使用Python ......
如何在一年内选择所有星期日（或任何一天）？
[ '01/03/2010','01/10/2010','01/17/2010','01/24/2010', ...]
这些日期代表2010年的星期日。这也适用于我想的一周中的任何一天。
答案 0 :(得分：42)
您可以使用datetime模块中的date查找一年中的第一个星期天，然后继续添加七天，从而产生新的星期日：
from datetime import date, timedelta
def allsundays(year):
d = date(year, 1, 1) # January 1st
d += timedelta(days = 6 - d.weekday()) # First Sunday
while d.year == year:
yield d
d += timedelta(days = 7)
for d in allsundays(2010):
print(d)
答案 1 :(得分：7)
使用dateutil module，您可以通过以下方式生成列表：
#!/usr/bin/env python
import dateutil.relativedelta as relativedelta
import dateutil.rrule as rrule
import datetime
year=2010
before=datetime.datetime(year,1,1)
after=datetime.datetime(year,12,31)
rr = rrule.rrule(rrule.WEEKLY,byweekday=relativedelta.SU,dtstart=before)
print rr.between(before,after,inc=True)
虽然如果没有dateutil，找到所有星期日并不难，但是如果您有更复杂或更多的日期计算，该模块非常方便。
如果您使用的是Debian / Ubuntu，则dateutil由python-dateutil包提供。
答案 2 :(得分：4)
您可以迭代该年的日历。 以下内容应返回给定年份的所有星期二和星期四。
# Returns all Tuesdays and Thursdays of a given year
from datetime import date
import calendar
year = 2016
c = calendar.TextCalendar(calendar.SUNDAY)
for m in range(1,13):
for i in c.itermonthdays(year,m):
if i != 0: #calendar constructs months with leading zeros (days belongng to the previous month)
day = date(year,m,i)
if day.weekday() == 1 or day.weekday() == 3: #if its Tuesday or Thursday
print "%s-%s-%s" % (year,m,i)
答案 3 :(得分：2)
Pandas凭借其date_range功能为此目的提供了强大的功能。
结果是pandas DatetimeIndex，但可以轻松转换为列表。
import pandas as pd
def allsundays(year):
return pd.date_range(start=str(year), end=str(year+1),
freq='W-SUN').strftime('%m/%d/%Y').tolist()
allsundays(2017)[:5] # First 5 Sundays of 2017
# ['01/01/2017', '01/08/2017', '01/15/2017', '01/22/2017', '01/29/2017']
答案 4 :(得分：2)
使用列表理解：
dds::sub::cond::ReadCondition rc(*mp_reader,
dds::sub::status::DataState::any(),
std::bind(&MyClass::do_stuff, this));
答案 5 :(得分：1)
这是一个完整的生成器函数，它基于@sth的解决方案。它包括他的解决方案评论中提到的关键修复。
您可以指定星期几（使用Python的索引，0 =星期一到6 =星期日），开始日期和枚举的周数。
def get_all_dates_of_day_of_week_in_year(day_of_week, start_year, start_month,
start_day, max_weeks=None):
'''
Generator function to enumerate all calendar dates for a specific day
of the week during one year. For example, all Wednesdays in 2018 are:
1/3/2018, 1/10/2018, 1/17/2018, 1/24/2018, 1/31/2018, 2/7/2018, etc.
Parameters:
----------
day_of_week : int
The day_of_week should be one of these values: 0=Monday, 1=Tuesday,
2=Wednesday, 3=Thursday, 4=Friday, 5=Saturday, 6=Sunday.
start_year : int
start_month : int
start_day : int
The starting date from which to list out all the dates
max_weeks : int or None
If None, then list out all dates for the rest of the year.
Otherwise, end the list after max_weeks number of weeks.
'''
if day_of_week < 0 or day_of_week > 6:
raise ValueError('day_of_week should be in [0, 6]')
date_iter = date(start_year, start_month, start_day)
# First desired day_of_week
date_iter += timedelta(days=(day_of_week - date_iter.weekday() + 7) % 7)
week = 1
while date_iter.year == start_year:
yield date_iter
date_iter += timedelta(days=7)
if max_weeks is not None:
week += 1
if week > max_weeks:
break
从2018年1月1日开始，为期10周的所有星期三的使用示例。
import calendar
day_of_week = 2
max_weeks = 10
for d in get_all_dates_of_day_of_week_in_year (day_of_week, 2018, 1, 1, max_weeks):
print "%s, %d/%d/%d" % (calendar.day_name[d.weekday()], d.year, d.month, d.day)
上面的代码产生：
Wednesday, 2018/1/3
Wednesday, 2018/1/10
Wednesday, 2018/1/17
Wednesday, 2018/1/24
Wednesday, 2018/1/31
Wednesday, 2018/2/7
Wednesday, 2018/2/14
Wednesday, 2018/2/21
Wednesday, 2018/2/28
Wednesday, 2018/3/7
答案 6 :(得分：1)
根据@sth的答案，我想给你一个没有功能的选择
from datetime import date, timedelta,datetime
sunndays = list()
year_var = datetime.now() #get current date
year_var = year_var.year #get only the year
d = date(year_var, 1, 1) #get the 01.01 of the current year = 01.01.2020
#now we have to skip 4 days to get to sunday.
#d.weekday is wednesday so it has a value of 2
d += timedelta(days=6 - d.weekday()) # 01.01.2020 + 4 days (6-2=4)
sunndays.append(str(d.strftime('%d-%m-%Y'))) #you need to catch the first sunday
#here you get every other sundays
while d.year == year_var:
d += timedelta(days=7)
sunndays.append(str(d.strftime('%d-%m-%Y')))
print(sunndays) # only for control
如果您想每个星期一为例
#for 2021 the 01.01 is a friday the value is 4
#we need to skip 3 days 7-4 = 3
d += timedelta(days=7 - d.weekday())
答案 7 :(得分：0)
根据@ sth的答案，它将失去星期天1日。这会更好：
d = datetime.date(year, month-1, 28)
for _ in range(5):
d = d + datetime.timedelta(days=-d.weekday(), weeks=1)
if d.month!=month:
break
date.append(d)
答案 8 :(得分：0)
如果要寻找一种更一般的方法（即不仅是星期天），我们可以基于sth的answer：
def weeknum(dayname):
if dayname == 'Monday': return 0
if dayname == 'Tuesday': return 1
if dayname == 'Wednesday':return 2
if dayname == 'Thursday': return 3
if dayname == 'Friday': return 4
if dayname == 'Saturday': return 5
if dayname == 'Sunday': return 6
这会将日期名称转换为int。
然后做：
从日期时间导入日期开始，时间增量
def alldays(year, whichDayYouWant):
d = date(year, 1, 1)
d += timedelta(days = (weeknum(whichDayYouWant) - d.weekday()) % 7)
while d.year == year:
yield d
d += timedelta(days = 7)
for d in alldays(2020,'Sunday'):
print(d)
请注意% 7中存在alldays()。输出：
2020-01-05
2020-01-12
2020-01-19
2020-01-26
2020-02-02
2020-02-09
2020-02-16
...
还可以：
for d in alldays(2020,'Friday'):
print(d)
这将为您提供：
2020-01-03
2020-01-10
2020-01-17
2020-01-24
2020-01-31
2020-02-07
2020-02-14
...