Merge lists A and B, shuffle row-wise, then split into shuffled lists A and B

Online experiments
1

Hi everyone,

I ran into some issues with list creation within the experiment, i.e., merging two lists:

stimlist A = [“1a.jpg”,“2a.jpg”,…,“192a.jpg”]
stimlistB = [“1b.jpg”,“2b.jpg”,…,“192b.jpg”]

into a table with “stimlistA” and “stimlistB” as variable/column names. This table should be shuffled row-wise/pair-wise, resulting in something like this:

grafik
grafik758×177 5.65 KB

Then I would like to split the shuffled table into two (now shuffled) lists:

stimlistA_shuffled = [“2a.jpg”,“192a.jpg”,…,“1a.jpg”]
stimlistB_shuffled = [“2b.jpg”,“192b.jpg”,…,“1b.jpg”]

When running on my local machine, the following Py code does the job:

stimlist = list(zip(stimlistA,stimlistB))
shuffle(stimlist)
shuffledstimlist = stimlist
stimlistA_shuffled, stimlistB_shuffled = map(list, zip(*stimlist))

However, I have not managed to translate this into JS to be used in an online experiment …

What I´ve tried so far:

stimlist = zip(stimlistA, stimlistB);
shuffle(stimlist);

after inserting the following code at the beginning of the experiment:

shuffle = util.shuffle;
Array.prototype.append = [ ].push;

const zip = (…arr) => {
const zipped = [ ];
arr.forEach((element, ind) => {
element.forEach((el, index) => {
if(!zipped[index]){
zipped[index] = [ ];
};
if(!zipped[index][ind]){
zipped[index][ind] = [ ];
}
zipped[index][ind] = el || ‘’;
})
});
return zipped;
};

URL of experiment: https://run.pavlovia.org/fapsySRHHD/mstps_20210628

Any help is greatly appreciated!

3

Hi @AKZ , for your JavaScript, you could use

// Create new arrays for shuffled lists
[newListA, newListB] = [[], []];
// Create new indexing array for getting stim list values by row
arrayIndex = [...Array(stimlistA.length).keys()];
// shuffle indexing array
shuffle(arrayIndex);

// Fill new lists with stimlist values shuffled by row
for (let i of arrayIndex) {
    newListA.push(stimListA[i]);
    newListB.push(stimListB[i]);
}
4

Btw, this does seem like something the trial handler can do, unless you have multiple pairs that need to be shuffled

5

Hi,

thank you so much for your reply!
This helped a lot and I can follow the logic of the code. However, it only seems to shuffle newlistB but not newlistA according to the arrayIndext.
This is the output I found in the console:
grafik

And this is what I inserted as code in the experiment:

var newlistA;
var newlistB;
// create empty table:
[newlistA, newlistB] = [[ ], [ ]];
// create indexing array for shuffling:
array_shufflingIndex = […Array(stimlistA.length).keys()];
console.log(“array_shufflingIndex”, array_shufflingIndex);
//shuffle indexing array:
shuffle(array_shufflingIndex)
console.log(“array_shufflingIndex_shuffled”, array_shufflingIndex);

for (let i of array_shufflingIndex) {
newlistA.push(stimlistA[i]);
newlistB.push(stimlistB[i]);
}

Does anyone have an idea?

I guess it would be difficult to use the trial handler here as I have 6 item sets with 384 items each and several lists need to be passed to further phases of the experiment, where they should be split again etc. … But if there is an easier way - this would be more than welcome!
Thanks again!

6

Hi @AKZ , you might want to check that your code is not shuffling those lists elsewhere, because this code should not shuffle those lists in different orders.

7

Thanks!
It turned out that, for some reason, swapping positions of lists A and B in the for loop solved the problem - now lists are shuffled pair-wise.

for (let i of array_shufflingIndex) {
newlistB.push(stimlistB[i]);
newlistA.push(stimlistA[i]);
}

Thanks a lot @dvbridges, your code was of great help!