Reading integers from binary file in Python

Solution 1:

The read method returns a sequence of bytes as a string. To convert from a string byte-sequence to binary data, use the built-in struct module: http://docs.python.org/library/struct.html.

import struct

print(struct.unpack('i', fin.read(4)))

Note that unpack always returns a tuple, so struct.unpack('i', fin.read(4))[0] gives the integer value that you are after.

You should probably use the format string '<i' (< is a modifier that indicates little-endian byte-order and standard size and alignment - the default is to use the platform's byte ordering, size and alignment). According to the BMP format spec, the bytes should be written in Intel/little-endian byte order.

Solution 2:

An alternative method which does not make use of 'struct.unpack()' would be to use NumPy:

import numpy as np

f = open("file.bin", "r")
a = np.fromfile(f, dtype=np.uint32)

'dtype' represents the datatype and can be int#, uint#, float#, complex# or a user defined type. See numpy.fromfile.

Personally prefer using NumPy to work with array/matrix data as it is a lot faster than using Python lists.

Solution 3:

As of Python 3.2+, you can also accomplish this using the from_bytes native int method:

file_size = int.from_bytes(fin.read(2), byteorder='big')

Note that this function requires you to specify whether the number is encoded in big- or little-endian format, so you will have to determine the endian-ness to make sure it works correctly.

Solution 4:

Except struct you can also use array module

import array
values = array.array('l') # array of long integers
values.read(fin, 1) # read 1 integer
file_size  = values[0]