使用Python代码更快的正交解码器循环

时间:2014-08-22 04:17:36

标签: python raspberry-pi beagleboard interrupt-handling beagleboneblack

我正在使用BeagleBone Black并使用Adafruit的IO Python库。写了一个简单的正交解码功能,当电机以大约1800 RPM的速度运行时,它可以很好地工作。 但是当电机以更高的速度运行时,代码开始丢失一些中断,编码器计数开始累积错误。 你们有什么建议可以让代码更高效,或者是否有能够以更高频率循环中断的函数。

谢谢, KEL

以下是代码:

# Define encoder count function
def encodercount(term):
global counts       
global Encoder_A
global Encoder_A_old
global Encoder_B
global Encoder_B_old
global error


Encoder_A = GPIO.input('P8_7')  # stores the value of the encoders at time of interrupt
Encoder_B = GPIO.input('P8_8')

if Encoder_A == Encoder_A_old and Encoder_B == Encoder_B_old:
# this will be an error
    error += 1
    print 'Error count is %s' %error

elif (Encoder_A == 1 and Encoder_B_old == 0) or (Encoder_A == 0 and Encoder_B_old == 1):
# this will be clockwise rotation
    counts += 1
    print 'Encoder count is %s' %counts
    print 'AB is %s %s' % (Encoder_A, Encoder_B)

elif (Encoder_A == 1 and Encoder_B_old == 1) or (Encoder_A == 0 and Encoder_B_old == 0):
# this will be counter-clockwise rotation
    counts -= 1
    print 'Encoder count is %s' %counts
    print 'AB is %s %s' % (Encoder_A, Encoder_B)

else:
#this will be an error as well
    error += 1
    print 'Error count is %s' %error

Encoder_A_old = Encoder_A     # store the current encoder values as old values to be used as comparison in the next loop
Encoder_B_old = Encoder_B       

# Initialize the interrupts - these trigger on the both the rising and falling 
GPIO.add_event_detect('P8_7', GPIO.BOTH, callback = encodercount)   # Encoder A
GPIO.add_event_detect('P8_8', GPIO.BOTH, callback = encodercount)   # Encoder B

# This is the part of the code which runs normally in the background
while True:
    time.sleep(1)

1 个答案:

答案 0 :(得分:1)

提高代码效率...

def encodercount(term):
global counts       
global Encoder_A
global Encoder_A_old
global Encoder_B
global Encoder_B_old
global error

Encoder_A,Encoder_B = GPIO.input('P8_7'),GPIO.input('P8_8')

if ((Encoder_A,Encoder_B_old) == (1,0)) or ((Encoder_A,Encoder_B_old) == (0,1)):
# this will be clockwise rotation
    counts += 1
    print 'Encoder count is %s\nAB is %s %s' % (counts, Encoder_A, Encoder_B)

elif ((Encoder_A,Encoder_B_old) == (1,1)) or ((Encoder_A,Encoder_B_old) == (0,0)):
# this will be counter-clockwise rotation
    counts -= 1
    print 'Encoder count is %s\nAB is %s %s' % (counts, Encoder_A, Encoder_B)

else:
#this will be an error
    error += 1
    print 'Error count is %s' %error

Encoder_A_old,Encoder_B_old = Encoder_A,Encoder_B

# Initialize the interrupts - these trigger on the both the rising and falling 
GPIO.add_event_detect('P8_7', GPIO.BOTH, callback = encodercount)   # Encoder A
GPIO.add_event_detect('P8_8', GPIO.BOTH, callback = encodercount)   # Encoder B

# This is the part of the code which runs normally in the background
while True:
    time.sleep(1)

最大的好处将来自print单一电话。打印到stdout通常很慢,这会限制程序的性能。你应该考虑每20次打印一次或者不经常打印。