targets.append(var.target) in another inline script. I am not sure I understand what you mean by initialize the variable before the loop to make the script to crash on the first trial. Thank you very much for your help!

best,

Mollie

• Hi,

I actually realized that tracking the targets still won't help as those are selected randomly as well. So, the only way to know what the correct side is would be to create a variable called side which I have done like this:

if var.target== var.left:

var.side= 1

if var.target==var.right:

var.side=0

and it works meaning when the target matches the card on the right I get a 0 in my output and when the target matches the card on the left I get a 1 in my output. But I still am having the problem where I don't want the correct answer(ie. when either the target matches the right or the target matches the left) to be to the same side more than twice in a row. I have tried to adapt the script you gave me to do so but I keep getting errors. I tried something like this:

import random

if len(var.previous_side)>1:

if sum(var.previous_side[-2:]) == 2:

var.side = 0

elif sum(var.previous_side[-2:]) == 0:

var.side = 1

I don't know how to define previous_side? I have tried many ways. I'm really struggling. Thank you so much for your help.

Best,

Mollie

• edited April 13

Hi Mollie,

Can you share your experiment here. I think it might be quicker if I directly edit there. Another thought that I have (maybe nonsense), but if your main concern is to control the side at which the target appears, it would make more sense to create the trial sequence in advance, and then access the sequence item by item to present the stimuli. That way, you are in control and you always know where the correct response is. For example:

```import random
list_to_shuffle = [0,1]*20
while True:
random.shuffle(list_to_shuffle﻿)
for i in range(2,len(list_to_shuffle):
break_while = True
seq_count = sumlist_to_shuffle[i-2:i+1])
if seq_count == 0 or seq_count == 3:
break_while = False
break

if break_while:
break
```

Unfortunately, I keep getting a syntax error

That is weird, the line that lead to the error is correct syntactically. So, probably something happened before that. Again, if I see the full script, it'll get clearer.

Also, I just wanted to clarify.

There are several things here. First, the important part of you script (that makes sure you won't get 2 of the same kind in a row), relies on information of the current and of the previous two trials. That means, for the first and second trial, information on the previous two trials are not available yet, unless you define them before you need it. This is what I mean with initializing. Before, you even enter the block, you make sure that the variables that you will need to do the computation exist. Does that make sense?

I put this script in an inline script in the run tab before the loop

I think prepare tab makes more sense. And just to be clear, only the part in which you initialize the variable that you will need has to come before the loop. The actual computation has to be in the loop (after a response has been given).

Hope this helps,

Eduard

• Hi Eduard,

Thank you so much for all of your help! I appreciate it so much. I have attached the project here.

Best,

Mollie

• Sorry Molly, I am having a look at your script right now, but I have a hard time understanding what is supposed to happen exactly. Could you maybe summarize again super briefly your final goal and your current approach to achieving (incl. which variables need to be controlled)?

As far as I understand, you have two images per trial (one left, one right), and one them is also presented on the top. What you want is that the target is never present on the same side for more than 2 trials in a row, right? So, as far as I can see, you can either control the target per trial, and then assign the stimuli to left/right sides according to your rules, or the opposite, you specify which stimuli go where, and then select which of them is the target according to your rules. (or of course both).

Eduard

• Hey Eduard,

Thank you so much. I have a left and right column with objects that are shuffled horizontally that means either object_4 or object_5 will end up on the right or the left. Then I have a top card (here I created two columns target and target_2 and shuffled them horizontally) so either object_4 or object_5 will end up in the top card position. Here the toddlers are supposed to find the match. So if the top card is object_4 and the bottom right card is also object_4 the correct answer is the bottom right card(so they should press the bottom right card). All of that is perfect. The only problem is I do not want the bottom right card to match the top card more than twice in a row. So I have created a variable side. Side= 1 when the bottom right card matches the top card and side=0 when the top card matches the bottom left. From here I think I could make sure that the match is not on the right more than twice in a row or on the left more than twice in a row. Does that make sense? Thanks again for taking a look for me.

Best,

Mollie

• Hi Mollie,

I finally got to implement something that might be useful to you. The exciting things happen in the inline script "counterbalancing", in which the trial sequence is implemented. In the inline_script "selectTrialItem", the item per trial is being selected. I tried to use clear comments, and not change too much of your code, but if anything is unclear, let me know.

Eduard

• amazing! Thank you so much. I just have one last concern. I can't have any of the object pairs repeat more than once. but right now they are repeating. Could you help me with this last thing? I really appreciate your detailed notes and help!!

Best,

Mollie

• edited April 22

• Hi Mollie,

The attached file is updated to make sure no stimuli are repeated.

• Thank you so so much! I learned so much!

• Hi Eduard,

So sorry to keep bothering you. I have now tried to implement this in the context of my entire experimental design but now I have problems. Because I have a training block that comes before this block, for some reason the first trial is using the images from the training block (possibly because we have defined var.target as trial_no -1). Additionally, I have added that the match is revealed when it is touched and that is showing the next match to come (I think for the same reason). I have attached the whole project this time. I really appreciate your input and time. Thank you so much again.

Best,

Mollie

• Just kidding! I figured it out!! Thank you!