通过单独的线程（并行）将数据写入磁盘

Question

我想在一个循环中多次启动一个功能，每次从相机获取并成像，然后将图像写入光盘，而无需循环等待此过程完成。因此，每次调用此函数时，它都会与启动该函数的循环并行运行，这样我就可以在此期间继续做其他对时间敏感的事情。

我已经制作了这个示例，该示例使函数的第一个“执行”与循环并行运行，然后第二次失败，因为我无法.start（）两次。可以通过其他方式实现吗？

示例（原始帖子-更新如下）

import numpy as np
import threading
import time

def imacq():
    print('acquiring image...')
    time.sleep(1.8)
    print('saved image...')
    return

# Start image acqusition and writing to disc thread
imacq_thread = threading.Thread(target=imacq)

starttime = time.time()
sig_arr = np.zeros(100)
tim_arr = np.zeros(100)
image_cycles = 5
running = True
flag = True
for cycles in range(1,20):
    print(cycles)
    if cycles%image_cycles == 0:
        if flag is True:
            imacq_thread.start() # this works well the first time as intended
            # imacq() # this does not work as everything is paused until imacp() returns
            flag = False
    else:
        flag = True
    time.sleep(0.4)

编辑：在西尔瓦斯反馈之后： 我已经制作了两种不同的版本来触发函数，最终将使用它们来确定驱动器的发送，存储和映像，以及与决定发送触发/执行函数的时间的主脚本并行。一个版本基于Sylvaus的答案（线程），而另一个版本则基于多重处理。

基于Sylvaus答案的示例（线程）：

import matplotlib.pyplot as plt
import numpy as np
import time
from concurrent.futures import ThreadPoolExecutor


def imacq():
    print('taking image')
    n = 10000
    np.ones((n, n))*np.ones((n, n))  # calculations taking time
    print('saving image')
    return


sig_arr = np.zeros(100)
tim_arr = np.zeros(100)
image_cycles = 20
max_cycles = 100
freq = 10
cycles = 1
sigSign = 1

running = True
flag = True
timeinc = []
tic = time.time()
tic2 = tic
timeinc = np.zeros(max_cycles)
starttime = time.time()
with ThreadPoolExecutor() as executor:
    while running:
        t = time.time()-starttime
        tim_arr[:-1] = tim_arr[1:]
        tim_arr[-1] = t
        signal = np.sin(freq*t*(2.0*np.pi))
        sig_arr[:-1] = sig_arr[1:]
        sig_arr[-1] = signal

        time.sleep(0.00001)
        # Calculate cycle number
        sigSignOld = sigSign
        sigSign = np.sign(sig_arr[-1]-sig_arr[-2])
        if sigSign == 1 and sigSignOld != sigSign:
            timeinc[cycles] = time.time()-tic
            cycles += 1
            print('cycles: ', cycles, ' time inc.: ', str(timeinc[cycles-1]))
            tic = time.time()

        if cycles%image_cycles == 0:
            if flag is True:
                # The function is submitted and will be processed by a
                # a thread as soon as one is available
                executor.submit(imacq)
                flag = False
        else:
            flag = True
        if cycles >= max_cycles:
            running = False

print('total time: ', time.time()-tic2)

fig = plt.figure()
ax = plt.axes()
plt.plot(timeinc)

基于多重处理的示例：

import matplotlib.pyplot as plt
import numpy as np
import time
from multiprocessing import Process, Value, Lock


def trig_resp(running, trigger, p_count, pt, lock):
    while running.value == 1:  # note ".value" on each sharedctype variable
        time.sleep(0.0001)  # sleeping in order not to load CPU too excessively
        if trigger.value == 1:
            with lock:  # lock "global" variable before wrtting to it
                trigger.value = 0  # reset trigger
            tic = time.time()
            # Do a calculation that takes a significant time
            n = 10000; np.ones((n, n))*np.ones((n, n))
            with lock:
                pt.value = time.time() - tic  # calculate process time
                p_count.value += 1  # count number of finished processes
    return


if __name__ == "__main__":
    # initialize shared values (global accross processes/sharedctype).
    # Type 'i': integer, type 'd': double.
    trigger = Value('i', 0)  # used to trigger execution placed in trig_resp()
    running = Value('i', 1)  # A way to break the loop in trig_resp()
    p_count = Value('i', 0)  # process counter and flag that process is done
    pt = Value('d', 0.0)  # process time of latest finished process
    lock = Lock() # lock object used to avoid raise conditions when changing "global" values.
    p_count_old = p_count.value
    p1 = Process(target=trig_resp, args=(running, trigger, p_count, pt, lock))
    p1.start()  # Start process

    # A "simulated" sinusiodal signal
    array_len = 50
    sig_arr = np.zeros(array_len)  # Signal array
    tim_arr = np.zeros(array_len)  # Correpsonding time array
    freq = 10  # frequency of signal

    # trigger settings
    im_int = 20  # cycle interval for triggering (acquiring images)
    max_cycles = 100  # max number of cycles before stopping main

