correct respon not match

Tanto

Hi All,
I have little problem with my coding to make correct respon from a list which contain same object:
my_stim = ['2','3','3']

experiment design goes here

pre_stim = my_stim[trial]
t_first = expyriment.stimuli.TextLine(text=pre_stim, text_bold = True, text_size=60)
t_trial = expyriment.design.Trial()
t_trial.add_stimulus(t_first)
if my_stim[trial] == '3' :
correct = 122
elif my_stim[trial] == '3' :
correct = 107
else:
correct = 122

It gave same answer in object '3' (122 not 107 in second row), then I try:

if my_stim[trial] == '3' and pre_stim == my_stim[0]:
correct = 122
elif my_stim[trial] == '3' and pre_stim == my_stim[1]:
correct = 107
else:
correct = 122
It also gave the exact same number in object '3'. Instead using list, do I have to use dict for delivering my stimuli?
Any suggest would be great!!
Thanks in advance

fladd

Hi,

it is not clear to me what exactly doesn't work. In your first example you will never reach the elif, because the expression to evaluate is the same for the if before.

In the second example they are different, though.

What do you mean with "it also gave the exact same number in object '3'"?

Tanto

Thanks for the respon!
Sorry, ignore the second example, thats just silly. What I am trying to reach is like first example but evaluate the elif statement. Just like this:

block_one = expyriment.design.Block(name="1")
trial_one = expyriment.design.Trial() #0
stim = expyriment.stimuli.TextLine(text='3', text_colour=expyriment.misc.constants.C_GREEN, text_bold = True, text_size=60)
trial_one.add_stimulus(stim)
correct = 109
label_id = "M"
trial_one.set_factor("correct", correct)
trial_one.set_factor("label_id", label_id)
trial_two = expyriment.design.Trial() #1
stim = expyriment.stimuli.TextLine(text='2', text_colour=expyriment.misc.constants.C_GREEN, text_bold = True, text_size=60)
trial_two.add_stimulus(stim)
correct = 109
label_id = "M"
trial_two.set_factor("correct", correct)
trial_two.set_factor("label_id", label_id)
trial_three = expyriment.design.Trial() #2
stim = expyriment.stimuli.TextLine(text='3', text_colour=expyriment.misc.constants.C_GREEN, text_bold = True, text_size=60)
trial_three.add_stimulus(stim)
correct = 120
label_id = "X"
trial_three.set_factor("label_id", label_id)
trial_three.set_factor("correct", correct)

Thanks for the answer, the time and the opensource Expyriment!!

fladd

It is still not entirely clear to me what you would like to achieve. The elif statement checks the same thing as the if statement before. If you want to check which trial it is, then you need to check for thing that make this trial unique (e.g. the combination of text and label_id). But why not just give each trial/stimulus a unique ID? Then you can simply check for this. Or maybe I still don't understand what you are trying to do. Maybe you can try to explain it in a bit more detail?

Tanto

Hi, Fladd

I am trying to make simple n-back task. Since I can't refers the same trial with different result (i.e, "2,3,2". "number 2" in first sequence is false, and "number 2" in third sequence is the answer) were my biggest issues because I will not get the correct key for the answer. I will try using unique ID, thanks!

And once again, it is possible to run dot probe task in expyriment?

fladd

And once again, it is possible to run dot probe task in expyriment?

I don't see why it wouldn't be.