Reading a binary file with python

Read the binary file content like this:

with open(fileName, mode='rb') as file: # b is important -> binary    fileContent = file.read()

then "unpack" binary data using struct.unpack:

The start bytes: struct.unpack("iiiii", fileContent[:20])

The body: ignore the heading bytes and the trailing byte (= 24); The remaining part forms the body, to know the number of bytes in the body do an integer division by 4; The obtained quotient is multiplied by the string 'i' to create the correct format for the unpack method:

struct.unpack("i" * ((len(fileContent) -24) // 4), fileContent[20:-4])

The end byte: struct.unpack("i", fileContent[-4:])

In general, I would recommend that you look into using Python's struct module for this. It's standard with Python, and it should be easy to translate your question's specification into a formatting string suitable for struct.unpack().

Do note that if there's "invisible" padding between/around the fields, you will need to figure that out and include it in the unpack() call, or you will read the wrong bits.

Reading the contents of the file in order to have something to unpack is pretty trivial:

import structdata = open("from_fortran.bin", "rb").read()(eight, N) = struct.unpack("@II", data)

This unpacks the first two fields, assuming they start at the very beginning of the file (no padding or extraneous data), and also assuming native byte-order (the @ symbol). The Is in the formatting string mean "unsigned integer, 32 bits".

To read a binary file to a bytes object:

from pathlib import Pathdata = Path('/path/to/file').read_bytes()  # Python 3.5+

To create an int from bytes 0-3 of the data:

i = int.from_bytes(data[:4], byteorder='little', signed=False)

To unpack multiple ints from the data:

import structints = struct.unpack('iiii', data[:16])

• pathlib
• int.from_bytes()
• struct