Python 'x days ago' to datetime

Question

I have lines that show a date in the following format:

x minutes/hours/days/months/years ago 

I need to parse this on datetime using python.

It seems datutil cannot do this.

Is there any way to do this?

Answer 1

Of course you can do it. You just need a timedelta .

 s = "3 days ago" parsed_s = [s.split()[:2]] time_dict = dict((fmt,float(amount)) for amount,fmt in parsed_s) dt = datetime.timedelta(**time_dict) past_time = datetime.datetime.now() - dt 

As an aside, it looks like a dateutil has a relativedelta , which acts like a timedelta, but the constructor also takes months and years in the arguments (and, apparently, the arguments should be integer).

Answer 2

This can be easily done with timedelta s:

 import datetime def string_to_delta(string_delta): value, unit, _ = string_delta.split() return datetime.timedelta(**{unit: float(value)}) 

Production:

 >>> string_to_delta("20 hours ago") datetime.timedelta(0, 72000) 

Although this will require additional work to work with months / years, since adding a month to a date is an ambiguous operation, it should be a simple addition if you know what you want it to mean.

To get the actual time, just distract the delta from datetime.datetime.now() .

Answer 3

Since your arguments are something like 2 days ago, 3 months ago, 2 years ago. The function below can help in getting the exact date for the arguments. First you need to import the following date utilities

 import datetime from dateutil.relativedelta import relativedelta 

Then we implement the function below

 def get_past_date(str_days_ago): TODAY = datetime.date.today() splitted = str_days_ago.split() if len(splitted) == 1 and splitted[0].lower() == 'today': return str(TODAY.isoformat()) elif len(splitted) == 1 and splitted[0].lower() == 'yesterday': date = TODAY - relativedelta(days=1) return str(date.isoformat()) elif splitted[1].lower() in ['hour', 'hours', 'hr', 'hrs', 'h']: date = datetime.datetime.now() - relativedelta(hours=int(splitted[0])) return str(date.date().isoformat()) elif splitted[1].lower() in ['day', 'days', 'd']: date = TODAY - relativedelta(days=int(splitted[0])) return str(date.isoformat()) elif splitted[1].lower() in ['wk', 'wks', 'week', 'weeks', 'w']: date = TODAY - relativedelta(weeks=int(splitted[0])) return str(date.isoformat()) elif splitted[1].lower() in ['mon', 'mons', 'month', 'months', 'm']: date = TODAY - relativedelta(months=int(splitted[0])) return str(date.isoformat()) elif splitted[1].lower() in ['yrs', 'yr', 'years', 'year', 'y']: date = TODAY - relativedelta(years=int(splitted[0])) return str(date.isoformat()) else: return "Wrong Argument format" 

You can call the function as follows:

 print get_past_date('5 hours ago') print get_past_date('yesterday') print get_past_date('3 days ago') print get_past_date('4 months ago') print get_past_date('2 years ago') print get_past_date('today') 

Answer 4

a completely exaggerated solution, but I needed something more flexible:

 def string_to_delta(relative): #using simplistic year (no leap months are 30 days long. #WARNING: 12 months != 1 year unit_mapping = [('mic', 'microseconds', 1), ('millis', 'microseconds', 1000), ('sec', 'seconds', 1), ('day', 'days', 1), ('week', 'days', 7), ('mon', 'days', 30), ('year', 'days', 365)] try: tokens = relative.lower().split(' ') past = False if tokens[-1] == 'ago': past = True tokens = tokens[:-1] elif tokens[0] == 'in': tokens = tokens[1:] units = dict(days = 0, seconds = 0, microseconds = 0) #we should always get pairs, if not we let this die and throw an exception while len(tokens) > 0: value = tokens.pop(0) if value == 'and': #just skip this token continue else: value = float(value) unit = tokens.pop(0) for match, time_unit, time_constant in unit_mapping: if unit.startswith(match): units[time_unit] += value * time_constant return datetime.timedelta(**units), past except Exception as e: raise ValueError("Don't know how to parse %s: %s" % (relative, e)) 

It can analyze things like:

• 2 days ago
• in 60 seconds
• 2 DAY and 4 Secs
• in 1 year, 1 Month, 2 days and 4 MICRO
• 2 Weeks 4 secs ago
• 7 millis ago

Huge, but: it simplifies the month and year to 30 and 365 days, respectively. Not always what you want, although this is enough for some cases.

Answer 5

Custom function to convert x hours ago to datetime , x hour, y mins ago to datetime , etc. In Python.

The function accepts a single parameter of type string, which is parsed using RegExp. RegExp can be customized according to the input function.

For use, see the examples below.

 import re from datetime import datetime, timedelta def convert_datetime(datetime_ago): matches = re.search(r"(\d+ weeks?,? )?(\d+ days?,? )?(\d+ hours?,? )?(\d+ mins?,? )?(\d+ secs? )?ago", datetime_ago) if not matches: return None date_pieces = {'week': 0, 'day': 0, 'hour': 0, 'min': 0, 'sec': 0} for i in range(1, len(date_pieces) + 1): if matches.group(i): value_unit = matches.group(i).rstrip(', ') if len(value_unit.split()) == 2: value, unit = value_unit.split() date_pieces[unit.rstrip('s')] = int(value) d = datetime.today() - timedelta( weeks=date_pieces['week'], days=date_pieces['day'], hours=date_pieces['hour'], minutes=date_pieces['min'], seconds=date_pieces['sec'] ) return d 

Usage example:

 dates = [ '1 week, 6 days, 11 hours, 20 mins, 13 secs ago', '1 week, 10 hours ago', '1 week, 1 day ago', '6 days, 11 hours, 20 mins ago', '1 hour ago', '11 hours, 20 mins ago', '20 mins 10 secs ago', '10 secs ago', '1 sec ago', ] for date in dates: print(convert_datetime(date)) 

Exit:

 2019-05-10 06:26:40.937027 2019-05-16 07:46:53.937027 2019-05-15 17:46:53.937027 2019-05-17 06:26:53.937027 2019-05-23 16:46:53.937027 2019-05-23 06:26:53.937027 2019-05-23 17:26:43.937027 2019-05-23 17:46:43.937027 2019-05-23 17:46:52.937027