Question

I have a .txt data where columns 6 and 7 are GPS position in the form:

50;18.5701400N,4;07.7693770E

When I read it by read_csv I try to convert it to cartesian coordinates by using converters. I wrote the function for converter

convertFunc = lambda x : float((x[0:5]+x[6:12]).replace(';','.'))
convert = {6:convertFunc,7:convertFunc}

when I use it on single value it works how I would like:

convertFunc(myData.Lat[1])
Out [159]:  55.187110250000003

when I try to use it in read_csv it does not work

myData = DataFrame(read_csv('~/data.txt', sep=',' names=['A', 'B', 'C', 'D', 'E', 'Lat', 'Long'],converters=convert))

I have an error:

...
convertFunc = lambda x : float((x[0:5] + x[6:12]).replace(';', '.'))
ValueError: invalid literal for float(): DGPS ongitu

I don’t know where do it wrong or what I misunderstand in converters?
Or maybe anyone knows good way (package) to work with GPS data in that form?

(I think it can be some problem with lambda When I want to apply my function to the column I have an error: TypeError: only length-1 arrays can be converted to Python scalars)

Answer 1

That converter is a bit hacky; might I recommend something more robust like this?

def convert_dmds(s):
    deg, min = s[:-1].split(';')
    sign = 1 if s[-1] in 'NE' else -1
    return sign * (float(deg) + float(min) / 60.0)

def convert_gps(s):
    lat, lon = s.split(',')
    return (convert_dmds(lat), convert_dmds(lon))

Also, the error indicates that you are trying to convert something that is clearly not a GPS string — a header row, perhaps?