将十六进制CANbus消息转换为GPS long / lat

Question

I´m looking for some help converting CAN packets from a PCAN-GPS unit to long/lat GPS coordinates.
For example i receive Data=90F98E400A0045 for longitude and the package is in the following format: imgur link

So I get the Degree and Indicator part which match my location from Google Maps but when I try to convert the hex value to a floating point I either get enormous or tiny float values that are not longitude values.

I wrote this code (and some other variations) in Python. My location is somewhere in germany ;)

s = ['E8', '5F', '8F', '40', '0A', '00', '45']
con = []
x = ""



def tohex(hex_string):
    return ''.join('{0:04b}'.format(int(c, 16)) for c in hex_string)


def long(str):
    min = str[:32]
    vorz = min[:1]
    mat = int(min[1:24], 2)
    exp = int(min[24:32], 2)
    l = (2**exp) * float("0."+mat.__str__())
    deg = str[32:48]

    print(int(deg,2))

    return l


for e in s:
    con.append(tohex(e))
    x += tohex(e)

print("%.100f" % long(x))

Answer 1

The more efficient solution is to use the struct.unpack() function.

Solution for Python 3.5.x

1- store the binary data recorded from CANbus in a bytes object.

tCanLng1=bytes([0xE8, 0x5F, 0x8F, 0x40, 0x0A, 0x00, 0x45])

2- define the format specifier of the GPS_PositionLongitude structure.

For GPS_LongitudeMinutes (float) is 'f' (4 bytes).

For GPS_LongitudeDegree (int16) is 'h' (2 bytes).

For GPS_IndicatorEW (char) is 'c' (1 byte).

End for the byte alignment (little-endian) add as first specifier '<'

3- Then perform the unpack of the CANbus binary data as follow:

>> vCanLng=struct.unpack('<fhc',tCanLng1)
>> vCanLng
(4.480457305908203, 10, b'E')

4- And the result for your position in Germany is:

GPS_LongitudeMinutes = vCanLng[0] = 4.480457305908203
GPS_LongitudeDegree = vCanLng[1] = 10
GPS_IndicatorEW = vCanLng[2] = 'E'

GPS_PositionLongitude = 10° 4.480457305908203 E = (10 + (4.480457305908203/60))° E = 10.0746742884318 ° East .