我认为你不需要线程,但我可能会误解你的要求,所以我将提出2个替代方案.
>没有线程和睡眠
假设您当前的程序流程如下：
while True:
data = read_sensors()
command = process(data)
motor_do(command)
time.sleep(delay)
然后你可以删除睡眠,如果最后一次呼叫至少延迟几秒钟,则只调用motor_do.
last_call_time = -delay # to be sure that motor_do can be called the first time
# "On Windows, [time.clock] returns wall-clock seconds elapsed since the first call to this
# function, as a floating point number, based on the Win32 function QueryPerformanceCounter()
# The resolution is typically better than one microsecond." (python 2.7 doc)
# i.e. close to 0 on first call of time.clock()
while True:
data = read_sensors()
command = process(data)
motor_try(command)
def motor_try(command):
global last_call_time
current_time = time.clock()
# on win that works, on unix... you may want to take a look at the NB
elapsed = current_time - last_call_time
if elapsed >= delay:
motor_do(command)
last_call_time = current_time
>使用线程(这是一个例子,我没有使用python 2.7进行线程/异步的经验,所以可能有更好的方法来做到这一点)
假设您当前的程序流程如下：
while True:
data = read_sensors()
command = process(data)
motor_do(command) // this function sleeps and you CANNOT change it
然后你必须启动1个线程,它只会异步地将命令推送到电机.
import thread
command = None
command_lock = thread.allocate_lock()
def main():
thread.start_new_thread(motor_thread)
global command
while True:
data = read_sensors()
with command_lock:
command = process(data)
def motor_thread():
while True:
while not command: # just to be sure
time.sleep(0.01)
# small delay here (for instance, the time consumed by read_sensors + process)
with command_lock:
motor_do(command)
command = None
time.sleep(delay)
注意：在Unix上,time.clock()返回处理器时间(=没有空闲时间),所以最好使用time.time()…除非更改系统时钟：“此函数通常返回非 – 如果在两次调用之间设置了系统时钟,则它可以返回比前一次调用更低的值.“ (python 2.7 doc)
我不知道time.sleep()对系统时钟变化的反应.
Python3：只使用time.monotonic()…或time.perf_counter().
转播
