我有一组股票信息,日期时间设置为指数,股市只在工作日开放,所以我所有的行都是工作日,这很好,我想确定一行是本周的开始还是本周的结束,由于假期的原因,这可能不总是在周一/周五。一个更好的想法是确定数据帧中是否有下一天/前一天的行条目(因为我的数据保证只存在于工作日(,但我不知道如何计算。以下是我的数据示例:
date day_of_week day_of_month day_of_year month_of_year
5/1/2017 0 1 121 5
5/2/2017 1 2 122 5
5/3/2017 2 3 123 5
5/4/2017 3 4 124 5
5/8/2017 0 8 128 5
5/9/2017 1 9 129 5
5/10/2017 2 10 130 5
5/11/2017 3 11 131 5
5/12/2017 4 12 132 5
5/15/2017 0 15 135 5
5/16/2017 1 16 136 5
5/17/2017 2 17 137 5
5/18/2017 3 18 138 5
5/19/2017 4 19 139 5
5/23/2017 1 23 143 5
5/24/2017 2 24 144 5
5/25/2017 3 25 145 5
5/26/2017 4 26 146 5
5/30/2017 1 30 150 5
这是我当前的代码
# Date fields
def DateFields(df_input):
dates = df_input.index.to_series()
df_input['day_of_week'] = dates.dt.dayofweek
df_input['day_of_month'] = dates.dt.day
df_input['day_of_year'] = dates.dt.dayofyear
df_input['month_of_year'] = dates.dt.month
df_input['isWeekStart'] = "No" #<--- Need help here
df_input['isWeekEnd'] = "No" #<--- Need help here
df_input['date'] = dates.dt.strftime('%Y-%m-%d')
return df_input
如何计算一行是周初还是周末?
我正在寻找的示例:
date day_of_week day_of_month day_of_year month_of_year isWeekStart isWeekEnd
5/1/2017 0 1 121 5 1 0
5/2/2017 1 2 122 5 0 0
5/3/2017 2 3 123 5 0 0
5/4/2017 3 4 124 5 0 1 # short week, Thursday is last work day
5/8/2017 0 8 128 5 1 0
5/9/2017 1 9 129 5 0 0
5/10/2017 2 10 130 5 0 0
5/11/2017 3 11 131 5 0 0
5/12/2017 4 12 132 5 0 1
5/15/2017 0 15 135 5 1 0
5/16/2017 1 16 136 5 0 0
5/17/2017 2 17 137 5 0 0
5/18/2017 3 18 138 5 0 0
5/19/2017 4 19 139 5 0 1
5/23/2017 1 23 143 5 1 0 # short week, Tuesday is first work day
5/24/2017 2 24 144 5 0 0
5/25/2017 3 25 145 5 0 0
5/26/2017 4 26 146 5 0 1
5/30/2017 1 30 150 5 1 0
编辑:我忘了有些假期是在周中,在这种情况下,如果它能把这些当作一个单独的"周";前后相应标记。尽管如果它还不够聪明,那么度过一个漫长的周末将是一个良好的开端。
以下是BusinessDay
:的一个想法
prev_working_day = df['date'] - pd.tseries.offsets.BusinessDay(1)
df['isFirstWeekDay'] = (df['date'].dt.isocalendar().week !=
prev_working_day.dt.isocalendar().week)
上一个工作日也是如此。请注意,默认的假日日历是"美国"。看看这篇不同的文章。
输出:
date day_of_week day_of_month day_of_year month_of_year isFirstWeekDay
0 2017-05-01 0 1 121 5 True
1 2017-05-02 1 2 122 5 False
2 2017-05-03 2 3 123 5 False
3 2017-05-04 3 4 124 5 False
4 2017-05-08 0 8 128 5 True
5 2017-05-09 1 9 129 5 False
6 2017-05-10 2 10 130 5 False
7 2017-05-11 3 11 131 5 False
8 2017-05-12 4 12 132 5 False
9 2017-05-15 0 15 135 5 True
10 2017-05-16 1 16 136 5 False
11 2017-05-17 2 17 137 5 False
12 2017-05-18 3 18 138 5 False
13 2017-05-19 4 19 139 5 False
14 2017-05-23 1 23 143 5 False
15 2017-05-24 2 24 144 5 False
16 2017-05-25 3 25 145 5 False
17 2017-05-26 4 26 146 5 False
18 2017-05-30 1 30 150 5 False
这里有一种使用每周分组的方法。
df['date'] = pd.to_datetime(df['date'])
business_days = df.assign(date_copy = df['date']).groupby(pd.Grouper(key='date_copy', freq='W'))['date'].apply(list).to_frame()
business_days['isWeekStart'] = business_days['date'].apply(lambda x: [1 if i == min(x) else 0 for i in x])
business_days['isWeekEnd'] = business_days['date'].apply(lambda x: [1 if i == max(x) else 0 for i in x])
business_days = business_days.apply(pd.Series.explode)
pd.merge(df, business_days, left_on='date', right_on='date')
输出:
date day_of_week day_of_month day_of_year month_of_year isWeekStart isWeekEnd
0 2017-05-01 0 1 121 5 1 0
1 2017-05-02 1 2 122 5 0 0
2 2017-05-03 2 3 123 5 0 0
3 2017-05-04 3 4 124 5 0 1
4 2017-05-08 0 8 128 5 1 0
5 2017-05-09 1 9 129 5 0 0
6 2017-05-10 2 10 130 5 0 0
7 2017-05-11 3 11 131 5 0 0
8 2017-05-12 4 12 132 5 0 1
9 2017-05-15 0 15 135 5 1 0
10 2017-05-16 1 16 136 5 0 0
11 2017-05-17 2 17 137 5 0 0
12 2017-05-18 3 18 138 5 0 0
13 2017-05-19 4 19 139 5 0 1
14 2017-05-23 1 23 143 5 1 0
15 2017-05-24 2 24 144 5 0 0
16 2017-05-25 3 25 145 5 0 0
17 2017-05-26 4 26 146 5 0 1
18 2017-05-30 1 30 150 5 1 1
请注意,2017-05-30
同时标记为WeekStart和WeekEnd,因为它是该周的唯一日期。