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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?
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Floating-point numbers use the IEEE (Institute of Electrical and Electronics Engineers) format. Single-precision values with float type have 4 bytes, consisting of a sign bit, an 8-bit excess-127 binary exponent, and a 23-bit mantissa.
4-byte float. float(18) defines a floating point type with at least 18 binary digits of precision in the mantissa. A 4‑byte floating point field is allocated for it, which has 23 bits of precision.
A byte value can be interchanged to an int value using the int. from_bytes() function. The int. from_bytes() function takes bytes, byteorder, signed, * as parameters and returns the integer represented by the given array of bytes.
>>> 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@' Just a little addition, if you want 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 >>> If you love us? You can donate to us via Paypal or buy me a coffee so we can maintain and grow! Thank you!
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