datetimeを用いて「年月日時間」のリストを作る

Question

python datetimeモジュールを使って最初と最後の日を

start_day = 2018020100 # (%Y:2018, %m:02 %d:02 %H:00)
end_day = 2018020723

のように指定することでdate_listに

date_list = [2018020100, 2018020101, 2018020102, ... , 2018020720, 2018020721, 2018020722, 2018020723]

のように１時間おきの日付リストを作成したいのですが，どのようにすればいいでしょうか？

Answer 1

普通にループさせる。

import datetime

# intは面倒だからとりあえず文字列にしとく
start = str(2018020100)
end = str(2018020723)

start_dt = datetime.datetime.strptime(start, "%Y%m%d%H")
end_dt = datetime.datetime.strptime(end, "%Y%m%d%H")

lst = []
t = start_dt
while t <= end_dt:
    lst.append(t)
    t += datetime.timedelta(hours=1)

print([int(x.strftime("%Y%m%d%H")) for x in lst])
""" =>
[2018020100, 2018020101, 2018020102, 2018020103, 2018020104, 2018020105, 2018020106, 2018020107, 2018020108, 2018020109, 2018020110, 2018020111, 2018020112, 2018020113, 2018020114, 2018020115, 2018020116, 2018020117, 2018020118, 2018020119, 2018020120, 2018020121, 2018020122, 2018020123, 2018020200, 2018020201, 2018020202, 2018020203, 2018020204, 2018020205, 2018020206, 2018020207, 2018020208, 2018020209, 2018020210, 2018020211, 2018020212, 2018020213, 2018020214, 2018020215, 2018020216, 2018020217, 2018020218, 2018020219, 2018020220, 2018020221, 2018020222, 2018020223, 2018020300, 2018020301, 2018020302, 2018020303, 2018020304, 2018020305, 2018020306, 2018020307, 2018020308, 2018020309, 2018020310, 2018020311, 2018020312, 2018020313, 2018020314, 2018020315, 2018020316, 2018020317, 2018020318, 2018020319, 2018020320, 2018020321, 2018020322, 2018020323, 2018020400, 2018020401, 2018020402, 2018020403, 2018020404, 2018020405, 2018020406, 2018020407, 2018020408, 2018020409, 2018020410, 2018020411, 2018020412, 2018020413, 2018020414, 2018020415, 2018020416, 2018020417, 2018020418, 2018020419, 2018020420, 2018020421, 2018020422, 2018020423, 2018020500, 2018020501, 2018020502, 2018020503, 2018020504, 2018020505, 2018020506, 2018020507, 2018020508, 2018020509, 2018020510, 2018020511, 2018020512, 2018020513, 2018020514, 2018020515, 2018020516, 2018020517, 2018020518, 2018020519, 2018020520, 2018020521, 2018020522, 2018020523, 2018020600, 2018020601, 2018020602, 2018020603, 2018020604, 2018020605, 2018020606, 2018020607, 2018020608, 2018020609, 2018020610, 2018020611, 2018020612, 2018020613, 2018020614, 2018020615, 2018020616, 2018020617, 2018020618, 2018020619, 2018020620, 2018020621, 2018020622, 2018020623, 2018020700, 2018020701, 2018020702, 2018020703, 2018020704, 2018020705, 2018020706, 2018020707, 2018020708, 2018020709, 2018020710, 2018020711, 2018020712, 2018020713, 2018020714, 2018020715, 2018020716, 2018020717, 2018020718, 2018020719, 2018020720, 2018020721, 2018020722, 2018020723]
"""

Comments

[loreeeee]

ありがとうございます。自分の中で読みやすいコードだったのでベストアンサーにさせていただきます！

Answer 2

# -*- coding: utf-8 -*-
from datetime import datetime, timedelta


def date_generator(start_day:int, end_day:int):
    assert start_day < end_day
    current_dt = datetime.strptime(str(start_day), "%Y%m%d%H")
    end_dt = datetime.strptime(str(end_day), "%Y%m%d%H")

    while current_dt <= end_dt:
        yield current_dt
        current_dt += timedelta(hours=1)


def main() -> None:
    date_list = list(map(lambda x: int(x.strftime("%Y%m%d%H")), date_generator(2018020100, 2018020723)))
    print(date_list)


if __name__ == "__main__":
    main()

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Answer 3

from datetime import datetime, timedelta

...

dst = []
tmp_time = start_time

while tmp_time <= end_time:
    dst.append(tmp_time)
    tmp_time += timedelta(hours=1)

start_timeやend_timeの作り方はご自身で考えましょう。

• Python 標準ライブラリ » datetime — 基本的な日付型および時間型

Answer 4

from datetime import datetime, timedelta

start_day = datetime(2018,2,2 ,0)
end_day = datetime(2018,2,  7, 23)
td=end_day-start_day
[start_day+timedelta(hours=i) for i in range(td.days*24+td.seconds//3600 +1)]