How to flip a byte in python?
I am trying to decode a file that is a set of backward bytes. Currently my code reads and copies the file, but I want to edit some of the bytes before I write out a copy. After printing a binary file in string format, it looks like this:
& r \ x00 \ x00 \ x00 \ x06 \ x00P \ x00 \ x14 \ x00 \ x10 \ x00 \ x0e \ x00P \ x00 \ x15 \ x00) 7 \ xf8y (= (\ XDB (\ x8e \ x08 \ x00
... etc. I would like to flip the bytes that count as \ x ** like this:
\ x01 → \ x10, \ x81 → \ x18, \ x40 → \ x04, \ xae → \ xea
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You want to swap the first 4 bits for the last 4 bits of a byte. Just rebuild the array bytes
in a list comprehension with some modifications and masking:
>>> b = b'r\x00\x00\x00\x06\x00P\x00\x14\x00\x10\x00\x0e\x00P\x00\x15\x00)7\xf8y(=(\xdb(\x8e\x08\x00'
>>> bytes(((x<<4 & 0xF0) + (x >> 4)) for x in b)
b"'\x00\x00\x00`\x00\x05\x00A\x00\x01\x00\xe0\x00\x05\x00Q\x00\x92s\x8f\x97\x82\xd3\x82\xbd\x82\xe8\x80\x00"
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Your question needs some clarification. You are replacing nibbles in string or binary. If in binary you want to also swap lumps on printable characters (e.g. for the ... \ x00 part ) 7 \ xf8 y (= ( \ xdb ..., bold characters not in \ x ?? format, so they should change pieces?)?
If you want to swap to string representation and only to \ x ?? parts, then you can use a regular expression like this:
import re
preswap = <set to your string>
swapped = re.sub(r'(\\x)(.)(.)',r'\1\3\2',preswap)
This creates 3 consecutive match elements: '\ x', any single char (first), any single char (second). When this pattern is found, it is swapped to match items 1, then 3 (second single char), then 2 (first single char), and the result is in the swapped variable.
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