Days/dates 在给定的一周内
Days/dates in a given week
我正在尝试获取包含给定一周中的天数的列表。
我可以得到年份和星期。
例如：
>>> year, week, dow = datetime.today().isocalendar()
>>> week
>>> 39
我想得到第 39 周的 7 天。对于今年，2015 年，我会得到
[21, 22, 23, 24, 25, 26, 27]
如图here
备注
我发现有很多问题要求：
- How to get number of week?
- Current week
这不是重复的，我不需要最后 7 天，而且我已经有了当前的周数。
我需要获得给定一周的 7 天，在本例中为第 39 周。感谢您的宝贵时间。
首先，找到一年开始的星期几。这在互联网上很容易找到。这为您提供了一年中第 1 周的偏移量。
请注意，星期将让您计算一年中的第几天，范围是 1-365。第 01 周从第 1 天开始；第 02 周从第 8 天开始，依此类推
start_day = 1 + (week-1) * 7 - 偏移量
首先，找到一年开始的星期几。这在互联网上很容易找到。这为您提供了一年中第 1 周的偏移量。
既然您有了开始日，就可以轻松找到该月的日期。在进入给定月份的范围之前减去每个月的天数有点乏味，但这很简单。
这让你继续吗？我已经用英文描述完成了这个，因为你没有给我们任何调试代码。
isocalendar 的困难在于它并没有真正计算每月的日期。因此，您必须翻译回来才能获得它。 strptime 可以帮助：
year, week, dow = datetime.today().isocalendar()
result = [datetime.strptime(str(year) + "-" + str(week-1) + "-" + str(x), "%Y-%W-%w").day for x in range(1,7)]
我们在这里做的是构建一个 striptime 可以理解的字符串，从一周前开始（考虑从 0 到 1 的计数）和一周的开始（星期一，即是 1) 并为未来 7 天的每一天构建一个日期时间。
通过在这两个语句之间使用 week（添加或删除周以获得月休假），我们可以看到它有效：
>>> year, week, dow = datetime.today().isocalendar()
>>> result = [datetime.strptime(str(year) + "-" + str(week-1) + "-" + str(x), "%Y-%W-%w").day for x in range(1,7)]
>>> result
[21, 22, 23, 24, 25, 26]
>>> year, week, dow = datetime.today().isocalendar()
>>> week = week + 1
>>> result = [datetime.strptime(str(year) + "-" + str(week-1) + "-" + str(x), "%Y-%W-%w").day for x in range(1,7)]
>>> result
[28, 29, 30, 1, 2, 3]
现在，为了解决评论中提出的两个非常实际的问题，我们必须稍微修改一下：
year, week, dow = datetime.today().isocalendar()
week_start = datetime.strptime(str(year) + "-" + str(week-2) + "-0", "%Y-%W-%w")
result = [(week_start + timedelta(days=x)).day for x in range(0,7)]
这里使用timedelta来递增。为了完成这项工作，我们必须跨越一周的时间间隔进行备份（因此 -2 而不是 -1）。然后 for comprehension 在我们遍历一周时增加一个越来越大的时间增量：
>>> year, week, dow = datetime.today().isocalendar()
>>> week_start = datetime.strptime(str(year) + "-" + str(week-2) + "-0", "%Y-%W-%w")
>>> result = [(week_start + timedelta(days=x)).day for x in range(0,7)]
>>> result
[20, 21, 22, 23, 24, 25, 26]
>>> year, week, dow = datetime.today().isocalendar()
>>> week = week + 1
>>> week_start = datetime.strptime(str(year) + "-" + str(week-2) + "-0", "%Y-%W-%w")
>>> result = [(week_start + timedelta(days=x)).day for x in range(0,7)]
>>> result
[27, 28, 29, 30, 1, 2, 3]
当然可以改进，但似乎有效....
