带反馈和多个if循环的psychopy构建实验

1 投票
1 回答
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提问于 2025-04-20 22:28

我正在尝试使用Coder视图来做一个实验,这个实验涉及反馈和多个条件判断。不过,我该怎么做呢?这个任务最终会包含4道数学题,参与者每道题最多可以尝试3次。结构大概是这样的……

循环1:遍历这4道题

循环2:每道题允许最多3次尝试

循环3:如果回答正确,就说“正确”,然后进入下一道题;如果回答错误,就说“错误”,并询问他们是否想再试一次或者继续下一道题

这是我第一次使用Python,我在代码中遇到了一些问题。没有返回错误信息,但代码没有记录到回答,因此任务在提示屏幕上卡住了。下面是我的代码。我没有包括库和其他设置。

t=0
nProblem=4
nAttempt=3

while currentProb <= nProblem:
    problemTimer.reset()
    attempt = 1

    # *prompt* updates
    prompt.setAutoDraw(True)
    prompt.setText(u'Problem prompt will go here.\n\nType in your answer and press enter to submit.', log=False)

    while attempt <= nAttempt:

        response = []
        # *response* updates
        core.wait(1) #LB - using this in place of that enormous if-statement

        event.clearEvents(eventType='keyboard')
        theseKeys = event.getKeys(keyList=['1', '2', '3', '4', '5', '6', '7', '8', '9', '0'])
        # check for quit:
        if "escape" in theseKeys:
            endExpNow = True
        if ('1', '2', '3', '4', '5', '6', '7', '8', '9', '0') in theseKeys: # subject responds with number value
            response.append(theseKeys) # will tack on responses to previous responses

        while event.getKeys(keyList=['return'])==[]:

            # *timer* updates
            if t <= 0.0:
                # keep track of start time/frame for later
                timer.setAutoDraw(True)
            #elif timer.status == STARTED and t >= (0.0 + (600-win.monitorFramePeriod*0.75)): #most of one frame period left
                #timer.setAutoDraw(False)
            #if timer.status == STARTED:  # only update if being drawn
                timer.setText(str(round((600+routineTimer.getTime())/60,1)) , log=False)

            # *minutesleft* updates
            if t >= 0.0:
                # keep track of start time/frame for later
                minutesleft.setAutoDraw(True)
            #elif minutesleft.status == STARTED and t >= (0.0 + (600-win.monitorFramePeriod*0.75)): #most of one frame period left
            #minutesleft.setAutoDraw(False)

            numberlist=event.getKeys(keyList=['1','2','3','4','5','6','7','8','9','0','backspace','return'])
            for number in numberlist:
                #if key isn't backspace, add key pressed to the string
                if number !='backspace':
                    response.append(number)
                #otherwise, take the last letter off the string
                elif len(text)>=0:
                    response.remove(numberlist[number-1])
            #continually redraw text onscreen until return pressed
            answer = visual.TextStim(win, text=response,color="black",pos=(0,-100))
            answer.draw()
            win.flip()
            event.clearEvents()

        if len(theseKeys) > 0:  # at least one key was pressed
            response.keys.extend(theseKeys)  # storing all keys
            response.rt.append(response.clock.getTime())

        #check for quit
        if "escape" in theseKeys:
            endExpNow = True

        if response == '9999': # was this correct?
            correctAns = 1
        else: 
            correctAns = 0
        if theseKeys == 'enter':
            response.keys.extend(theseKeys) # storing all keys
            response.rt.append(attemptresponse.clock.getTime())
            if correctAns == 1:
                attempt += 888 #ends and goes to next problem
                currentProb += 1
                dataFile.write(attempt,attemptresponse,theseKeys,response,correctAns) #output separated by commas
                #dataFile.write('PID    COND    PROB    ATT TIME    RESP\n')
                response_correct.draw()
                win.flip()
                event.waitKeys()
            if correctAns == 0:
                attempt += 1 #LB = was previously setting to 1 forever
                dataFile.write(attempt-1,attemptresponse,theseKeys,response,correctAns) #output separated by commas
                response_incorrect.draw()
                win.flip()
                theseKeys = event.getKeys(keyList=['q','space'])
                if theseKeys == 'q':
                    continueRoutine = False
                if theseKeys == 'space':
                    prompt.draw()
                    win.flip()
                    event.waitKeys()

1 个回答

1

我不能给你写完整的代码,但希望能指出一些要点,帮助你接近目标。

获取响应

theseKeys = event.getKeys(keyList=['1', '2', '3', '4', '5', '6', '7', '8', '9', '0'])
if "escape" in theseKeys:

在这里,theseKeys 只能包含 keyList 中的内容,所以“escape”这个键是不会出现的。你可以在 keyList 中添加“escape”键,可能还要加上你后面会用到的“enter”键。

if ('1', '2', '3', '4', '5', '6', '7', '8', '9', '0') in theseKeys:

theseKeys 是一个列表,当你用 "(x,y,z) in list" 这样的方式检查时,它会寻找一个完整的元素 (x,y,z),而不是单独检查 x、y 和 z。如果你的 keyList 是上面的样子,你就知道任何不是“escape”的响应都是其中之一,所以其实不需要这样检查。因此可以这样做:

theseKeys = event.getKeys(keyList=['1', '2', '3', '4', '5', '6', '7', '8', '9', '0', 'escape'])
if "escape" in theseKeys:
    core.quit()  # assuming that psychopy.core is imported
elif "enter" in theseKeys:
    # Something something
else:
   response.append(theseKeys[0])  # OBS: pick the first response

循环

看起来你更想用 for 循环,而不是 while 循环。使用 for 循环时,你不需要自己去跟踪当前的循环次数。所以,不用这样:

while currentProb <= nProblem:

这样做:

for currentProb in range(nProblem):

这样 currentProb 的增加会自动进行,循环会在应该结束的时候自动停止。这样做会更优雅。

等待响应

我有点不确定你是否想在 while 循环中使用 event.waitKeys() 而不是 event.getKeys(),如果你想控制等待时间,可以使用 maxWait 参数,并且如果你想跟踪时间,可以用 core.Clock()。如果你想动画某些东西,while 循环是可以的,但如果不是,使用 event.waitKeys() 会简单且安全得多。

最后,在你尝试让这个工作的时候,多用 print 语句来检查 theseKeys 和其他变量的实际内容。这是调试和捕捉问题的最好方法。

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