Ensuring continuously updated image file is safe to open

Question

So here's the situation: I've got two Python programs, one to control a uEye camera module, making use of the SimpleCV library, and another to do a bit of analysis on the image. The reason for them being separate is that SimpleCV is 2.7, while a few modules I need to use in the analysis stage are for 3.X only.

The camera program will continuously capture and save an image to a location (rewriting the old image), which I've timed to be around every 30 ms. The analysis program takes in an image every 100 ms or so. Now, the issue I'm concerned with is that if the analysis program tries to read in the image while the camera program happens to be writing it, it will spring an error.

I'm fairly certain placing an exception statement to catch the OSError and have it simply try again would suffice, but I feel that is a bit forceful. I've also thought about having the camera program write a number (say, 100) of images, to lesson the odds that the two will happen to be working on the same file at once, but that seems unreliable. In a perfect world, I could ditch SimpleCV and go with a 3.X module, allowing the writing and reading to happen in sequence only, but I've yet to find a suitable replacement that works with the camera.

Any thoughts on the most efficient, robust way of avoiding this issue?

Here is the (simplified) camera program:

from SimpleCV import *

cam = Camera(0)
while True:
    img = cam.getImage()
    img.save("nav.jpg")

And the important part of the analysis program:

from PIL import Image

img = Image.open("nav.jpg")

You'd open an image file in binary mode (there's no difference in UNIX, but still). — ivan_pozdeev, Aug 04 '14 at 15:53
I do have the problem of operating a uEye camera on python under windows and it doesn't really work while apparently you have successfully managed to do it. I posted this question (http://stackoverflow.com/questions/40563139/ueye-camera-with-python-on-windows) so maybe you can help out there. — ImportanceOfBeingErnest, Nov 12 '16 at 13:13

score 0 · Answer 1 · edited May 23 '17 at 12:06

The easiest way is to open the file with exclusive access so no-one can have it open for the duration of you working with it. See What is the best way to open a file for exclusive access in Python? for implementation details.

Be sure to file.close() or with <file_open> as f to close the file as soon as you can to minimize interference with the agents that "continuously update" it. Yes, and be sure to handle the file locked case in those agents.

Ensuring continuously updated image file is safe to open

1 Answers1