Python Multithreading an Image-Rendering

Question

I am generating rather large images, and so I want to thread it to render different parts of the image at once.

for startX in xrange(0, width, threadChunk):
    t = Thread(target = threaded_chunk, args = (thread_num, startX, startX + threadChunk, 1, 0, height, 1))
    t.start()
    t.join()
    print 'next thread'
    thread_num += 1

However, this doesn't seem to do what I think it should --- it seems to me that it waits for one thread to finish before going onto the next one. I found this out because my actual threaded function (threaded_chunk), has a print statement at the end, which notifies me that the thread has finished. The shell looks like this when I run the program:

Thread 1 completed
next thread
Thread 2 completed
next thread
# etc

Therefore, it is waiting for one to finish before running the next.

I am using t.join() because I want to execute a final step after all of the threads have finished: image.save(imagepath).

Is there a way to run numerous threads at once but then wait until all of the threads have finished to execute a task? Thanks!

EDIT: My current method works on Ubuntu 12.04 LTS, but not Windows 8.1.

If I'm not mistaken, based on answers to a similar question I had, I think in order for the multiple rendering threads to run simultaneously you have to make sure that the function you're sending to be threaded calls C code directly. That is, it should call PIL.Image.polygon (or whatever rendering module/operation you're using) instead of threadedchunk which I assume is a function you defined on your own using Python code. I could be wrong though, so curious to hear your results. — Karim Bahgat
– Karim Bahgat, Commented Jan 18, 2014 at 19:45

BartoszKP · Accepted Answer · 2014-01-18 17:59:16Z

2

You should first start all threads, and then wait for them. You can do it like this:

threads = []
for startX in xrange(0, width, threadChunk):
    t = Thread(target = threaded_chunk, args = (thread_num, startX, startX + threadChunk, 1, 0, height, 1))
    threads.append(t)
    print 'next thread'

[t.start() for t in threads]
[t.join() for t in threads]

answered Jan 18, 2014 at 17:59

BartoszKP

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