From string to datetime, looking at the current (local) time

Question

Is there any function, in python, that extracts time information from a string to get the time from the current local one? For example:

I have a column with time information in the past compared to the current time (let's say UK time, approximately 20:00)

Time

10 hours # ago
6 hours # ago
12 hours # ago
2 days # ago
1 day # ago

I would like to have, looking at the current UK time, (i.e., approximately 20:00):

New_Time

10:00 
14:00 
8:00
48:00
24:00

The column above comes from the current local time minus the number of hours (from string to numerical) in Time column.

Is it something that I might do with a function? The New_Time is the output that I will need.

In case there might not be a function in python, my approach would be:

• use something to detect the current time (to save it in a variable)
• consider hours as 60 min and 1 day as 24 hours ... in order to run some mathematical operations, once transformed strings into numerical
• subtract from the current time the amount of minutes/hours to go back in time

Answer 1

You can create a method that would return the time that it was a few hours/days/... ago. For example, two methods that will respectively return what day it was x days ago and what hour it was x hours ago :

def getTimeDaysAgo(x):
    today = datetime.datetime.now()
    value = today.day - x
    previousMonth = today.month - 1
    previousMonthDuration = 30
    if previousMonth == 2:
        if today.year % 4 == 0:
            previousMonthDuration = 29
        else:
            previousMonthDuration = 28
    if previousMonth % 2 == 1:
        previousMonthDuration = 31
    return value if value >= 0 else value + previousMonthDuration


def getTimeHoursAgo(x):
    value = datetime.datetime.now().hour - x
    return value if value >= 0 else value + 24

I'm not sure why you want to get "48:00" as a result of 2 days ago and "10:00" as a result of 10 hours ago at the same time. Feel free to ask for more infos, I'd be glad to help you out.

Answer 2

You can make use of the dateparser package (some explanation in comments):

import datetime
import dateparser # pip install dateparser

# a function to format timedelta to string needed below,
# see https://stackoverflow.com/q/538666/10197418
def timedelta2str(sec):
    hours, remainder = divmod(sec, 3600)
    return f'{int(hours):02}:{int(remainder//60):02}'

def formatTimeAgo(string, reftime):
    # try to parser... if successful, you get a datetime object
    dtobj = dateparser.parse(string, settings={'RELATIVE_BASE': reftime})
    if isinstance(dtobj, datetime.datetime):
        # calculate a timedelta object against reference time
        td = reftime - dtobj
        # now format output based on input (days or hours..)
        if td >= datetime.timedelta(1):
            return timedelta2str(td.total_seconds())
        else:
            return (reftime-td).strftime('%H:%M')
    else:
        return "N/A"

# exemplary input
t = ("10 hours", "6 hours", "2 days", "1 day")
# a reference time
reftime = datetime.datetime(2021,4,10,20)
for elem in t:
    print(elem, '->', formatTimeAgo(elem, reftime))
       
# giving you
>>> 10 hours -> 10:00
>>> 6 hours -> 14:00
>>> 2 days -> 48:00
>>> 1 day -> 24:00

Now you can adjust that to be used on pandas.Series:

import pandas as pd

df = pd.DataFrame({'Time': ["10 hours", "6 hours", "12 hours", "2 days", "1 day"]})

df['New_Time'] = df['Time'].apply(formatTimeAgo, args=(reftime,))

df['New_Time']
0    10:00
1    14:00
2    08:00
3    48:00
4    24:00
Name: New_Time, dtype: object