I'm trying to read a BMP file in Python. I know the first two bytes indicate the BMP firm. The next 4 bytes are the file size. When I execute:
fin = open("hi.bmp", "rb")
firm = fin.read(2)
file_size = int(fin.read(4))
I get:
ValueError: invalid literal for int() with base 10: 'F#\x13'
What I want to do is reading those four bytes as an integer, but it seems Python is reading them as characters and returning a string, which cannot be converted to an integer. How can I do this correctly?
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.
i = struct.unpack(...)[0] I often write i, = struct.unpack(...)
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.
a = np.fromfile('file.bin', dtype=np.uint32)
np.fromfile( file, dtype='>i2') , > or < determine big or little endian. Depending on the number of byte you can go with i2 or i4
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.
int's are dynamically sized (it's the same implementation as Python2's long; this is also sometimes referred to as "big int's"), which I believe is the motivation for adding this int method in the first place. docs.python.org/3/library/….
n and it'd work all the sameExcept struct you can also use array module
import array
values = array.array('l') # array of long integers
values.fromfile(fin, 1) # read 1 integer
file_size = values[0]
array is an efficient way to read a binary file but not very flexible when we have to deal with structure, as you correctly mentioned.As you are reading the binary file, you need to unpack it into a integer, so use struct module for that
import struct
fin = open("hi.bmp", "rb")
firm = fin.read(2)
file_size, = struct.unpack("i",fin.read(4))
When you read from a binary file, a data type called bytes is used. This is a bit like list or tuple, except it can only store integers from 0 to 255.
Try:
file_size = fin.read(4)
file_size0 = file_size[0]
file_size1 = file_size[1]
file_size2 = file_size[2]
file_size3 = file_size[3]
Or:
file_size = list(fin.read(4))
Instead of:
file_size = int(fin.read(4))
Here's a late solution but I though it might help.
fin = open("hi.bmp", "rb")
firm = fin.read(2)
file_size = 0
for _ in range(4):
(file_size << 8) += ord(fin.read(1))
