Float to bytearray conversion for sigfox payload

[robert-hh]

@tttadam You use struct.pack() to put these two floats into an 8 byte array without converting them to a string. That results in the shortest possible message, not readable for human eyes withou doing an unpack.

import ustruct
lat =47.497132 
lon =19.063296
msg = ustruct.pack("ff", lat, lon)
# ..
# and unpack
lat, lon = ustruct.unpack("ff", msg)

[tttadam]

@robert-hh said in Float to bytearray conversion for sigfox payload:

ustruct.unpack("ff", msg)

Nice, thank you.
I think is still posible save some more space if cut the first caracter and the dot. (And then possible to include more information)
Which I try and fail to doing here. I think the unpacked value is rounded somehow.

>>> lat
47.49713
>>> lon
19.0633
>>> msg2 =int((str(lat*100000)[1:-2]+str(lon*100000)[1:-2]))
>>> msg = ustruct.pack("ff", msg2)
>>> ret = ustruct.unpack("ff", msg)
>>> print(ret)
(7.497139e+11, 0.0)

[jcaron]

@tttadam What you are doing is covering two floats to strings, concatenating them, converting the total to an int, packing that in as a float, and unpacking.

Forget about str. Just use pack directly with the original values: ustruct.pack("ff",lat,lon)

Or if you want to avoid sending floats over the air, you may convert them to ints and pack than as such instead.

[robert-hh]

@tttadam msg2 contains an single int. So you must not use "ff" for pack, since that tells, that two float numbers are supplied. You have to use "i", which will pack (and unpack) it into a 4 byte format.
Please have a look at the documentation of struct: https://docs.python.org/3/library/struct.html

[tttadam]

@robert-hh said in Float to bytearray conversion for sigfox payload:

I see now. Okay. But I think I am still missing some information. Because I can't get back the same number.

>>> lat
47.49713
>>> lon
19.0633
>>> msg2 =int(str(lat*100000)[1:-2]+str(lon*100000)[1:-2])
>>> msg2
749713906330
>>> msg = ustruct.pack("i", msg2)
>>> msg
b'\x9aZn\x8e'
>>> ret = ustruct.unpack("i", msg)
>>> ret
(-1905370470,)

[robert-hh]

@tttadam The numer is too long for the i format, then you have to use the "q" format (long long). But that results in 8 bytes, and then you can also directly pack the two floats and have no further need for reformatting or scaling (see my post below).
Using the "q" format, you could eventually drop the last three bytes, they are zero, and add them before unpacking. But that does not seem to be worth the effort.

[robert-hh]

@tttadam Or you could split the number:

>>> msg2
749713906330
>>> msg = ustruct.pack("IB", msg2 // 256, msg2 % 256)
>>> msg
b'Zn\x8e\xae\x9a'
>>> msb,lsb = struct.unpack("IB", msg)
>>> ret=msb * 256 + lsb
>>> ret
749713906330

But still that is overkill compared to the float packing, unless you fight for every single byte.

[tttadam]

@robert-hh So basicly your implamentation is the best. :)
Thank you.

[jcaron]

@tttadam why do you keep doing the same stuff you have already been told twice is not correct?

You have been told twice that the solution is very simply ustruct.pack(“ff”,lat,long). What’s wrong with that? Stop with the str and the concatenation and all that. Just pack the two values, and you’re done.

[robert-hh]

@jcaron I think that this was @tttadam's conclusion.