How to define a non ISO date parser in Python?

Question

I need to parse strings like "28-05-2014_02:44:32.630" from the Pandas lib.

The following code works like a charm with standard ISO date "28-05-2014T02:44:32.630":

In [1]: import dateutil.parser

In [3]: dateutil.parser.parse("28-05-2014T02:44:32.630", dayfirst=True)
Out[3]: datetime.datetime(2014, 5, 28, 2, 44, 32, 630000)

But not with my input string:

In [4]: dateutil.parser.parse("28-05-2014_02:44:32.630", dayfirst=True)
...
ValueError: unknown string format

How can i define my own parser ?

Thk in advance !

(edit)

Here is my working code (thanks to roippi):

import pandas
from datetime import datetime

def my_date_parser(d):
    return datetime.strptime(d, '%d-%m-%Y_%H:%M:%S.%f')

i = pandas.read_csv('test.tsv', sep='\t', index_col=0, parse_dates=True, date_parser=my_date_parser)
...

roippi · Accepted Answer · 2014-09-19 15:24:44Z

3

Use datetime.strptime directly:

parseme = "28-05-2014_02:44:32.630"

from datetime import datetime

datetime.strptime(parseme, '%d-%m-%Y_%H:%M:%S.%f')
Out[34]: datetime.datetime(2014, 5, 28, 2, 44, 32, 630000)

answered Sep 19, 2014 at 15:24

roippi

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mjallday · Accepted Answer · 2014-09-19 15:25:43Z

2

Check out strptime https://docs.python.org/2/library/datetime.html#strftime-strptime-behavior

The method definition is datetime.strptime(date_string, format)

So we could use datetime.strptime('28-05-2014_02:44:32.630', '%d-%m-%Y_%H:%M:%d.f')

Not sure about the microsecond, you may been to strip that. You could use '28-05-2014_02:44:32.630'.split('.')[0] to remove it.

answered Sep 19, 2014 at 15:25

mjallday

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Kasravnd · Accepted Answer · 2014-09-19 15:28:17Z

0

you can use regex for crate your parser !

import re
def my_parser(string):
 return re.split(r':|-|_',string)

Demo:

>>> s="28-05-2014_02:44:32.630"
>>> import re
>>> s="28-05-2014_02:44:32.630"
>>> re.split(r':|-|_',s)
['28', '05', '2014', '02', '44', '32.630']

edited Sep 19, 2014 at 15:28

answered Sep 19, 2014 at 15:20

Kasravnd

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