Python 使用MovieStim2跳过基于PsychoPy中按键的视频(在循环中)
我用《心理变态》中的MovieStim2播放了3个视频。不过,我怀疑这个问题源于没有真正理解循环 当参与者按键时,我想跳到下一个视频。我理解当参与者按下“q”时我的代码是如何退出的,但我不知道如何让它在他们按下“b”时跳到下一个视频,例如。这是我目前的代码:Python 使用MovieStim2跳过基于PsychoPy中按键的视频(在循环中),python,loops,video,break,psychopy,Python,Loops,Video,Break,Psychopy,我用《心理变态》中的MovieStim2播放了3个视频。不过,我怀疑这个问题源于没有真正理解循环 当参与者按键时,我想跳到下一个视频。我理解当参与者按下“q”时我的代码是如何退出的,但我不知道如何让它在他们按下“b”时跳到下一个视频,例如。这是我目前的代码: vidNum = 1 for f in ['clip1.mpg', 'clip2.mpg', 'clip3.mpg']: clock.reset() #logfile.write("AfterForLoopTime,%s,V
vidNum = 1
for f in ['clip1.mpg', 'clip2.mpg', 'clip3.mpg']:
clock.reset()
#logfile.write("AfterForLoopTime,%s,Video %s\n" % (core.getTime(), vidNum))
# Create your movie stim.
mov = visual.MovieStim2(win, videopath+f,
size=640,
# pos specifies the /center/ of the movie stim location
pos=[0, 0],
flipVert=False,
flipHoriz=False,
) # loop=False - need to comment this to use a loop
# Start the movie stim by preparing it to play
shouldflip = mov.play()
logfile.write("AfterShouldflipLine58,%s, Video %s\n" % (clock.getTime(), vidNum))
while mov.status != visual.FINISHED:
# Only flip when a new frame should be displayed. Can significantly reduce
# CPU usage. This only makes sense if the movie is the only /dynamic/ stim
# displayed.
if shouldflip:
# Movie has already been drawn , so just draw text stim and flip
#text.draw()
win.flip()
else:
# Give the OS a break if a flip is not needed
time.sleep(0.001)
# Drawn movie stim again. Updating of movie stim frames as necessary
# is handled internally.
shouldflip = mov.draw()
# Check for action keys.....
for key in event.getKeys():
if key in ['escape', 'q']:
win.close()
core.quit()
我尝试添加类似于在q之后退出的最后一个块的代码:
elif key in ['b']:
break
但我意识到我真的想打破这个循环:
for f in ['clip1.mpg', 'clip2.mpg', 'clip3.mpg']:
然而,这似乎也不起作用
for f in ['clip1.mpg', 'clip2.mpg', 'clip3.mpg']:
for key in event.getKeys():
if key in ['b']:
break
实际上,您想要打破的循环是:
while mov.status != visual.FINISHED:
我能想到的最简单的方法是,如果用户按键,只需将您的电影状态设置为-1
(或visual.FINISHED
)
例如:
if key in ['b']:
mov.status = -1
这将打破你目前的视频从我可以告诉你