Question

I have a binary file that I have to parse and I'm using Python. Is there a way to take 4 bytes and convert it to a single precision floating point number?

Was it helpful?

Solution

>>> import struct
>>> struct.pack('f', 3.141592654)
b'\xdb\x0fI@'
>>> struct.unpack('f', b'\xdb\x0fI@')
(3.1415927410125732,)
>>> struct.pack('4f', 1.0, 2.0, 3.0, 4.0)
'\x00\x00\x80?\x00\x00\x00@\x00\x00@@\x00\x00\x80@'

OTHER TIPS

Just a little addition, if you want the a float number as output from the unpack method instead of a tuple just write

>>> [x] = struct.unpack('f', b'\xdb\x0fI@')
>>> x
3.1415927410125732

If you have more floats then just write

>>> [x,y] = struct.unpack('ff', b'\xdb\x0fI@\x0b\x01I4')
>>> x
3.1415927410125732
>>> y
1.8719963179592014e-07
>>> 
Licensed under: CC-BY-SA with attribution
Not affiliated with StackOverflow
scroll top