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I have string period = 2022-09-12/2022-09-15

How can i get arr = [2022-09-12, 2022-09-13, 2022-09-14, 2022-09-15]

Thanks

Diepdang
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  • What part of the problem are you having trouble with? Splitting the string? Turning the parts into dates? Getting a range of dates from that start and end? Putting the result in a list? What actually is in that list? Strings and you forgot the quotes, or datetime values? Share your code and be specific, have a look at https://stackoverflow.com/help/how-to-ask – Grismar Sep 16 '22 at 22:58
  • I'm newbie in python And i don't have solution for this problem Sorry if my question is unclear – Diepdang Sep 16 '22 at 23:09
  • Oh, that's one way of going about it. thanks – Diepdang Sep 16 '22 at 23:29

2 Answers2

2

You can do it like this:

start_date = datetime.datetime(2022, 9, 12)
end_date = datetime.datetime(2022, 9, 15)

involved_dates = []
for i in range((end_date - start_date).days):
    involved_dates.append(start_date + datetime.timedelta(days=i))
Oleksandr Lisovyi
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  • You may have problems if the months are different. Edit: no sorry :( – Ahmet Burak Sep 16 '22 at 23:01
  • i run this and get this value ```[datetime.datetime(2022, 9, 12, 0, 0), datetime.datetime(2022, 9, 13, 0, 0), datetime.datetime(2022, 9, 14, 0, 0)]``` But you can format value in array like 2022-09-22. Thanks so much – Diepdang Sep 16 '22 at 23:13
1

Since you didn't specify what part of the problem you're struggling with, here's a solution for the whole thing:

from datetime import datetime, timedelta

text = '2022-09-12/2022-09-15'


def date_range(start, end):
    # this function takes two dates and generates all the dates in the range
    for d in range((end - start).days + 1):
        yield start + timedelta(days=d)


# defining the format your dates come in (and in which you may want them)
format = '%Y-%m-%d'

# this is the answer, using the above function
dates = list(date_range(*map(lambda d: datetime.strptime(d, format), text.split('/'))))
print(dates)

# taking the `dates` result, you can of course format them as strings:
print([datetime.strftime(d, format) for d in dates])

Output:

[datetime.datetime(2022, 9, 12, 0, 0), datetime.datetime(2022, 9, 13, 0, 0), datetime.datetime(2022, 9, 14, 0, 0), datetime.datetime(2022, 9, 15, 0, 0)]
['2022-09-12', '2022-09-13', '2022-09-14', '2022-09-15']

On the key expression:

list(date_range(*map(lambda d: datetime.strptime(d, format), text.split('/'))))

The function is a generator, so there's a list() around it to exhaust the generator into a list.

The map() takes a function, in this case a lambda and applies it to all elements of an iterable, in this case text.split('/'), which is just the two date strings in your original text.

The lambda takes those two parts and parses them into actual datetime using datetime.strptime(d, format), so the map() yields the two dates.

The * then spreads those into the function call to date_range(), providing the dates as arguments.

Note: if this also needs to work for a descending range of dates, for example '2022-09-15/2022-09-12', you can use this instead (which works for either):

def date_range(start, end):
    days = (end - start).days
    sign = -1 if days < 0 else 1
    for d in range(0, days + sign, sign):
        yield start + timedelta(days=d)
Grismar
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