python,如何知道一年中的周数,周六作为一周的第一天

Question

我正在使用python/pandas,并想知道如何获得一周中的星期数,而星期六作为一周的第一天.我做了很多搜索,但一直是周一或周日作为一周的第一天...请帮助...谢谢

谢谢大家!真的很感谢你所有的快速答案..但我必须道歉,我没有清楚地提出我的问题.

我想知道一年中的周数.例如,2015-08-09是第32周,而星期一是星期的第一天,但​​是第33周,而星期六是第一天.

谢谢@Cyphase和大家,我改变了一些Cyphase的代码并且它有效.

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

Answer 1

如果我理解正确,以下是您想要的:

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

上面的使用dt.week和shifts乘2行,然后向前填充NaN值.

Answer 2

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))

第 4 天（星期四）的输出：

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))