def days_of_the_current_numbered_week():
import datetime
import calendar
# dictionary of days of the week
days = { 0 : "Sunday",
1 : "Monday",
2 : "Tuesday",
3 : "Wednesday",
4 : "Thursday",
5 : "Friday",
6 : "Saturday",
7 : "Sunday" }
allYearDates = []
w_days_numbers = []
# today
now = datetime.datetime.now()
# get current values for year, mon, day
year = int(now.year)
mon = int(now.month)
day = int(now.day)
# get this week number
# thisWeekN = datetime.date(year, mon, day).isocalendar()[1]
thisWeekN = datetime.datetime.utcnow().isocalendar()[1]
# get Calendar obect
c = calendar.Calendar()
# get all the days for this year in a list
for i in range(1, 13):
for d in c.itermonthdays2(year, i):
allYearDates.append(d)
# in the first seven days could be 0's as days, to continue week #'s
# these tuples need to be removed to produce accurate mapping between
# weeks and 7 days chunks
first_seven = allYearDates[:7]
no_zeros = [ d for d in first_seven if d[0] != 0]
allYearDates = no_zeros + allYearDates[8:]
# divid all days of year list into one week chuncks
lt7 = listChunks(allYearDates,7)
# get days for this week number
thisWeekDays = lt7[thisWeekN]
# remove right part of the days tuple
sevenDaysL = [x for x,y in thisWeekDays]
# get week day numbers 1-7 for this week
for d in sevenDaysL:
w_days_numbers.append(datetime.date(year, mon, d).isocalendar()[2])
# zip week day numbers with the dates
zl = zip(sevenDaysL, w_days_numbers)
# month number prefix
if thisWeekN == 1:
prefix = 'st'
elif thisWeekN == 2:
prefix = 'nd'
elif thisWeekN == 3:
prefix = 'rd'
else:
prefix= "th"
# print heading
print("\n 7 days of this {}{} week: of the year {}"
.format(thisWeekN,prefix, year))
print("-------------------------------------------")
# print results
for el in list(zl):
if el[0] == day:
print("* {} * {}".format(el[0],days[el[1]]))
else:
print(" {} {}".format(el[0],days[el[1]]))
mytime= datetime.datetime.now().strftime("%Y-%m-%d %H-%M")
print("================")
print(mytime)
days_of_the_current_numbered_week()
输出
第 39 周的 7 天：2015 年
17 星期四
18 星期五
19 周六
20 周日
21 星期一
22 星期二
23 星期三
我正在尝试获取包含给定一周中的天数的列表。 我可以得到年份和星期。
例如：
>>> year, week, dow = datetime.today().isocalendar()
>>> week
>>> 39
我想得到第 39 周的 7 天。对于今年，2015 年，我会得到
[21, 22, 23, 24, 25, 26, 27]
如图here
备注
我发现有很多问题要求：
- How to get number of week?
- Current week
这不是重复的，我不需要最后 7 天，而且我已经有了当前的周数。
我需要获得给定一周的 7 天，在本例中为第 39 周。感谢您的宝贵时间。
首先，找到一年开始的星期几。这在互联网上很容易找到。这为您提供了一年中第 1 周的偏移量。
请注意，星期将让您计算一年中的第几天，范围是 1-365。第 01 周从第 1 天开始；第 02 周从第 8 天开始，依此类推
start_day = 1 + (week-1) * 7 - 偏移量
首先，找到一年开始的星期几。这在互联网上很容易找到。这为您提供了一年中第 1 周的偏移量。
既然您有了开始日，就可以轻松找到该月的日期。在进入给定月份的范围之前减去每个月的天数有点乏味，但这很简单。
这让你继续吗？我已经用英文描述完成了这个，因为你没有给我们任何调试代码。
isocalendar 的困难在于它并没有真正计算每月的日期。因此，您必须翻译回来才能获得它。 strptime 可以帮助：
year, week, dow = datetime.today().isocalendar()
result = [datetime.strptime(str(year) + "-" + str(week-1) + "-" + str(x), "%Y-%W-%w").day for x in range(1,7)]
我们在这里做的是构建一个 striptime 可以理解的字符串，从一周前开始（考虑从 0 到 1 的计数）和一周的开始（星期一，即是 1) 并为未来 7 天的每一天构建一个日期时间。
通过在这两个语句之间使用 week（添加或删除周以获得月休假），我们可以看到它有效：
>>> year, week, dow = datetime.today().isocalendar()
>>> result = [datetime.strptime(str(year) + "-" + str(week-1) + "-" + str(x), "%Y-%W-%w").day for x in range(1,7)]
>>> result
