Mos*_*man 14 python group-by dataframe pandas pandas-groupby
我有一个数据框,其中包含日期和公共假期
Date WeekNum Public_Holiday
1/1/2015 1 1
2/1/2015 1 0
3/1/2015 1 0
4/1/2015 1 0
5/1/2015 1 0
6/1/2015 1 0
7/1/2015 1 0
8/1/2015 2 0
9/1/2015 2 0
10/1/2015 2 0
11/1/2015 2 0
12/1/2015 2 0
13/1/2015 2 0
我必须创建一个名为Public_Holiday_Week的条件列,如果该特定周有公共假日,则该列应返回1
我希望看到这样的输出
Date WeekNum Public_Holiday Public_Holiday_Week
1/1/2015 1 1 1
2/1/2015 1 0 1
3/1/2015 1 0 1
4/1/2015 1 0 1
5/1/2015 1 0 1
6/1/2015 1 0 1
7/1/2015 1 0 1
8/1/2015 2 0 0
9/1/2015 2 0 0
10/1/2015 2 0 0
11/1/2015 2 0 0
12/1/2015 2 0 0
13/1/2015 2 0 0
我尝试使用np.where
df['Public_Holiday_Week'] = np.where(df['Public_Holiday']==1,1,0)
但是,如果不是公众假期,它适用于本周其他日子的0.
我必须在这里申请滚动吗？感谢您的帮助
为了提高性能,不要使用groupby,而是使用WeekNum至少一个1然后选择值isin,最后将布尔掩码转换为ints:
weeks = df.loc[df['Public_Holiday'].eq(1), 'WeekNum']
df['Public_Holiday_Week'] = df['WeekNum'].isin(weeks).astype(int)
print (df)
Date WeekNum Public_Holiday Public_Holiday_Week
0 1/1/2015 1 1 1
1 2/1/2015 1 0 1
2 3/1/2015 1 0 1
3 4/1/2015 1 0 1
4 5/1/2015 1 0 1
5 6/1/2015 1 0 1
6 7/1/2015 1 0 1
7 8/1/2015 2 0 0
8 9/1/2015 2 0 0
9 10/1/2015 2 0 0
10 11/1/2015 2 0 0
11 12/1/2015 2 0 0
12 13/1/2015 2 0 0
正如@Mohamed Thasin指出的那样,如果有必要可以按周分组,但随后得到不同的输出,因为不同的week数字:
df['weeks'] = pd.to_datetime(df['Date'], dayfirst=True).dt.week
weeks = df.loc[df['Public_Holiday'].eq(1), 'weeks']
df['Public_Holiday_Week'] = df['weeks'].isin(weeks).astype(int)
print (df)
Date WeekNum Public_Holiday weeks Public_Holiday_Week
0 1/1/2015 1 1 1 1
1 2/1/2015 1 0 1 1
2 3/1/2015 1 0 1 1
3 4/1/2015 1 0 1 1
4 5/1/2015 1 0 2 0
5 6/1/2015 1 0 2 0
6 7/1/2015 1 0 2 0
7 8/1/2015 2 0 2 0
8 9/1/2015 2 0 2 0
9 10/1/2015 2 0 2 0
10 11/1/2015 2 0 2 0
11 12/1/2015 2 0 3 0
12 13/1/2015 2 0 3 0
resample并跳过WeekNum列的使用.df.assign(
Public_Holiday_Week=
df.resample('W-Wed', on='Date').Public_Holiday.transform('max')
)
Date WeekNum Public_Holiday Public_Holiday_Week
0 2015-01-01 1 1 1
1 2015-01-02 1 0 1
2 2015-01-03 1 0 1
3 2015-01-04 1 0 1
4 2015-01-05 1 0 1
5 2015-01-06 1 0 1
6 2015-01-07 1 0 1
7 2015-01-08 2 0 0
8 2015-01-09 2 0 0
9 2015-01-10 2 0 0
10 2015-01-11 2 0 0
11 2015-01-12 2 0 0
12 2015-01-13 2 0 0