在下面的代码中,我显示pic1持续3秒,然后我显示一个固定十字架1秒钟,然后我显示pic2直到用户按下一个键。据我所知,它应该只是在我的第3个'for'循环中收集按键,因为这是我创建键列表并检查键等的地方。但是,如果我在pic1或固定十字架中按键,一旦pic2出现,它立即继续使用代码。它似乎是在我的第三个'for循环'之前注册了按键,然后产生效果就像我的第三个循环开始发挥作用一样。我不知道这是怎么回事,因为我在pic1和固定显示期间没有检查任何键。有人可以在这里启发我吗?也许我误解了关于getKeys的一些基本信息。
- 如果我没有按任何东西,则会发生预期的行为..它显示pic2并等待一个键。如果按下一个键,它只会继续使用代码,或者如果超过60秒(我将图像设置为显示60秒,用户应该在前5秒内响应,因此60只是安全)。
def block1():
running = 1
while running ==1:
for frames in range(image_frames): #3 seconds
pic1[0].draw()
window.flip()
for frame in range(fixation): #1 second
fix.draw()
window.flip()
for frames in range(stim_Frame): #only moves on with keypress (or 60 secs)
pic2[0].draw()
start = window.flip()
if frames == 0:
stim_time = start
print "stim time: "
print stim_time
allKeys = event.getKeys(keyList = ('f','h','escape'))
for thisKey in allKeys:
if thisKey == 'escape':
print "you quit"
window.close()
core.quit()
if thisKey == 'f':
keyTime=core.getTime()
thisResp = 1
print "keytime is: "
print keyTime
elif thisKey == 'h':
keyTime=core.getTime()
thisResp = 0
print "keytime is: "
print keyTime
if thisResp == 1 or thisResp == 0:
break
running = 2
window.flip()
干杯, 史蒂夫
答案 0 :(得分:2)
event.getKeys()
返回其内存缓冲区中的所有密钥。在第三次循环之前清除缓冲区。
def block1():
running = 1
while running ==1:
for frames in range(image_frames): #3 seconds
pic1[0].draw()
window.flip()
for frame in range(fixation): #1 second
fix.draw()
window.flip()
event.clearEvents() # Clear the previously pressed keys.
for frames in range(stim_Frame): #only moves on with keypress (or 60 secs)
pic2[0].draw()
start = window.flip()
if frames == 0:
stim_time = start
print "stim time: "
print stim_time
allKeys = event.getKeys(keyList = ('f','h','escape'))
for thisKey in allKeys:
if thisKey == 'escape':
print "you quit"
window.close()
core.quit()
if thisKey == 'f':
keyTime=core.getTime()
thisResp = 1
print "keytime is: "
print keyTime
elif thisKey == 'h':
keyTime=core.getTime()
thisResp = 0
print "keytime is: "
print keyTime
if thisResp == 1 or thisResp == 0:
break
running = 2
window.flip()