为什么我的 event.getKeys() 内存缓冲区没有清除？
Why is my event.getKeys() memory buffer not clearing?
背景：
我有一项任务是使用 VLCMovieStim 播放视频，并且每一帧都在连续的双极尺度上收集评分。给出它的外观的想法。
event.getKeys() 跟踪按下 2 或 3 按钮但尚未释放的时间。当满足这些条件时，变量 keys 会更改值。对于 keys 由于按键而采用 'LEFT' 或 'RIGHT' 值的每一帧，比例将在相应方向上增加 1%。释放键时，event.clearEvents() 清除内存缓冲区，有效地将 keys 的值更改回 'STOP'。
问题：
当等级量表震荡次数足够多时，它就会断裂。从秤当前位于的任何位置，它都会前进到两个极点之一并停留在该终端位置，无论之后存在什么按钮状态。 print() 函数已经证明 keys 的值确实在每一帧都成功重置，但似乎一旦到达条件，感觉将条件 keys 更改为 'LEFT' 或 'RIGHT' 已经满足，尽管没有按下任何按钮。看起来虽然我正在更改 keys 的值但我实际上并没有清除事件内存，但我不知道为什么会这样。我假设我的逻辑一定有缺陷，但经过大约一周的讨论后，我不知所措：
kb = keyboard.Keyboard()
if kb.status == STARTED:
keys = 'STOP'
# When the left key is pressed and not released, keys = LEFT
# If either boolean is 'True', simply holding down the button will not yield continuous incrementing in the rating
if len(kb.getKeys([keyLeft], waitRelease = True, clear = True)) == 0 or len(kb.getKeys([keyRight], waitRelease = True, clear = True)) == 0:
keys = 'STOP'
if len(kb.getKeys([keyLeft], waitRelease = False, clear = False)) > 0:
keys = 'LEFT'
if len(kb.getKeys([keyLeft], waitRelease = True, clear = False)) > 0:
kb.clearEvents()
elif len(kb.getKeys([keyRight], waitRelease = False, clear = False)) > 0:
keys = 'RIGHT'
if len(kb.getKeys([keyRight], waitRelease = True, clear = False)) > 0:
kb.clearEvents()
if keys == 'LEFT':
y1 -= round(0.005,3)
y1 = round(y1,3)
y3 = ((y1)/2)
if y1 < 0:
keyCount += 1
if keyCount > 100:
keyCount = 100
if y1 >0:
keyCount -= 1
if y1 == 0.000:
keyCount = 0
if y1 <= -0.5:
y1 = -0.5
keyCount = 100
if keys == 'RIGHT':
y1 += round(0.005,3)
y1 = round(y1,3)
y3 = ((y1)/2)
if y1 > 0:
keyCount += 1
if keyCount > 100:
keyCount = 100
if y1 <0:
keyCount -= 1
if y1 == 0.000:
keyCount = 0
if y1 >= 0.5:
y1 = 0.5
keyCount = 100
为了共享可复制的代码而不是发送垃圾邮件 post，我已经隔离了必要的代码并将其与必要的刺激一起提供，here.
提前感谢您的宝贵时间！
我想出了自己的解决方案。我只是简单地更改了条件逻辑，以指定如果当前仅按下一个键，则键只能采用方向值。这个问题似乎总是在改变方向时产生，所以通过限制一个人快速改变方向的能力，我似乎已经解决了这个问题。为了额外的保险，我指定如果刻度当前处于任一极点，则键不能采用方向值。几个小时以来，我一直在尝试破解脚本，但一直没有成功。当然，在它再次发生之前我无法确定，但祈祷吧。
if kb.status == STARTED:
# At the start of the loop, we check whether the scale is at either of its poles,
# and if so, we reset the keys variable and clear the event log.
if abs(y1) >= 0.5:
keys = 'STOP'
kb.clearEvents()
# If the left button is being pressed, and the right button is not being
# pressed, and the scale is not currently at the left pole, change the
# value of keys to LEFT. These additional conditionals are crucial for me.
if len(kb.getKeys([keyLeft], waitRelease = False, clear = False)) == 1 and len(kb.getKeys([keyRight], waitRelease = False, clear = False)) != 1 and y1 > -0.5:
keys = 'LEFT'
# Once the key is released, clear the memory buffer and change
# the value of keys to STOP
if len(kb.getKeys([keyLeft], waitRelease = True, clear = False)) > 0:
keys = 'STOP'
kb.clearEvents()
