pandas python，如何知道一年中的周数，星期六作为一周的第一天
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python, how to know the week number in the year of the day, Saturday as the first day of the week
提问by fuji4900
I am using python/pandas, and want to know how to get the week number in the year of one day while Saturday as the first day of the week. i did search a lot, but all the way takes either Monday or Sunday as the first day of week... Please help...thanks
我正在使用 python/pandas，并想知道如何在星期六作为一周的第一天时获取某一年的周数。我确实搜索了很多，但一直将周一或周日作为一周的第一天...请帮忙...谢谢
Thanks all! really appreciated all your quick answers..but i have to apology that i am not making my question clearly.
谢谢大家！真的很感谢你所有的快速回答......但我必须道歉，我没有清楚地提出我的问题。
I want to know the week number in the year. for example, 2015-08-09 is week 32 while Monday as first day of week, but week 33 while Saturday as first day of week.
我想知道一年中的周数。例如，2015-08-09 是第 32 周，而星期一是一周的第一天，但第 33 周而星期六是一周的第一天。
Thanks @Cyphase and everyone, I changed a bit the code of Cyphase and it works.
感谢@Cyphase 和大家，我对 Cyphase 的代码进行了一些更改，并且可以正常工作。
def week_number(start_week_on, date_=None):
assert 1 <= start_week_on <= 7 #Monday=1, Sunday=7
if not date_:
date_ = date.today()
__, normal_current_week, normal_current_day = date_.isocalendar()
print date_, normal_current_week, normal_current_day
if normal_current_day >= start_week_on:
week = normal_current_week + 1
else:
week = normal_current_week
return week
回答by EdChum
If I understand correctly the following does what you want:
如果我理解正确，以下内容可以满足您的需求：
In [101]:
import datetime as dt
import pandas as pd
?
df = pd.DataFrame({'date':pd.date_range(start=dt.datetime(2015,8,9), end=dt.datetime(2015,9,1))})
df['week'] = df['date'].dt.week.shift(-2).ffill()
df['orig week'] = df['date'].dt.week
df['day of week'] = df['date'].dt.dayofweek
df
Out[101]:
date week orig week day of week
0 2015-08-09 33 32 6
1 2015-08-10 33 33 0
2 2015-08-11 33 33 1
3 2015-08-12 33 33 2
4 2015-08-13 33 33 3
5 2015-08-14 33 33 4
6 2015-08-15 34 33 5
7 2015-08-16 34 33 6
8 2015-08-17 34 34 0
9 2015-08-18 34 34 1
10 2015-08-19 34 34 2
11 2015-08-20 34 34 3
12 2015-08-21 34 34 4
13 2015-08-22 35 34 5
14 2015-08-23 35 34 6
15 2015-08-24 35 35 0
16 2015-08-25 35 35 1
17 2015-08-26 35 35 2
18 2015-08-27 35 35 3
19 2015-08-28 35 35 4
20 2015-08-29 36 35 5
21 2015-08-30 36 35 6
22 2015-08-31 36 36 0
23 2015-09-01 36 36 1
The above uses dt.weekand shifts by 2 rows and then forward fills the NaNvalues.
回答by KiDo
import datetime
datetime.date(2015, 8, 9).isocalendar()[1]
回答by Cyphase
You could just do this:
你可以这样做：
from datetime import date
def week_number(start_week_on, date_=None):
assert 0 <= start_week_on <= 6
if not date_:
date_ = date.today()
__, normal_current_week, normal_current_day = date_.isocalendar()
if normal_current_day >= start_week_on:
week = normal_current_week
else:
week = normal_current_week - 1
return week
print("Week starts We're in")
for start_week_on in range(7):
this_week = week_number(start_week_on)
print(" day {0} week {1}".format(start_week_on, this_week))
Output on day 4 (Thursday):
第 4 天（星期四）的输出：
Week starts We're in
day 0 week 33
day 1 week 33
day 2 week 33
day 3 week 33
day 4 week 33
day 5 week 32
day 6 week 32
