Python - reading 10 bit integers from a binary file

Question

I have a binary file containing a stream of 10-bit integers. I want to read it and store the values in a list.

It is working with the following code, which reads my_file and fills pixels with integer values:

file = open("my_file", "rb")

pixels = []
new10bitsByte = ""

try:
    byte = file.read(1)
    while byte:
        bits = bin(ord(byte))[2:].rjust(8, '0')
        for bit in reversed(bits):
            new10bitsByte += bit
            if len(new10bitsByte) == 10:
                pixels.append(int(new10bitsByte[::-1], 2))
                new10bitsByte = ""             
    byte = file.read(1)

finally:
    file.close()

It doesn't seem very elegant to read the bytes into bits, and read it back into "10-bit" bytes. Is there a better way to do it?

With 8 or 16 bit integers I could just use file.read(size) and convert the result to an int directly. But here, as each value is stored in 1.25 bytes, I would need something like file.read(1.25)...

Answer 1

Here's a generator that does the bit operations without using text string conversions. Hopefully, it's a little more efficient. :)

To test it, I write all the numbers in range(1024) to a BytesIO stream, which behaves like a binary file.

from io import BytesIO

def tenbitread(f):
    ''' Generate 10 bit (unsigned) integers from a binary file '''
    while True:
        b = f.read(5)
        if len(b) == 0:
            break
        n = int.from_bytes(b, 'big')

        #Split n into 4 10 bit integers
        t = []
        for i in range(4):
            t.append(n & 0x3ff)
            n >>= 10
        yield from reversed(t)

# Make some test data: all the integers in range(1024),
# and save it to a byte stream
buff = BytesIO()

maxi = 1024
n = 0
for i in range(maxi):
    n = (n << 10) | i
    #Convert the 40 bit integer to 5 bytes & write them
    if i % 4 == 3:
        buff.write(n.to_bytes(5, 'big'))
        n = 0

# Rewind the stream so we can read from it
buff.seek(0)

# Read the data in 10 bit chunks
a = list(tenbitread(buff))

# Check it 
print(a == list(range(maxi)))    

output

True

---

Doing list(tenbitread(buff)) is the simplest way to turn the generator output into a list, but you can easily iterate over the values instead, eg

for v in tenbitread(buff):

or

for i, v in enumerate(tenbitread(buff)):

if you want indices as well as the data values.

---

Here's a little-endian version of the generator which gives the same results as your code.

def tenbitread(f):
    ''' Generate 10 bit (unsigned) integers from a binary file '''
    while True:
        b = f.read(5)
        if not len(b):
            break
        n = int.from_bytes(b, 'little')

        #Split n into 4 10 bit integers
        for i in range(4):
            yield n & 0x3ff
            n >>= 10

We can improve this version slightly by "un-rolling" that for loop, which lets us get rid of the final masking and shifting operations.

def tenbitread(f):
    ''' Generate 10 bit (unsigned) integers from a binary file '''
    while True:
        b = f.read(5)
        if not len(b):
            break
        n = int.from_bytes(b, 'little')

        #Split n into 4 10 bit integers
        yield n & 0x3ff
        n >>= 10
        yield n & 0x3ff
        n >>= 10
        yield n & 0x3ff
        n >>= 10
        yield n 

This should give a little more speed...

Answer 2

As there is no direct way to read a file x-bit by x-bit in Python, we have to read it byte by byte. Following MisterMiyagi and PM 2Ring's suggestions I modified my code to read the file by 5 byte chunks (i.e. 40 bits) and then split the resulting string into 4 10-bit numbers, instead of looping over the bits individually. It turned out to be twice as fast as my previous code.

file = open("my_file", "rb")

pixels = []
exit_loop = False

try:
    while not exit_loop:
        # Read 5 consecutive bytes into fiveBytesString
        fiveBytesString = ""
        for i in range(5):
            byte = file.read(1)
            if not byte:
                exit_loop = True
                break
            byteString = format(ord(byte), '08b')
            fiveBytesString += byteString[::-1]
        # Split fiveBytesString into 4 10-bit numbers, and add them to pixels
        pixels.extend([int(fiveBytesString[i:i+10][::-1], 2) for i in range(0, 40, 10) if len(fiveBytesString[i:i+10]) > 0])

finally:
    file.close()

Answer 3

Adding a Numpy based solution suitable for unpacking large 10-bit packed byte buffers like the ones you might receive from AVT and FLIR cameras.

This is a 10-bit version of @cyrilgaudefroy's answer to a similar question; there you can also find a Numba alternative capable of yielding an additional speed increase.

import numpy as np

def read_uint10(byte_buf):
    data = np.frombuffer(byte_buf, dtype=np.uint8)
    # 5 bytes contain 4 10-bit pixels (5x8 == 4x10)
    b1, b2, b3, b4, b5 = np.reshape(data, (data.shape[0]//5, 5)).astype(np.uint16).T
    o1 = (b1 << 2) + (b2 >> 6)
    o2 = ((b2 % 64) << 4) + (b3 >> 4)
    o3 = ((b3 % 16) << 6) + (b4 >> 2)
    o4 = ((b4 % 4) << 8) + b5

    unpacked =  np.reshape(np.concatenate((o1[:, None], o2[:, None], o3[:, None], o4[:, None]), axis=1),  4*o1.shape[0])
    return unpacked

Reshape can be omitted if returning a buffer instead of a Numpy array:

unpacked =  np.concatenate((o1[:, None], o2[:, None], o3[:, None], o4[:, None]), axis=1).tobytes()

Or if image dimensions are known it can be reshaped directly, e.g.:

unpacked =  np.reshape(np.concatenate((o1[:, None], o2[:, None], o3[:, None], o4[:, None]), axis=1), (1024, 1024))

If the use of the modulus operator appears confusing, try playing around with:

np.unpackbits(np.array([255%64], dtype=np.uint8))

Edit: It turns out that the Allied Vision Mako-U cameras employ a different ordering than the one I originally suggested above:

o1 = ((b2 % 4) << 8) + b1
o2 = ((b3 % 16) << 6) + (b2 >> 2)
o3 = ((b4 % 64) << 4) + (b3 >> 4)
o4 = (b5 << 2) + (b4 >> 6)

So you might have to test different orders if images come out looking wonky initially for your specific setup.