convert image to byte literal in python

Question

I\'m trying to store an image as text, so that I can do something like this example of a transparent icon for a Tk gui:

import tempfile

# byte literal code for          


        

Answer 1

Well I did figure out you can do it this way:

import tempfile, base64, io

# byte literal code for a transparent icon, I think
ICON = (b'\x00\x00\x01\x00\x01\x00\x10\x10\x00\x00\x01\x00\x08\x00h\x05\x00\x00'
        b'\x16\x00\x00\x00(\x00\x00\x00\x10\x00\x00\x00 \x00\x00\x00\x01\x00'
        b'\x08\x00\x00\x00\x00\x00@\x05\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00'
        b'\x00\x01\x00\x00\x00\x01') + b'\x00'*1282 + b'\xff'*64

# makes a temp file for the transparent icon and saves it
_, ICON_PATH = tempfile.mkstemp()
with open(ICON_PATH, 'wb') as icon_file:
    icon_file.write(ICON)

with open(ICON_PATH, 'rb') as imgFile:
    a = imgFile.read()

b = base64.b64encode(a)

print b # output does not match what ICON equals above

with open('test.png','wb') as writeFile:
    writeFile.write(b.decode('base64'))

but I still want to know how to get the same format as 'ICON = (...' at the top

Answer 2

I think you're looking for

print(repr(a))

For a as defined in your code, this will print b'\x00\x00\x01\x00\x01\x00\x10\x10\x00\x00\x01\x00\x08\x00h\x05\x00\x00\x16\x00\x00\x00(\x00\x00\x00\x10\x00\x00\x00 \x00\x00\x00\x01\x00\x08 and so on, similar to your original ICON definition, but quite large because all the \x00s and \xffs at the end are written out.

---

In your code, you have included some ad hoc compression (namely + b'\x00'*1282 + b'\xff'*64). To get compression automatically, so the ICON definition in your source file doesn't have to be so large, leverage an existing compression library, like zlib:

import zlib
print(repr(zlib.compress(a)))

On my machine, this prints 'x\x9cc``\x04B\x01\x01\x06 \xc9\xc1\x90\xc1\xca\xc0 \xc6\xc0\xc0\xa0\x01\xc4@!\x06\x05\x06\x888\x088\xb02 \x00#\x14\x8f\x82Q0\nF\xc1\x08\x05\xff)\x04\x00U\xf1A\x17', which is quite small. To decompress, use zlib.decompress:

import zlib

ICON = zlib.decompress(b'x\x9cc``\x04B\x01\x01\x06 \xc9\xc1\x90\xc1\xca\xc0 '
    b'\xc6\xc0\xc0\xa0\x01\xc4@!\x06\x05\x06\x888\x088\xb02 \x00#\x14\x8f\x82'
    b'Q0\nF\xc1\x08\x05\xff)\x04\x00U\xf1A\x17')

ICON now has the same value as in your original example.

---

If you now want a representation that's still more compact in your source file, it is time to apply base 64 encoding, which gets rid of the verbose binary encoding in python (the \x..-format).

To encode:

import base64, zlib

print(repr(base64.b64encode(zlib.compress(a))))

This gives me 'eJxjYGAEQgEBBiDJwZDBysAgxsDAoAHEQCEGBQaIOAg4sDIgACMUj4JRMApGwQgF/ykEAFXxQRc='

To decode:

import base64, zlib

ICON = zlib.decompress(base64.b64decode('eJxjYGAEQgEBBiDJwZDBy'
    'sAgxsDAoAHEQCEGBQaIOAg4sDIgACMUj4JRMApGwQgF/ykEAFXxQRc='))

And again, ICON has the same value as originally specified.

---

The final strategy as presented is good for ico files. I see that you also mention png files. These already have compression applied, so you should probably prefer to use only base 64 encoding:

import base64

print(base64.b64encode(png_icon))

and

PNG_ICON = base64.b64decode( ** insert literal here ** )

---

As it turns out, these encodings are also available through the str.encode and str.decode APIs. This lets you off without writing the imports. For completeness, here they are:

Encoding:

print(repr(a.encode('zlib').encode('base64')))

Decoding:

ICON = ('eJxjYGAEQgEBBiDJwZDBysAgxsDAoAHEQCEGBQaIOAg4sDIgACMUj4J'
    'RMApGwQgF/ykEAFXxQRc=').decode('base64').decode('zlib')

Answer 3

I think you're just not printing out the data properly — there doesn't seem to be any need to mess around with base64 doing this.

Here's proof:

from itertools import izip
import tempfile

# byte literal code for a transparent icon, I think
ICON = (b'\x00\x00\x01\x00\x01\x00\x10\x10\x00\x00\x01\x00\x08\x00h\x05\x00\x00'
        b'\x16\x00\x00\x00(\x00\x00\x00\x10\x00\x00\x00 \x00\x00\x00\x01\x00'
        b'\x08\x00\x00\x00\x00\x00@\x05\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00'
        b'\x00\x01\x00\x00\x00\x01') + b'\x00'*1282 + b'\xff'*64

# make a temp file from ICON data for testing
_, ICON_PATH = tempfile.mkstemp()
with open(ICON_PATH, 'wb') as icon_file:
    icon_file.write(ICON)

# Convert raw data in the file into a valid Python string literal.

# helper function
def grouper(n, seq):
    "s -> (s0,s1,...sn-1), (sn,sn+1,...s2n-1), (s2n,s2n+1,...s3n-1), ..."
    for i in xrange(0, len(seq), n):
        yield seq[i:i+n]

# read data file in binary mode
a = open(ICON_PATH, 'rb').read()

# create Python code to define string literal
code = '\n'.join(['ICON2 = ('] +
                 ['    '+repr(group) for group in grouper(16, a)] +
                 [')'])

print 'len(ICON): {}'.format(len(ICON))
print 'len(a): {}'.format(len(a))
print code
exec(code)
print
print 'len(ICON2): {}'.format(len(ICON2))
print 'ICON2 == ICON: {}'.format(ICON2 == ICON)

Output:

len(ICON): 1406
len(a): 1406
ICON2 = (
    '\x00\x00\x01\x00\x01\x00\x10\x10\x00\x00\x01\x00\x08\x00h\x05'
    '\x00\x00\x16\x00\x00\x00(\x00\x00\x00\x10\x00\x00\x00 \x00'
    '\x00\x00\x01\x00\x08\x00\x00\x00\x00\x00@\x05\x00\x00\x00\x00'
    '\x00\x00\x00\x00\x00\x00\x00\x01\x00\x00\x00\x01\x00\x00\x00\x00'
    '\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00'
       ...
    '\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00'
    '\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00'
    '\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\xff\xff'
    '\xff\xff\xff\xff\xff\xff\xff\xff\xff\xff\xff\xff\xff\xff\xff\xff'
    '\xff\xff\xff\xff\xff\xff\xff\xff\xff\xff\xff\xff\xff\xff\xff\xff'
    '\xff\xff\xff\xff\xff\xff\xff\xff\xff\xff\xff\xff\xff\xff\xff\xff'
    '\xff\xff\xff\xff\xff\xff\xff\xff\xff\xff\xff\xff\xff\xff'
)

len(ICON2): 1406
ICON2 == ICON: True