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