我目前不確定用于以下問題的邏輯以及編程的新手。（目前正在學習 python）嘗試遍歷給定月份的每個日期 - 比如 05/01 - 05/31 并以下面的格式打印出來。周一到周五的日期要單獨打印。星期六和星期日的日期要單獨打印。如果這個月從星期五開始 - 05/01/2020，輸出應該是 那個星期的最后一個工作日。對于 2020 年 4 月，產量將如下所示，因為 4 月的第一周從周三開始。我設法想出以下嘗試，但不確定如何進一步進行。import sysfrom datetime import date, datetime, timedeltayear = int(sys.argv[1])month = int(sys.argv[2])st_dt = int(sys.argv[3])en_dt = int(sys.argv[4])first_date = datetime(year, month, st_dt).date()get_first_day = datetime(year, month, st_dt).isoweekday()def daterange(startDate, endDate, delta=timedelta(days=1)): currentDate = startDate while currentDate <= endDate: yield currentDate currentDate += deltafor date in daterange(date(year, month, st_dt), date(year, month, en_dt), delta=timedelta(days=1)): print(date) date.py 2020 5 1 31 # script提出了一個獨立的“if 循環”，正如我之前所說，不確定如何構建更大的圖景:(if get_first_day == 1: #print("Monday") sec_d = first_date + timedelta(days=4)elif get_first_day == 2: sec_d = first_date + timedelta(days=3)elif get_first_day == 3: sec_d = first_date + timedelta(days=2)elif get_first_day == 4: sec_d = first_date + timedelta(days=2)elif get_first_day == 5: sec_d = first_date #print("Friday")else: passprint(f"Second date:{sec_d} ") -- which gave -- > Second date:2020-05-01
1 回答
慕妹3146593
TA貢獻1820條經驗 獲得超9個贊
您可以將日期保存在字典中，字典鍵是日歷周和日期類型（周末、星期幾）的元組。
每一天都保存在allDays字典中，按周數和日期類型的組合作為鍵進行分組：
('18', 'weekend'): [datetime.date(2020, 5, 2), datetime.date(2020, 5, 3)],
('18', 'working'): [datetime.date(2020, 5, 1)],
('19', 'weekend'): [datetime.date(2020, 5, 9), datetime.date(2020, 5, 10)],
('19', 'working'): [datetime.date(2020, 5, 4), ...
所以你只需要取出每個 dict 項目的第一個和最后一個項目：
import sys
from datetime import date, datetime, timedelta
year, month, st_dt, en_dt = 2020, 5, 1, 31
first_date = datetime(year, month, st_dt).date()
get_first_day = datetime(year, month, st_dt).isoweekday()
def daterange(startDate, endDate, delta=timedelta(days=1)):
currentDate = startDate
while currentDate <= endDate:
yield currentDate
currentDate += delta
allDays = {}
_lastDayType = None
for dte in daterange(date(year, month, st_dt), date(year, month, en_dt), delta=timedelta(days=1)):
if 0 <= dte.weekday() < 5:
_dayType = 'working'
else:
_dayType = 'weekend'
_weeknum = dte.strftime("%V") # number of calendar week
_key = (_weeknum, _dayType)
if _key not in allDays: # create an empty list if unique key doesnt exist
allDays[_key] = []
allDays[_key].append(dte) # add the dates ...
for k,v in allDays.items():
if len(v) == 1:
first, last = v[0], v[0]
else:
first, last = v[0], v[-1]
print("%s >> %s" % (first, last))
輸出：
2020-05-01 >> 2020-05-01
2020-05-02 >> 2020-05-03
2020-05-04 >> 2020-05-08
2020-05-09 >> 2020-05-10
2020-05-11 >> 2020-05-15
2020-05-16 >> 2020-05-17
2020-05-18 >> 2020-05-22
2020-05-23 >> 2020-05-24
2020-05-25 >> 2020-05-29
2020-05-30 >> 2020-05-31
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