# Just as above, except for the right key.
elif len(kb.getKeys([keyRight], waitRelease = False, clear = False)) == 1 and len(kb.getKeys([keyLeft], waitRelease = False, clear = False)) != 1 and y1 < 0.5 :
keys = 'RIGHT'
if len(kb.getKeys([keyRight], waitRelease = True, clear = False)) > 0:
keys = 'STOP'
kb.clearEvents()
背景：
我有一项任务是使用 VLCMovieStim 播放视频，并且每一帧都在连续的双极尺度上收集评分。给出它的外观的想法。
event.getKeys() 跟踪按下 2 或 3 按钮但尚未释放的时间。当满足这些条件时，变量 keys 会更改值。对于 keys 由于按键而采用 'LEFT' 或 'RIGHT' 值的每一帧，比例将在相应方向上增加 1%。释放键时，event.clearEvents() 清除内存缓冲区，有效地将 keys 的值更改回 'STOP'。
问题：
当等级量表震荡次数足够多时，它就会断裂。从秤当前位于的任何位置，它都会前进到两个极点之一并停留在该终端位置，无论之后存在什么按钮状态。 print() 函数已经证明 keys 的值确实在每一帧都成功重置，但似乎一旦到达条件，感觉将条件 keys 更改为 'LEFT' 或 'RIGHT' 已经满足，尽管没有按下任何按钮。看起来虽然我正在更改 keys 的值但我实际上并没有清除事件内存，但我不知道为什么会这样。我假设我的逻辑一定有缺陷，但经过大约一周的讨论后，我不知所措：
kb = keyboard.Keyboard()
if kb.status == STARTED:
keys = 'STOP'
# When the left key is pressed and not released, keys = LEFT
# If either boolean is 'True', simply holding down the button will not yield continuous incrementing in the rating
if len(kb.getKeys([keyLeft], waitRelease = True, clear = True)) == 0 or len(kb.getKeys([keyRight], waitRelease = True, clear = True)) == 0:
keys = 'STOP'
if len(kb.getKeys([keyLeft], waitRelease = False, clear = False)) > 0:
keys = 'LEFT'
if len(kb.getKeys([keyLeft], waitRelease = True, clear = False)) > 0:
kb.clearEvents()
elif len(kb.getKeys([keyRight], waitRelease = False, clear = False)) > 0:
keys = 'RIGHT'
if len(kb.getKeys([keyRight], waitRelease = True, clear = False)) > 0:
kb.clearEvents()
if keys == 'LEFT':
y1 -= round(0.005,3)
y1 = round(y1,3)
y3 = ((y1)/2)
if y1 < 0:
keyCount += 1
if keyCount > 100:
keyCount = 100
if y1 >0:
keyCount -= 1
if y1 == 0.000:
keyCount = 0
if y1 <= -0.5:
y1 = -0.5
keyCount = 100
if keys == 'RIGHT':
y1 += round(0.005,3)
y1 = round(y1,3)
y3 = ((y1)/2)
if y1 > 0:
keyCount += 1
if keyCount > 100:
keyCount = 100
if y1 <0:
keyCount -= 1
if y1 == 0.000:
keyCount = 0
if y1 >= 0.5:
y1 = 0.5
keyCount = 100
为了共享可复制的代码而不是发送垃圾邮件 post，我已经隔离了必要的代码并将其与必要的刺激一起提供，here.
提前感谢您的宝贵时间！
我想出了自己的解决方案。我只是简单地更改了条件逻辑，以指定如果当前仅按下一个键，则键只能采用方向值。这个问题似乎总是在改变方向时产生，所以通过限制一个人快速改变方向的能力，我似乎已经解决了这个问题。为了额外的保险，我指定如果刻度当前处于任一极点，则键不能采用方向值。几个小时以来，我一直在尝试破解脚本，但一直没有成功。当然，在它再次发生之前我无法确定，但祈祷吧。
if kb.status == STARTED:
# At the start of the loop, we check whether the scale is at either of its poles,
# and if so, we reset the keys variable and clear the event log.
if abs(y1) >= 0.5:
keys = 'STOP'
kb.clearEvents()
# If the left button is being pressed, and the right button is not being
# pressed, and the scale is not currently at the left pole, change the
# value of keys to LEFT. These additional conditionals are crucial for me.
if len(kb.getKeys([keyLeft], waitRelease = False, clear = False)) == 1 and len(kb.getKeys([keyRight], waitRelease = False, clear = False)) != 1 and y1 > -0.5:
keys = 'LEFT'
# Once the key is released, clear the memory buffer and change
# the value of keys to STOP
if len(kb.getKeys([keyLeft], waitRelease = True, clear = False)) > 0:
keys = 'STOP'
kb.clearEvents()
# Just as above, except for the right key.
elif len(kb.getKeys([keyRight], waitRelease = False, clear = False)) == 1 and len(kb.getKeys([keyLeft], waitRelease = False, clear = False)) != 1 and y1 < 0.5 :
keys = 'RIGHT'
if len(kb.getKeys([keyRight], waitRelease = True, clear = False)) > 0:
keys = 'STOP'
kb.clearEvents()