LED chase using python

Question

I am using a simple while loop and an array to chase LEDs on a strip.

while True:
    for i in range(nLEDs):
        R = [ 255 ] * nLEDs
        G = [ 255 ] * nLEDs
        B = [ 255 ] * nLEDs
        intensity = [ 0 ] * nLEDs
        intensity[i] = 1
        setLEDs(R, G, B, intensity)
        time.sleep(0.05)

What would be the most elegant way to chase to the end and back repeatedly (kind of like a bouncing ball)?

John La Rooy · Accepted Answer · 2015-08-12 07:16:13Z

1

Very simply, you can duplicate your for loop. Making it reversed the second time.

It appears that there is no need to redefine R, G, B over and over, so those could be moved out of the loop, but maybe you are planning to change those, so I left them in for now

while True:
    for i in range(nLEDs):
        R = [ 255 ] * nLEDs
        G = [ 255 ] * nLEDs
        B = [ 255 ] * nLEDs
        intensity = [ 0 ] * nLEDs
        intensity[i] = 1
        setLEDs(R, G, B, intensity)
        time.sleep(0.05)

    for i in reversed(range(nLEDs)):
        R = [ 255 ] * nLEDs
        G = [ 255 ] * nLEDs
        B = [ 255 ] * nLEDs
        intensity = [ 0 ] * nLEDs
        intensity[i] = 1
        setLEDs(R, G, B, intensity)
        time.sleep(0.05)

Ideally your API has a setLED function that you can call, Then you don't need to set the state of all of the LEDs when only 2 ever change at a time.

answered Aug 12, 2015 at 7:16

John La Rooy

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1 Comment

Zachary Russell Heineman Over a year ago

Thanks, makes perfect sense. As a follow up, what would be an easy way to add some physics (to make the bounce of the ball feel more real)?

Verma · Accepted Answer · 2015-08-12 07:28:55Z

0

Instead of defining intensity as a list you could define it as a collections.deque and then rotate

R = [ 255 ] * nLEDs
G = [ 255 ] * nLEDs
B = [ 255 ] * nLEDs
intensity = collections.deque([ 0 ] * nLEDs)
intensity[0] = 1
while True:
    for i in range(nLEDs):
        intensity.rotate(1)
        setLEDs(R, G, B, list(intensity))
        time.sleep(0.05)

and then add in an additional for loop to go back

answered Aug 12, 2015 at 7:28

Verma

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