[21, 22, 23, 24, 25, 26]
>>> year, week, dow = datetime.today().isocalendar()
>>> week = week + 1
>>> result = [datetime.strptime(str(year) + "-" + str(week-1) + "-" + str(x), "%Y-%W-%w").day for x in range(1,7)]
>>> result
[28, 29, 30, 1, 2, 3]
现在，为了解决评论中提出的两个非常实际的问题，我们必须稍微修改一下：
year, week, dow = datetime.today().isocalendar()
week_start = datetime.strptime(str(year) + "-" + str(week-2) + "-0", "%Y-%W-%w")
result = [(week_start + timedelta(days=x)).day for x in range(0,7)]
这里使用timedelta来递增。为了完成这项工作，我们必须跨越一周的时间间隔进行备份（因此 -2 而不是 -1）。然后 for comprehension 在我们遍历一周时增加一个越来越大的时间增量：
>>> year, week, dow = datetime.today().isocalendar()
>>> week_start = datetime.strptime(str(year) + "-" + str(week-2) + "-0", "%Y-%W-%w")
>>> result = [(week_start + timedelta(days=x)).day for x in range(0,7)]
>>> result
[20, 21, 22, 23, 24, 25, 26]
>>> year, week, dow = datetime.today().isocalendar()
>>> week = week + 1
>>> week_start = datetime.strptime(str(year) + "-" + str(week-2) + "-0", "%Y-%W-%w")
>>> result = [(week_start + timedelta(days=x)).day for x in range(0,7)]
>>> result
[27, 28, 29, 30, 1, 2, 3]
当然可以改进，但似乎有效....
def days_of_the_current_numbered_week():
import datetime
import calendar
# dictionary of days of the week
days = { 0 : "Sunday",
1 : "Monday",
2 : "Tuesday",
3 : "Wednesday",
4 : "Thursday",
5 : "Friday",
6 : "Saturday",
7 : "Sunday" }
allYearDates = []
w_days_numbers = []
# today
now = datetime.datetime.now()
# get current values for year, mon, day
year = int(now.year)
mon = int(now.month)
day = int(now.day)
# get this week number
# thisWeekN = datetime.date(year, mon, day).isocalendar()[1]
thisWeekN = datetime.datetime.utcnow().isocalendar()[1]
# get Calendar obect
c = calendar.Calendar()
# get all the days for this year in a list
for i in range(1, 13):
for d in c.itermonthdays2(year, i):
allYearDates.append(d)
# in the first seven days could be 0's as days, to continue week #'s
# these tuples need to be removed to produce accurate mapping between
# weeks and 7 days chunks
first_seven = allYearDates[:7]
no_zeros = [ d for d in first_seven if d[0] != 0]
allYearDates = no_zeros + allYearDates[8:]
# divid all days of year list into one week chuncks
lt7 = listChunks(allYearDates,7)
# get days for this week number
thisWeekDays = lt7[thisWeekN]
# remove right part of the days tuple
sevenDaysL = [x for x,y in thisWeekDays]
# get week day numbers 1-7 for this week
for d in sevenDaysL:
w_days_numbers.append(datetime.date(year, mon, d).isocalendar()[2])
# zip week day numbers with the dates
zl = zip(sevenDaysL, w_days_numbers)
# month number prefix
if thisWeekN == 1:
prefix = 'st'
elif thisWeekN == 2:
prefix = 'nd'
elif thisWeekN == 3:
prefix = 'rd'
else:
prefix= "th"
# print heading
print("\n 7 days of this {}{} week: of the year {}"
.format(thisWeekN,prefix, year))
print("-------------------------------------------")
# print results
for el in list(zl):
if el[0] == day:
print("* {} * {}".format(el[0],days[el[1]]))
else:
print(" {} {}".format(el[0],days[el[1]]))
mytime= datetime.datetime.now().strftime("%Y-%m-%d %H-%M")
print("================")
print(mytime)
days_of_the_current_numbered_week()
输出
第 39 周的 7 天：2015 年
17 星期四
18 星期五
19 周六
20 周日
21 星期一
22 星期二
23 星期三