Parse date and time form a csv file

Question

I've some csv files with the following format:

330913;23;2;2013;0;0;6;8;7
330914;23;2;2013;0;5;25;8;7
330915;23;2;2013;0;10;11;8;7
330916;23;2;2013;0;15;30;8;7
330917;23;2;2013;0;20;17;8;7
330918;23;2;2013;0;25;4;8;7

I read them into a pandas DataFrame and need to specify a column (say) 'dt' with the date and time. My best try so far is the following:

df = pd.read_csv( './cucu.csv', sep=';', \
    header=None, dtype='str' )
df[ 'dt' ] = pd.to_datetime(\
    df[3]+df[2]+df[1]+df[4]+df[5]+df[6], \
    format='%Y%m%d%H%M%S')

My question is, how do I do that without handling strings? I'm pretty sure I've done this in the past using something like:

df = pd.read_csv( './cucu.csv', sep=';', header=None, \
    parse_dates={'dt': [3,2,1,4,5,6]} )

but it's not working right now: I get a column dt with strings like 2013 2 23 0 0 6

What am I missing?

Answer 1

Check out the read_csv method. Specifically, the date_parser kwarg is what you are looking for. It takes the resulting string created by the parse_date columns and processes it.

df = pd.read_csv('./cucu.csv', sep=';', header=None, parse_dates={'dt': [3,2,1,4,5,6]}, date_parser=lambda dts: pd.to_datetime(dts, format='%Y %m %d %H %M %S'))