我正在尝试通过Python打开和关闭电磁阀。可以在实验开始之前使用信息框定义此切换模式。例如,我可以定义切换的时间(切换的次数)或切换的时间(切换的时间,以秒为单位)。我向Arduino发送了两个字节,一个字节用于通道选择(1至8),第二个字节用于状态(0或1)。
切换次数非常有效。在while循环中,我给出了myTime> 0的指令,并且在每次切换过程中都将其减小了1。因此,while循环将一直运行,直到myTime或切换数变为零为止。
但是,当我尝试从信息框中分配时间时,它将变成一个无限循环。我可以从信息框中提供时间,然后将其添加到time.time()中,并尝试在while循环中进行比较,并想在time.time()大于所需时间时终止循环。
如何确保可以同时使用两种条件终止while循环?
Python code:
from __future__ import absolute_import, division, print_function
import serial
from time import sleep
import struct
from psychopy import core, data, event, gui, visual
import time
try:
arduino = serial.Serial('COM8',19200)
sleep(2)
print("Connection to " + 'COM8' + " established succesfully!\n")
except Exception as e:
print(e)
global command
## create a DlgFromDict
info = {'Observer':'jwp', 'Channel':['1','2','3','4','5','6','7','8'],
'BreathingCycle':4, 'Timer': 0, 'ExpVersion': 1.1, 'Debug Mode': True}
infoDlg = gui.DlgFromDict(dictionary=info, title='TestExperiment',
order=['ExpVersion', 'Observer'],
tip={'Observer': 'trained visual observer, initials'},
fixed=['ExpVersion'])
myChannel = info['Channel']
myTime = info ['BreathingCycle']
myTimer = info ['Timer']
win = visual.Window(fullscr=True, size=(1536, 864), monitor='laptop')
#TODO: handle in a different way the screen resolution
instruction1 = visual.TextStim(win, text=u"""Valve on!""")
instruction2 = visual.TextStim(win, text=u"""Valve off!""")
instruction3 = visual.TextStim(win, text=u"""Thank you!""")
if infoDlg.OK:
Mytimer = time.time() + myTimer
while (myTime > 0 or time.time() < Mytimer):
if myChannel == '1':
if u'i' in event.waitKeys():
command = arduino.write(struct.pack(u'>H',513))
myTime = myTime - 1
instruction1.draw()
win.flip()
if u'e' in event.waitKeys():
command = arduino.write(struct.pack(u'>H',512))
myTime = myTime - 1
instruction2.draw()
win.flip()
elif myChannel == '2':
if u'i' in event.waitKeys():
command = arduino.write(struct.pack(u'>H',257))
myTime = myTime - 1
instruction1.draw()
win.flip()
if u'e' in event.waitKeys():
command = arduino.write(struct.pack(u'>H',256))
myTime = myTime - 1
instruction2.draw()
win.flip()
elif myChannel == '3':
if u'i' in event.waitKeys():
command = arduino.write(struct.pack(u'>H',2049))
myTime = myTime - 1
instruction1.draw()
win.flip()
if u'e' in event.waitKeys():
command = arduino.write(struct.pack(u'>H',2048))
myTime = myTime - 1
instruction2.draw()
win.flip()
elif myChannel == '4':
if u'i' in event.waitKeys():
command = arduino.write(struct.pack(u'>H',1025))
myTime = myTime - 1
instruction1.draw()
win.flip()
if u'e' in event.waitKeys():
command = arduino.write(struct.pack(u'>H',1024))
myTime = myTime - 1
instruction2.draw()
win.flip()
elif myChannel == '5':
if u'i' in event.waitKeys():
command = arduino.write(struct.pack(u'>H',4097))
myTime = myTime - 1
instruction1.draw()
win.flip()
if u'e' in event.waitKeys():
command = arduino.write(struct.pack(u'>H',4096))
myTime = myTime - 1
instruction2.draw()
win.flip()
elif myChannel == '6':
if u'i' in event.waitKeys():
command = arduino.write(struct.pack(u'>H',8193))
myTime = myTime - 1
instruction1.draw()
win.flip()
if u'e' in event.waitKeys():
command = arduino.write(struct.pack(u'>H',8192))
myTime = myTime - 1
instruction2.draw()
win.flip()
elif myChannel == '7':
if u'i' in event.waitKeys():
command = arduino.write(struct.pack(u'>H',32769))
myTime = myTime - 1
instruction1.draw()
win.flip()
if u'e' in event.waitKeys():
command = arduino.write(struct.pack(u'>H',32768))
myTime = myTime - 1
instruction2.draw()
win.flip()
elif myChannel == '8':
if u'i' in event.waitKeys():
command = arduino.write(struct.pack(u'>H',16385))
myTime = myTime - 1
instruction1.draw()
win.flip()
if u'e' in event.waitKeys():
command = arduino.write(struct.pack(u'>H',16384))
myTime = myTime - 1
instruction2.draw()
win.flip()
win.close ()
arduino.close()
答案 0 :(得分:0)
我找到了解决方案,因为有event.waitKeys()函数,它基本上在无限长的时间内等待键,这就是为什么time.time()