Get weekday/day-of-week for Datetime column of DataFrame
You can get with this way:
import datetime
df['weekday'] = pd.Series(df.index).dt.day_name()
If the Timestamp
column is a datetime
value, then you can just use:df['weekday'] = df['Timestamp'].apply(lambda x: x.weekday())
or
df['weekday'] = pd.to_datetime(df['Timestamp']).apply(lambda x: x.weekday())
Use the new dt.dayofweek
property:
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
In the situation where the Timestamp
is your index you need to reset the index and then call the dt.dayofweek
property:
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
Strangely if you try to create a series from the index in order to not reset the index you get NaN
values as does using the result of reset_index
to call the dt.dayofweek
property without assigning the result of reset_index
back to the original 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
EDIT
As pointed out to me by user @joris you can just access the weekday
attribute of the index so the following will work and is more compact:
df['Weekday'] = df.index.weekday