CODEWITHSUNDEEP
pythonlistdatetimetime
Jarwin
Jarwin

Reputation: 1133

Convert a 12 hour time format to 24 hour time format (keeping record of the day) in python

Following is a list in python which contains a list of time for a train with each index as a stoppage for the train

Train_arrival_time = ['Source', '09:30 PM', '09:56 PM', '11:23 PM', '12:01 AM', '12:12 AM', '12:44 AM', '01:55 AM', '03:18 AM', '04:58 AM', '06:18 AM', '06:33 AM', '07:23 AM', '08:45 AM', '09:14 AM', '10:17 AM', '10:59 AM', '12:15 PM', '01:30 PM', '01:49 PM', '02:55 PM', '03:16 PM', '03:58 PM', '05:15 PM', '05:38 PM', '06:45 PM', '07:20 PM', '08:07 PM', '08:38 PM', '09:25 PM', '11:28 PM', '12:50 AM', '01:21 AM', '01:53 AM', '02:45 AM', '02:57 AM', '03:45 AM', '05:20 AM', '06:00 AM', '06:30 AM']

As evident from the list, this train runs for three consecutive days (source is the starting station). I am trying to get the time difference between two stations , and since using 12-hour format will create chaos i tried using the following script to convert the whole list to 24-hour format

from datetime import *
t12 = '09:35 PM'
t24 =  datetime.strptime(t12, '%I:%M %p')

which is giving

1900-01-01 21:35:00

as an output. Is there any way to avoid getting the UTC date? Additionally, I am stuck on how shall I get the time difference between two stations on different days; viz, 09:30PM (day 1) and 12:15PM (day2)

Upvotes: 2

Views: 1785

Answers (2)

jfs
jfs

Reputation: 414207

I am trying to get the time difference between two stations , and since using 12-hour format will create chaos i tried using the following script to convert the whole list to 24-hour format

The day doesn't matter if all you need is to find a time difference for adjacent items:

#!/usr/bin/env python
from datetime import datetime, timedelta

ZERO, DAY = timedelta(0), timedelta(days=1)
times = (datetime.strptime(time12h_string, '%I:%M %p')
         for time12h_string in Train_arrival_time[1:])
previous_time = next(times)
time_in_transit = [ZERO]
for time24h in times:
    time24h = datetime.combine(previous_time, time24h.time())
    while time24h < previous_time: # time on the next station should not be ealier
        time24h += DAY
    time_in_transit.append(time24h - previous_time + time_in_transit[-1])
    previous_time = time24h

Here's time_in_transit is the accumulative time between stations starting with the first station that has the arrival time i.e., time_in_transit[i] is a time in transit for the i-th station (the time difference between the station and the first station). It is computed as a series of partial sums (like itertools.accumulate()) of time differences between adjacent stations, namely:

  • stations are identified by their index in the Train_arrival_time list (shifted by one compared to the time_in_transit list) and/or their arrival times -- look at the Station column in the output below
  • (time24h - previous_time) is the time difference between adjacent stations -- look at the Adjacent column in the output below
  • time_in_transit[-1] is the previous item in the series (the last one) -- it is the same as time_in_transit[len(time_in_transit)-1] in Python
  • the current item is the sum "the current difference + the accumulated sum" -- look at the Total column in the output below.

Is there any way to avoid getting the UTC date?

The result of datetime.strptime() is a naive datetime object that does not correspond to any time zone. There is no point to talk about time zones if you don't know a specific date (year, month, day).

I am stuck on how shall I get the time difference between two stations on different days; viz, 09:30PM (day 1) and 12:15PM (day2)

It is easy to find the time between adjacent stations and the total time in transit:

print("Station  | Adjacent | Total")
print("\n".join(["{} | {:>8s} | {}".format(time12h, str(curr-prev), str(curr))
                 for time12h, curr, prev in zip(Train_arrival_time[1:],
                                                time_in_transit,
                                                [ZERO]+time_in_transit)]))

Output

Station  | Adjacent | Total
09:30 PM |  0:00:00 | 0:00:00
09:56 PM |  0:26:00 | 0:26:00
11:23 PM |  1:27:00 | 1:53:00
12:01 AM |  0:38:00 | 2:31:00
12:12 AM |  0:11:00 | 2:42:00
...
05:20 AM |  1:35:00 | 1 day, 7:50:00
06:00 AM |  0:40:00 | 1 day, 8:30:00
06:30 AM |  0:30:00 | 1 day, 9:00:00

To find the time difference between i-th and j-th stations:

time_difference = time_in_transit[j] - time_in_transit[i]

Upvotes: 2

tobias_k
tobias_k

Reputation: 82899

First, convert all the time strings to dates. The day does not matter, but we will need it later.

from datetime import datetime, timedelta
times = [datetime.strptime(x, '%I:%M %p') for x in Train_arrival_time[1:]]

Now, you can keep a running count of days to add, and add those to all the datetimes in the list. Whenever the current time is smaller than the previous, add another day.

days_to_add = 0
for i in range(1, len(times)):
    times[i] += timedelta(days=days_to_add)
    if times[i] < times[i-1]:
        days_to_add += 1
        times[i] += timedelta(days=1)

Now, you can get the adjusted times and the total time difference between the first and last time. To get an output list "day 1, time x", just strip the year and month from the output.

for x in times:
    print(x.strftime("Day %d, %H:%M"))
print(times[-1] - times[0])

Output:

Day 01, 21:30
...
Day 03, 06:30
1 day, 9:00:00

Of course, if the time difference between two consecutive times is greater than one whole day, then you are out of luck.

Upvotes: 1

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