9
\$\begingroup\$

I've done a script that will write things as images, which can then be read elsewhere (eg. you could share an mp3 through imgur). This is the first one I've done where I've been trying to improve my writing style and make stuff run as efficient as possible, so if there is anything you see here that I can improve it'd be really helpful to know.

For the record, this is an old version, but after this part it got a lot more complicated. I kept the same writing style anyway, so this is a lot easier to understand. This version will make a grey noisy image, but I added some code to use custom images after (which is where it got more complicated).

from PIL import Image
import cPickle, base64

class ImageStore:

    def __init__( self, imageName="ImageDataStore" ):
        self.imageDataPadding = [116, 64, 84, 123, 93, 73, 106]
        self.imageName = imageName

    def __repr__( self ):
        return "Use this to store or read data from an image.\nUsage:\n - ImageStore().write(input), ImageStore().read()\nYou can also define the name and location of the image.\n - ImageStore( 'C:\filename )'"

    def write( self, input, widthRatio=0.52 ):

        encodedData = base64.b64encode( cPickle.dumps( input ) )
        pixelData = [int( format( ord( letter ) ) ) for letter in encodedData]
        pixelData += imageDataPadding
        #Pad to end with multiple of 3
        for i in xrange( 3-len( pixelData ) ):
            rawData += [255]

        #Set image info
        minimumWidth = 8
        minimumHeight = 8
        currentPixels = len( pixelData )/3

        #Calculate width and height
        if currentPixels <= minimumWidth*minimumHeight:
            #Set to minimum values
            width = minimumWidth
            height = minimumHeight
        else:
            #Calculate based on ratio
            width = int( round( pow( currentPixels, ratioWidth ), -1 ) )
            #Make sure it is divisable by 3
            width /= 3
            width *= 3
            #Note: height is now calculated to be as small as possible while still containing all the data, this is just the old way
            height = int( round( pow( width, 1/( 1/( 1-ratioWidth )-1 ) ), -1 ) )

        #Draw image
        imageOutput = Image.new("RGB", ( width, height ) )
        imageData = imageOutput.load()

        #Assign pixel colours
        numbersToAddIncrement = 0
        for y in range( height ):
            for x in range( width ):
                try:
                    for i in xrange( 3 ):
                        dataRGB[i] = inputData[numbersToAddIncrement]
                        numbersToAddIncrement += 1
                except:
                    for i in xrange( 3 ):
                        dataRGB[i] = rd.randint( 52, 128 )

                dataRGB = [number[1] for number in dataRGB.items()]

                imageData[x,y] = tuple( dataRGB )

        #Save image
        imageOutput.save( str( self.imageName ) + ".png", "PNG" )
        return "Saved file: " + str( self.imageName ) + ".png"

    def read( self ):

        #Read image
        try:
            imageInput = Image.open( str( self.imageName ) + ".png" )
        except:
            return "No image found."

        #Store pixel info
        rawData = []
        for pixels in imageInput.getdata():
            for rgb in xrange( 3 ):
                rawData.append( pixels[rgb] )

        #Truncate end of file
        try:
            for i in xrange( len( rawData ) ):
                j = 0
                while rawData[i+j] == imageDataPadding[j]:
                    j += 1
                    if j == len( imageDataPadding ):
                        rawData = rawData[0:i]
                        break
                if j == len( imageDataPadding ):
                    break
        except:
            print "File is probably corrupted."

        #Decode data    
        encodedData = "".join( [chr( pixel ) for pixel in rawData] )
        outputData = cPickle.loads( base64.b64decode( encodedData ) )

        return outputData

Also, with custom images, I'm not quite happy with the way it works. Basically, I modify the pixels on an image to store the data, and to read it, you compare the modified image to the original one. Could anyone think of a way that wouldn't require the original image?

And here's the whole thing in it's current state (despite my best efforts it got messy) - http://pastebin.com/e1VUTPb5

\$\endgroup\$

1 Answer 1

7
\$\begingroup\$
  1. First off, you seem very confused with your whitespace. In Python, you shouldn't have "padding" around the contents of braces, {[()]}. For example, the line, xrange( 3-len( pixel_data ) ) would become xrange(3 - len(pixelData)).
  2. Secondly, you need whitespace between your operators. For example, len( pixelData )/3 would become len(pixel_data) / 3.
  3. Your naming violates the official style guidelines from PEP8. Variables and functions should be in snake_case, and classes should be in PascalCase.
  4. Never use except without knowing which specific error to catch. There could easily be an error that goes un-noticed because you weren't trying to catch a specific error.
  5. Finally, the code below the comment #Store pixel info, in the function read, can be shortened to this:

    rawData = [[pixels[rgb] for rgb in xrange(3)] for pixels in imageInput.getdata()]
    
\$\endgroup\$
1
  • \$\begingroup\$ Thanks, I've recently tried my best to sort out 1, 3 and 4, and with the whitespace between operators, I try to not use it unless it's a larger calculation where spaces make it easy to understand the different sections (simple example would be like (input**2)/5 + input/2 as opposed to input ** 2 / 5 + input / 2). It did dawn on me the #Store pixel info part seemed a bit clunky, so thanks haha. The code got up to 3500 lines with no documentation though, so instead of going through and updating it all, I'll probably try rewrite it from scratch :) \$\endgroup\$
    – Peter
    Jun 30, 2015 at 8:13

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.