Hello, I still couldnt figure this out. If we cannot equally distribute, can we restrict the amount of trials for allincong condition that has 120 possible combinations? If we randomly select 30 of them, the problem ewill be solvers hopefully
We can equally distribute across the five conditions if you delete thisImage_color, thisWord_color and thisImagery_color from your spreadsheet and all of the duplicate rows so you only have 5 rows.
In Begin Experiment define
colors = ['red','blue','green','yellow','orange']
or whatever.
Then in Begin Routine
shuffle(colors)
if condType == 'allincong':
thisImage_color = colors[0]
thisWod_color = colors[1]
thisImagery_color = colors[2]
elif condType == 'allcong'
thisImage_color = colors[0]
thisWod_color = colors[0]
thisImagery_color = colors[0]
# You will need to add the rest yourself
If it ādoesnāt workā please show what you tried and exactly how it failed.
Delete code rows 76-80 from your final code component above
above you can see the code without these lines as well. It again gives empty rows
Reattach the spreadsheet ensuring it doesnāt have those as column titles. Iām wondering whether you just deleted the contents of the columns, not the columns themselves from your conditions file.
there are no columns in the xlsx file with such titles only images words and positions. even if i update the data file is still empty
Isnāt there a column called condType?
Sorry i missed that part. Would you mind telling me what should be under this column should it be emptyb
allincong
allcong
imagewordcong
imageimagerycong
wordimagerycong
Just a handy shortcut: Itās possible to multiply a list in order to repeat its elements:
>>> list(range(1, 3+1)) * 5 #note: the upper end of range is exclusive, so add 1
[1, 2, 3, 1, 2, 3, 1, 2, 3, 1, 2, 3, 1, 2, 3]
This is especially helpful if you have many elements within and/or repetitions of the list and may be a reason to deviate from @Becca 's solution.
I try to avoid solutions that will fail online. However, I could have used:
conditionsChoice1 = []
conditionsChoice2 = []
conditionsChoice3 = []
for Jdx in range(3):
Idx in range(1,6):
conditionChoice1.append(Idx)
conditionChoice2.append(Idx)
conditionChoice3.append(Idx)
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Dear all,
After struggling for weeks, I am sharing the solution in case someone else needs.
Begin Exp:
trial_order = [1, 2, 3, 4, 5, 1, 2, 3, 4, 5, 1, 2, 3, 4, 5, 1, 2, 3, 4, 5, 1, 2, 3, 4, 5, 1, 2, 3, 4, 5, 1, 2, 3, 4, 5, 1, 2, 3, 4, 5, 1, 2, 3, 4, 5, 1, 2, 3, 4, 5, 1, 2, 3, 4, 5, 1, 2, 3, 4, 5, 1, 2, 3, 4, 5, 1, 2, 3, 4, 5, 1, 2, 3, 4, 5, 1, 2, 3, 4, 5, 1, 2, 3, 4, 5, 1, 2, 3, 4, 5, 1, 2, 3, 4, 5, 1, 2, 3, 4, 5, 1, 2, 3, 4, 5, 1, 2, 3, 4, 5, 1, 2, 3, 4, 5, 1, 2, 3, 4, 5, 1, 2, 3, 4, 5, 1, 2, 3, 4, 5, 1, 2, 3, 4, 5, 1, 2, 3, 4, 5, 1, 2, 3, 4, 5, 1, 2, 3, 4, 5] #based on number of trials
shuffle(trial_order)
Begin Routine
objectChoice = [1,2,3,4,5,6]
wordChoice = [1,2,3,4,5,6]
imageryChoice = [1,2,3,4,5,6]
if cond_n == 1: #allincongruent
shuffle(objectChoice)
thisExp.addData('imageN',objectChoice[0])
if objectChoice [0] == 1:
thisImage = image1
thisImage_color = 'Kırmızı'
thisExp.addData('Image',1)
wordChoice.remove(1) #so that word will not be equal to the image
imageryChoice.remove(1) #so that imagery will not be equal to the image
elif objectChoice [0] == 2:
thisImage = image2
thisImage_color = 'Mor'
thisExp.addData('Image',2)
wordChoice.remove(2)
imageryChoice.remove(2)
elif objectChoice [0] == 3:
thisImage = image3
thisImage_color = 'Pembe'
wordChoice.remove(3)
imageryChoice.remove(3)
thisExp.addData('Image',3)
elif objectChoice [0] == 4:
thisImage = image4
thisImage_color = 'Turuncu'
wordChoice.remove(4)
imageryChoice.remove(4)
thisExp.addData('Image',4)
elif objectChoice [0] == 5:
thisImage = image5
thisImage_color = 'YeÅil'
wordChoice.remove(5)
imageryChoice.remove(5)
thisExp.addData('Image',5)
elif objectChoice [0] == 6:
thisImage = image6
thisImage_color = 'Mavi'
wordChoice.remove(6)
imageryChoice.remove(6)
thisExp.addData('Image',6)
shuffle(wordChoice)
thisExp.addData('wordN',wordChoice[0])
if wordChoice [0] == 1:
thisWord = word1
thisWord_color = 'Kırmızı'
imageryChoice.remove(1) #so that word will not be equal to the imagery
elif wordChoice [0] == 2:
thisWord = word2
thisWord_color = 'Mor'
imageryChoice.remove(2)
elif wordChoice [0] == 3:
thisWord = word3
thisWord_color = 'Pembe'
imageryChoice.remove(3)
elif wordChoice [0] == 4:
thisWord = word4
thisWord_color = 'Turuncu'
imageryChoice.remove(4)
elif wordChoice [0] == 5:
thisWord = word5
thisWord_color = 'YeÅil'
imageryChoice.remove(5)
elif wordChoice [0] == 6:
thisWord = word6
thisWord_color = 'Mavi'
imageryChoice.remove(6)
shuffle(imageryChoice)
thisExp.addData('imageryN',imageryChoice[0])
if imageryChoice [0] == 1:
thisImagery = imagery1
thisImagery_color = 'Kırmızı'
thisExp.addData('Imagery',1)
elif imageryChoice [0] == 2:
thisImagery = imagery2
thisImagery_color = 'Mor'
thisExp.addData('Imagery',2)
elif imageryChoice [0] == 3:
thisImagery = imagery3
thisImagery_color = 'Pembe'
thisExp.addData('Imagery',3)
elif imageryChoice [0] == 4:
thisImagery = imagery4
thisImagery_color = 'Turuncu'
thisExp.addData('Imagery',4)
elif imageryChoice [0] == 5:
thisImagery = imagery5
thisImagery_color = 'YeÅil'
thisExp.addData('Imagery',5)
elif imageryChoice [0] == 6:
thisImagery = imagery6
thisImagery_color = 'Mavi'
thisExp.addData('Imagery',6)
# save
thisExp.addData('imageColor', thisImage_color)
thisExp.addData('wordColor', thisWord_color)
thisExp.addData('imageryColor', thisImagery_color)
you can repeat this logic for other conditions
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