我想打印这些日期范围之间的所有星期四
from datetime import date, timedelta
sdate = date(2015, 1, 7) # start date
edate = date(2015, 12, 31) # end date
最好的pythonic方法是什么？
【问题讨论】：
标签： python python-3.x python-2.7
使用您的sdate.weekday() # returns int between 0 (mon) and 6 (sun)：
sdate = ...
while sdate < edate:
if sdate.weekday() != 3: # not thursday
sdate += timedelta(days=1)
continue
# It is thursday
print(sdate)
sdate += timedelta(days=7) # next week
【讨论】：
计算从开始日期到星期四的天数：
days_to_thursday = (3 - sdate.weekday()) % 7
计算两个日期之间的星期四数：
week_diff = ((edate - sdate).days - days_to_thursday ) // 7
获取日期范围内的所有星期四：
thursdays = [sdate + timedelta(days=days_to_thursday + 7 * more_weeks) \
for more_weeks in range(week_diff + 1) ]
如果需要，请打印：
for t in thursdays:
print(t)
【讨论】：
大多数人每天都在迭代，这是一种浪费。也可能有助于延迟计算星期四，直到您真正需要它们。为此，您可以使用生成器。
def get_days(start, day_index, end=None):
# set the start as the next valid day
start += timedelta(days=(day_index - start.weekday()) % 7)
week = timedelta(days=7)
while end and start < end or not end:
yield start
start += week
这会延迟获得第二天，直到您需要它，如果您不指定和结束日期，则允许无限天。
thursday_generator = get_days(date(2015, 1, 7), 3, date(2015, 12, 31))
print(list(thursday_generator))
"""
[datetime.date(2015, 1, 8), datetime.date(2015, 1, 15), datetime.date(2015, 1, 22), ...]
"""
您可以轻松地转储为字符串：
print("\n".join(map(str, thursday_generator)))
"""
2015-01-08
2015-01-15
2015-01-22
...
"""
您还可以使用 f-strings 进行自定义字符串格式化：
print("\n".join(f"{day:%A %x}" for day in thursday_generator))
"""
Thursday 01/08/15
Thursday 01/15/15
Thursday 01/22/15
...
"""
如果您不指定结束日期，它将永远持续下去。
In [28]: thursday_generator = get_days(date(2015, 1, 7), 3)
...: print(len(list(thursday_generator)))
---------------------------------------------------------------------------
OverflowError Traceback (most recent call last)
<ipython-input-28-b161cdcccc75> in <module>
1 thursday_generator = get_days(date(2015, 1, 7), 3)
----> 2 print(len(list(thursday_generator)))
<ipython-input-16-0691db329606> in get_days(start, day_index, end)
5 while end and start < end or not end:
6 yield start
----> 7 start += week
8
OverflowError: date value out of range
【讨论】：
您可以尝试列表推导来获取两个日期之间的星期四。
此代码实际上将日期输出为格式化字符串，但您可以通过删除 strftime 来获取实际日期。
from datetime import date, timedelta
sdate = date(2015, 1, 7) # start date
edate = date(2015, 12, 31) # end date
thursdays = [(sdate+timedelta(days=d)).strftime('%A %Y-%m-%d') for d in range(0, (edate-sdate).days+1)
if (sdate+timedelta(days=d)).weekday() ==3]
print('\n'.join(thursdays))
""" Example output
Thursday 2015-01-08
Thursday 2015-01-15
Thursday 2015-01-22
Thursday 2015-01-29
Thursday 2015-02-05
Thursday 2015-02-12
""""
【讨论】：
import datetime
import calendar
def weekday_count(start, end, day):
start_date = datetime.datetime.strptime(start, '%d/%m/%Y')
end_date = datetime.datetime.strptime(end, '%d/%m/%Y')
day_count = []
for i in range((end_date - start_date).days):
if calendar.day_name[(start_date + datetime.timedelta(days=i+1)).weekday()] == day:
print(str(start_date + datetime.timedelta(days=i+1)).split()[0])
weekday_count("01/01/2017", "31/01/2017", "Thursday")
# prints result
# 2017-01-05
# 2017-01-12
# 2017-01-19
# 2017-01-26
【讨论】：
简单的解决方案：
from datetime import date, timedelta
sdate = date(2015, 1, 7) # start date
edate = date(2015, 12, 31) # end date
delta = edate - sdate
for day in range(delta.days + 1):
day_obj = sdate + timedelta(days=day)
if day_obj.weekday() == 3: # Thursday
print(day_obj)
# 2015-01-08
# 2015-01-15
# ...
# 2015-12-24
# 2015-12-31
最有效的解决方案：
from datetime import date, timedelta
sdate = date(2015, 1, 7) # start date
edate = date(2015, 12, 31) # end date
day_index = 3 # Thursday
delta = (day_index - sdate.weekday()) % 7
match = sdate + timedelta(days=delta)
while match <= edate: # Change this to `<` to ignore the last one
print(match) # Can be easily converted to a generator with `yield`
match += timedelta(days=7)
# 2015-01-08
# 2015-01-15
# ...
# 2015-12-24
# 2015-12-31
文档：
【讨论】：
abs 函数使此解决方案对于任何工作日大于 3 的日期都不正确。它分别给出1, 2 or 3 的增量而不是6, 5 or 4。