pandas python pandas的转换器
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converters for python pandas
提问by tomasz74
I have a .txt data where columns 6 and 7 are GPS position in the form:
我有一个 .txt 数据，其中第 6 列和第 7 列是以下形式的 GPS 位置：
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
当我通过 read_csv 读取它时，我尝试使用转换器将其转换为笛卡尔坐标。我为转换器编写了函数
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:
当我在单个值上使用它时，它的工作方式是我想要的：
convertFunc(myData.Lat[1])
Out [159]: 55.187110250000003
when I try to use it in read_csv it does not work
当我尝试在 read_csv 中使用它时它不起作用
myData = DataFrame(read_csv('~/data.txt', sep=',' names=['A', 'B', 'C', 'D', 'E', 'Lat', 'Long'],converters=convert))
I have an error:
我有一个错误：
...
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?
我不知道哪里做错了或者我对转换器有什么误解？或者也许有人知道以这种形式处理 GPS 数据的好方法（包）？
(I think it can be some problem with lambdaWhen I want to apply my function to the column I have an error: TypeError: only length-1 arrays can be converted to Python scalars)
（我认为这可能是有些问题，lambda当我想要我的功能应用到专栏中，我有一个错误：TypeError: only length-1 arrays can be converted to Python scalars）
回答by nneonneo
That converter is a bit hacky; might I recommend something more robust like this?
那个转换器有点笨拙；我可以推荐一些像这样更强大的东西吗？
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?
此外，该错误表明您正在尝试转换显然不是 GPS 字符串的内容——也许是标题行？
回答by Wouter Overmeire
Your converter is not ok.
你的转换器不行。
In [67]: convertFunc = lambda x : float((x[0:5]+x[6:12]).replace(';','.'))
In [68]: convertFunc('4;07.7693770E')
---------------------------------------------------------------------------
ValueError Traceback (most recent call last)
...
ValueError: invalid literal for float(): 4.07.693770
On top of a dodgy converter, i think you apply the converter to the wrong column (look at the exception you get).
在狡猾的转换器之上，我认为您将转换器应用于错误的列（看看您得到的异常）。
