2
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This checks surrounding pixels to my object's pixel:

self.surr = [None, None, None, None, None, None, None, None]

for i in range(9):
#for x in range(-1, 2):
    #for y in range(-1, 2):
        if i != 5:
            x = i % 3 - 2
            y = int((i % 3) / 3) - 1

        if x == 0 and y == 0:
            pass
        else:
            PAI = allPixels[(self.x + x) % width][(self.y + y) % height] if allPixels[(self.x + x) % width][(self.y + y) % height] != None else None

        self.surr[(y * 3) + x] = (PAI)

return self.surr

This returns a list of length 8 that holds either a Pixel object or None. allPixels is a 2D array that also holds either a Pixel object or None. I've tried then nested loops that are commented out but they run just a bit slower that the method I'm currently using. This however, is still too slow as if there are 3000 pixels on the screen, which is the lower bounds of the total pixels there will be on the screen in the end, do the maths and you have a LOT going on every frame.

Any help somehow making this run faster/more efficiently would be much appreciated.

If you want to see the whole code, it can be found here.

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closed as off-topic by Gareth Rees, Billal Begueradj, Stephen Rauch, t3chb0t, Toby Speight May 28 '18 at 10:40

This question appears to be off-topic. The users who voted to close gave this specific reason:

  • "Code not implemented or not working as intended: Code Review is a community where programmers peer-review your working code to address issues such as security, maintainability, performance, and scalability. We require that the code be working correctly, to the best of the author's knowledge, before proceeding with a review." – Gareth Rees, Billal Begueradj, Stephen Rauch, t3chb0t
If this question can be reworded to fit the rules in the help center, please edit the question.

  • 2
    \$\begingroup\$ I assume allPi is a typo and is meant to be allPixels. Please edit. \$\endgroup\$ – vnp May 20 '18 at 17:59
  • \$\begingroup\$ @vnp Yes, it was a typo, I've edited it now \$\endgroup\$ – Thomas Ayling May 20 '18 at 18:04
  • 2
    \$\begingroup\$ Investigate: scipy-lectures.org/advanced/image_processing \$\endgroup\$ – Stephen Rauch May 20 '18 at 22:06

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