Convert array of single integer pixels to RGB triplets in Python

Question

I'm playing around with a camera for a microscope using Micro-Manager 1.4. Using the Python interface, I've managed to access the camera, change exposure time etc, and I can capture individual images.

However, each image is returned as NumPy arrays where each pixel is represented as a single integer, e.g. "7765869". As far as I can find online, this is known as a "BufferedImage" in Java, and it means that the RGB values are encoded as:

BufferedImage = R * 2^16 + G * 2^8 + B

My question is: How can I, using e.g. Numpy or OpenCV, convert this kind of array into a more handy array where each pixel is a RGB triplet of uint8 values? Needless to say, the conversion should be as efficient as possible.

To expand on the theory behind this a bit, what you have is a linear diophantine equation of the form a * x + b * y + c * z == n, where in our case a = 2**16, b = 2**8, c = 1. Typically, the solutions to these equations involve finding gcd(a, b, c) and hoping it divides n; if it doesn't, then there are no solutions. Given that a solution exists, there are infinitely many solutions. However, we are restricted in this case to find a solution 0 <= x, y, z < 256, so there will be only one solution given that gcd(a, b, c) | n. — wflynny
– wflynny, Commented Oct 17, 2013 at 17:05

Jaime · Accepted Answer · 2013-10-17 17:10:14Z

9

The easiest is to let numpy do the conversion for you. Your numpy array will probably be of type np.uint32. If you view it as an array of np.uint8, you will have an RGB0 format image, i.e. the values of R, G, and B for each pixel, plus an empty np.uint8 following. It's easy to reshape and discard that zero value:

>>> img = np.array([7765869, 16777215], dtype=np.uint32)
>>> img.view(np.uint8)
array([109, 127, 118,   0, 255, 255, 255,   0], dtype=uint8)
>>> img.view(np.uint8).reshape(img.shape+(4,))[..., :3]
array([[109, 127, 118],
       [255, 255, 255]], dtype=uint8)

Best thing is there is no calculation or copying of data, just a reinterpretation of the contents of your original image: I don't think you can get much more efficient than that!

I recall that for some operations OpenCV requires a contiguous array, so you may have to add a .copy() to the end of that expression to really get rid of the column of zeros, not simply ignore it, although this would of course trigger the copy of data that the code above had avoided.

edited Oct 17, 2013 at 17:10

answered Oct 17, 2013 at 17:02

Jaime

67.7k19 gold badges128 silver badges164 bronze badges

Sign up to request clarification or add additional context in comments.

3 Comments

Happy Over a year ago

Works perfectly! Defining "img2" as "img.view(np.uint8).reshape(img.shape+(4,))[..., :3]" and then using "cv2.imwrite" perfectly saves the image. Thanks!

Jaime Over a year ago

@Bjarke You may want to make sure that you are saving an RGB image, and not a BGR one... It will be dependent on the endianess of your system. If it is wrong, simply stick [::-1] at the end of your definition of img2 and it will reverse the order in which they get read.

Jaime Over a year ago

@tom10 Yes it's clever, although in all fairness it is someone else's cleverness: I make a living designing printer pipelines, and this kind of byte fiddling is more or less standard procedure in our (HP) products.

Happy · Accepted Answer · 2013-10-17 16:45:28Z

2

One way is

Red = BufferedImage / 2**16
Green = (BufferedImage % 2**16) / 2**8
Blue = (BufferedImage % 2**8)

However, I doubt it's the most elegant (Pythonic?) or the fastest way.

answered Oct 17, 2013 at 16:45

Happy

4693 gold badges8 silver badges20 bronze badges

Comments

Joran Beasley · Accepted Answer · 2013-10-17 16:52:49Z

0

rgbs = [((x&0xff0000)>>16,(x&0xff00)>>8,x&0xff) for x in values]

at least I think ...

afaik the formula above can also be written as

BufferedRGB = RED<<16 + GREEN << 8 + BLUE 

red,green,blue = 0xFF,0x99,0xAA
red<<16 + green << 8 + blue  #= 0xFF99AA (buffered into one value)

#apply a bitmask to get colors back
red = (0xFF99AA & 0xFF0000) >> 16 # = 0xFF
green = (0xFF99AA & 0xFF00) >> 8  # = 0x99
blue = 0xFF99AA & 0xFF          # = 0xAA

which is somewhat more readable to me and clear what is going on

edited Oct 17, 2013 at 16:52

answered Oct 17, 2013 at 16:42

Joran Beasley

114k13 gold badges167 silver badges187 bronze badges

Comments

tom10 · Accepted Answer · 2013-10-17 16:56:45Z

0

The fastest approach would probably be to keep this in numpy:

from numpy import *
x = array([211*2**16 + 11*2**8 + 7])  # test data

b, g, r = bitwise_and(x, 255), bitwise_and(right_shift(x, 8), 255), bitwise_and(right_shift, 16), 255)

print r, g, b
(array([211]), array([11]), array([7]))

answered Oct 17, 2013 at 16:56

tom10

69.5k11 gold badges133 silver badges143 bronze badges

Collectives™ on Stack Overflow

Convert array of single integer pixels to RGB triplets in Python

4 Answers 4

3 Comments

Comments

Comments

Comments

Your Answer

Hot Network Questions

Collectives™ on Stack Overflow

4 Answers 4

3 Comments

Comments

Comments

Comments

Your Answer

Sign up or log in

Post as a guest

Related