Weekly Countdown Timer in Python

Question

I am trying to write a script that runs continuously in the background to countdown to a repeated weekly event. For example, it should tell me how many days, hours, and minutes it will take to reach the specified time.

I know how to do it if I had a specific date and time.

import datetime
delta = datetime.datetime(2018, 5, 5, 8) - datetime.datetime.now()

But what if I don't have a specific date and time? Can datetime let me choose the day of the week?

EDIT:

ie Some pseudocode like this is what I need.

delta = datetime(Thursday 8 PM) - datetime.datetime.now()
#returns datetime or timedelta in days, hours, minutes

Answer 1

EDIT: Thanks Ethan, i appreciate your constructive advice. I wrote a small script which should do what you want:

import datetime
import time
wanted_day = 'thursday'
wanted_time = 8

list = [['monday', 0],['tuesday', 1],['wednesday', 2],['thursday', 3],['friday', 4],['saturday', 5],['sunday', 6]]

for i in list:
    if wanted_day == i[0]:
        number_wanted_day = i[1]

# today delivers the actual day
today = datetime.datetime.today().weekday()

# delta_days describes how many days are left until the wanted day
delta_days = number_wanted_day - today

# time delivers the actual time
time = time.localtime(time.time())

if wanted_time > time[3]:
    delta_hours = wanted_time - time[3]
    delta_mins = 59 - time[4]
    delta_secs = 59 - time[5]

else:
    delta_days = delta_days - 1
    delta_hours = 23 - time[3] + wanted_time
    delta_mins = 59 - time[4]
    delta_secs = 59 - time[5]

print [delta_days, delta_hours, delta_mins, delta_secs]

The output looks like this then:

[2, 21, 3, 49]

2 is the number of days, 21 the number of hours, 3 the number of mins and 49 the number of secs (I used thursday 8 am as wanted time). You just need to input the time in the format 0-23, with am and pm you would need to adapt it a bit