    # initializing counters etc.
    cycles = 1  # number of cycles counted
    sigSign = 1  # sign of signal gradient
    flag = 1  # used to only set trigger once for the current cycle count
    trigger_count = 0  # counts how many times a trigger has been set

    tic = time.time()
    tic2 = tic
    timeinc = np.zeros(max_cycles) # Array to keep track of time used for each main loop run
    starttime = time.time()
    while running.value == 1:
        time.sleep(0.00001)  # mimics sample time (real world signal)
        t = time.time()-starttime  # local time
        signal = np.sin(freq*t*(2.0*np.pi))  # simulated signal
        # Keeping the latest array_len values (FIFO) of t and signal.
        tim_arr[:-1] = tim_arr[1:]
        tim_arr[-1] = t
        sig_arr[:-1] = sig_arr[1:]
        sig_arr[-1] = signal

        if p_count.value == p_count_old + 1:  # process have finished
            print('Process counter: ', p_count.value,  'process_time: ', pt.value)
            p_count_old = p_count.value

        # Calculate cycle number by monotoring sign of the gradient
        sigSignOld = sigSign  # Keeping track of previous signal gradient sign
        sigSign = np.sign(sig_arr[-1]-sig_arr[-2])  # current gradient sign
        if sigSign == 1 and sigSignOld == -1:  # a local minimum just happened
            timeinc[cycles] = time.time()-tic
            cycles += 1
            print('cycles: ', cycles, ' time inc.: ', str(timeinc[cycles-1]))
            tic = time.time()
            flag = 1

        if cycles % im_int == 0 and flag == 1:
            if cycles > 0:
                if trigger_count > p_count.value:
                    print('WARNING: Process: ', p_count.value,
                          'did not finish yet. Reduce freq or increase im_int')
                trigger.value = 1
                trigger_count += 1
                print('Trigger number: ', trigger_count)
                flag = 0

        if cycles >= max_cycles:
            running.value = 0

    print('total cycle time: ', time.time()-tic2)

    # Print the process time of the last run
    if p_count.value < max_cycles//im_int:
        if p_count.value == p_count_old + 1:
            print('process counter: ', p_count.value,  'process_time: ', pt.value)
            p_count_old = p_count.value

    print('total process time: ', time.time()-tic2)

    fig = plt.figure()
    ax = plt.axes()
    plt.plot(timeinc)

我在Windows 10笔记本电脑上，因此时间安排（主while循环“正在运行...：”的每个循环中的时间增量）取决于计算机上正在发生的其他事情，但是该版本基于多处理似乎不如基于线程的敏感。但是，基于多处理的解决方案不是很优雅，我怀疑可以实现相同或更好的解决方案（更简单，更不容易犯错误）的更智能解决方案（一致的时间增量和较低的CPU负载）。

我在此处分别为“多进程”和“线程”示例附加了时间增量图：

非常感谢您对改进这两种解决方案的任何反馈。

Answer 1

您可以使用Executor。这样，您只需提交您的任务，它们就会根据您使用的执行器的类型进行处理。

我不知道您的imacq中有什么，因此您可能不得不尝试ThreadPoolExecutor和ProcessPoolExecutor来找出最适合您的应用程序的一个。

示例：

import numpy as np
import time
from concurrent.futures import ThreadPoolExecutor

def imacq():
    print('acquiring image...')
    time.sleep(1.8)
    print('saved image...')
    return

starttime = time.time()
sig_arr = np.zeros(100)
tim_arr = np.zeros(100)
image_cycles = 5
running = True
flag = True

with ThreadPoolExecutor() as executor:
    for cycles in range(1,20):
        print(cycles)
        if cycles%image_cycles == 0:
            if flag is True:
                # The function is submitted and will be processed by a 
                # a thread as soon as one is available
                executor.submit(imacq)
                flag = False
        else:
            flag = True
        time.sleep(0.4)

Answer 2

您的采集设备，数据速率和数据量的细节似乎并不十分清楚，但我的印象是，问题在于您想要尽可能快地采集一个信号并想要捕获图像并只要该信号“有趣” 就会尽快写入磁盘，但不会延迟信号的下一次采集。

因此，似乎在主信号采集过程和图像捕获过程之间需要进行最少的数据交换。恕我直言，这建议在两个进程之间进行多处理（因此没有GIL）并使用队列（没有大量数据可供腌制）。

因此，我将研究这种类型的设置：

#!/usr/bin/env python3

from multiprocessing import Process, Queue, freeze_support

def ImageCapture(queue):
    while True:
        # Wait till told to capture image - message could contain event reference number
        item = queue.get()
        if item == -1:
           break
        # Capture image and save to disk

def main():
    # Create queue to send image capture requests on
    queue = Queue(8)

    # Start image acquisition process
    p = Process(target=ImageCapture, args=(queue,))
    p.start()

    # do forever
    #    acquire from DAQ
    #    if interesting
    #       queue.put(event reference number or filename)

    # Stop image acquisition process
    queue.put(-1)
    p.join()

if __name__ == "__main__":

    # Some Windows thing
    freeze_support()
    main()

如果ImageCapture()进程无法跟上，请启动两个或多个。

在Mac上，我测量了队列中平均邮件传递时间为32微秒，并且对100万条邮件的最大延迟为120微秒。