获取DataFrame的Datetime列的工作日/星期几

Question

我有一个df像以下一样的DataFrame (摘录,'Timestamp'是索引):

Timestamp              Value
2012-06-01 00:00:00     100
2012-06-01 00:15:00     150
2012-06-01 00:30:00     120
2012-06-01 01:00:00     220
2012-06-01 01:15:00      80
...and so on.

我需要一个新列df['weekday'],其中包含时间戳的相应工作日/星期几.

我怎么能得到这个？

Answer 1

使用新dt.dayofweek属性:

In [2]:

df['weekday'] = df['Timestamp'].dt.dayofweek
df
Out[2]:
            Timestamp  Value  weekday
0 2012-06-01 00:00:00    100        4
1 2012-06-01 00:15:00    150        4
2 2012-06-01 00:30:00    120        4
3 2012-06-01 01:00:00    220        4
4 2012-06-01 01:15:00     80        4

在Timestamp您的索引的情况下,您需要重置索引,然后调用dt.dayofweek属性:

In [14]:

df = df.reset_index()
df['weekday'] = df['Timestamp'].dt.dayofweek
df
Out[14]:
            Timestamp  Value  weekday
0 2012-06-01 00:00:00    100        4
1 2012-06-01 00:15:00    150        4
2 2012-06-01 00:30:00    120        4
3 2012-06-01 01:00:00    220        4
4 2012-06-01 01:15:00     80        4

奇怪的是,如果您尝试从索引创建一个系列以便不重置索引,您将获得NaN值,使用reset_index调用dt.dayofweek属性的结果而不将结果reset_index返回到原始df:

In [16]:

df['weekday'] = pd.Series(df.index).dt.dayofweek
df
Out[16]:
                     Value  weekday
Timestamp                          
2012-06-01 00:00:00    100      NaN
2012-06-01 00:15:00    150      NaN
2012-06-01 00:30:00    120      NaN
2012-06-01 01:00:00    220      NaN
2012-06-01 01:15:00     80      NaN
In [17]:

df['weekday'] = df.reset_index()['Timestamp'].dt.dayofweek
df
Out[17]:
                     Value  weekday
Timestamp                          
2012-06-01 00:00:00    100      NaN
2012-06-01 00:15:00    150      NaN
2012-06-01 00:30:00    120      NaN
2012-06-01 01:00:00    220      NaN
2012-06-01 01:15:00     80      NaN

编辑

正如用户@joris所指出的那样,您只需访问weekday索引的属性,以便以下工作并且更紧凑:

df['Weekday'] = df.index.weekday

Answer 2

您可以通过以下方式获得：

import datetime
df['weekday'] = pd.Series(df.index).dt.day_name()

Answer 3

如果Timestamp为datatime，则可以使用：
df['weekday'] = df['Timestamp'].apply(lambda x: x.weekday())

要么

df['weekday'] = pd.to_datetime(df['Timestamp']).apply(lambda x: x.weekday())