python 'x days ago' to datetime

Question

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

x minutes/hours/days/months/years ago

I need to parse that to a datetime using python.

It seems dateutil can't do it.

Is there a way to do it?

Answer 1

Sure 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 dateutil has a relativedelta which acts like a timedelta, but the constructor also accepts months and years in the arguments (and apparently the arguments need to be integers).

Answer 2

Easiest way is to use dateparser

import dateparser
date_ago = '4 months ago'
date     = dateparser.parse(date_ago)
date     = date.strftime("%Y-%m-%d")

Output:

'2020-01-08'

---

you may need install dateparser pip package

pip install dateparser

Answer 3

Since your arguments are something like 2 days ago, 3 months ago, 2 years ago. The function below could be of help in getting the exact date for the arguments. You first need to import the following date utils

import datetime
from dateutil.relativedelta import relativedelta

Then 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 then call the function like this:

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

This can be done easily with timedeltas:

import datetime

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

Producing:

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

Although this will require some extra work to deal with months/years - as adding a month to a date is an ambiguous operation, but it should be a simple addition if you know what you want it to mean.

To get an actual time, simply take the delta away from datetime.datetime.now().

Answer 5

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))

This can parse 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

A huge but: It simplifies month and year to 30 and 365 days respectively. Not always what you want, though it's enough for some cases.

Answer 6

@mgilson answer didn't work for me where dateutil.relativedelta did the job:

Edited following @mosc9575 advise

import datetime
from dateutil.relativedelta import relativedelta

time_ago = "1 month ago"
val, unit = time_ago.split()[:2]
past_time = datetime.datetime.now() - relativedelta(**{unit:int(val)})

Answer 7

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

Function takes single parameter of type string which is parsed using RegExp. RegExp can be customized to match function input.

For usage see 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

Example usage:

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))

Output:

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

Answer 8

make sure to install dependencies using pip3

from datetime import date
from dateutil.relativedelta import relativedelta
import re

baseDate = date.today() #date(2020, 4, 29)

hoursPattern = re.compile(r'(\d\d?\d?) hours? ago')
daysPattern = re.compile(r'(\d\d?\d?) days? ago')
weeksPattern = re.compile(r'(\d\d?\d?) weeks? ago')
monthsPattern = re.compile(r'(\d\d?\d?) months? ago')
yearsPattern = re.compile(r'(\d\d?\d?) years? ago')


days = 0
daysMatch = daysPattern.search(ago)
if daysMatch:
    days += int(daysMatch.group(1))

hours = 0
hoursMatch = hoursPattern.search(ago)
if hoursMatch:
    hours += int(hoursMatch.group(1))

weeks = 0
weeksMatch = weeksPattern.search(ago)
if weeksMatch:
    weeks += int(weeksMatch.group(1))

months = 0
monthsMatch = monthsPattern.search(ago)
if monthsMatch:
    months += int(monthsMatch.group(1))

years = 0
yearsMatch = yearsPattern.search(ago)
if yearsMatch:
    years += int(yearsMatch.group(1))

yourDate = baseDate - relativedelta(hours=hours, days=days, weeks=weeks, months=months, years=years)

Answer 9

Any chance for datetime to work with other languages (italian) 'x-days ago to date'?

import locale\
locale.setlocale(locale.LC_ALL,'it_IT.UTF-8') 

doesn't make any difference.