Python: Read hex from file into list?

Question

Is there a simple way to, in Python, read a file's hexadecimal data into a list, say hex?

So hex would be this:

hex = ['AA','CD','FF','0F']

I don't want to have to read into a string, then split. This is memory intensive for large files.

Answer 1

s = "Hello"
hex_list = ["{:02x}".format(ord(c)) for c in s]

Output

['48', '65', '6c', '6c', '6f']

Just change s to open(filename).read() and you should be good.

with open('/path/to/some/file', 'r') as fp:
    hex_list = ["{:02x}".format(ord(c)) for c in fp.read()]

---

Or, if you do not want to keep the whole list in memory at once for large files.

hex_list = ("{:02x}".format(ord(c)) for c in fp.read())

and to get the values, keep calling

next(hex_list)

to get all the remaining values from the generator

list(hex_list)

Answer 2

Using Python 3, let's assume the input file contains the sample bytes you show. For example, we can create it like this

>>> inp = bytes((170,12*16+13,255,15)) # i.e. b'\xaa\xcd\xff\x0f'
>>> with open(filename,'wb') as f:
...     f.write(inp)

Now, given we want the hex representation of each byte in the input file, it would be nice to open the file in binary mode, without trying to interpret its contents as characters/strings (or we might trip on the error UnicodeDecodeError: 'utf-8' codec can't decode byte 0xaa in position 0: invalid start byte)

>>> with open(filename,'rb') as f:
...     buff = f.read() # it reads the whole file into memory
...
>>> buff
b'\xaa\xcd\xff\x0f'
>>> out_hex = ['{:02X}'.format(b) for b in buff]
>>> out_hex
['AA', 'CD', 'FF', '0F']

If the file is large, we might want to read one character at a time or in chunks. For that purpose I recommend to read this Q&A