Getting the average of a certain hour on weekdays over several years in a pandas dataframe

Note: Now that Series have the dt accessor it's less important that date is the index, though Date/Time still needs to be a datetime64.

Update: You can do the groupby more directly (without the lambda):

In [21]: df.groupby([df["Date/Time"].dt.year, df["Date/Time"].dt.hour]).mean()
Out[21]:
                     Value
Date/Time Date/Time
2010      0             60
          1             50
          2             52
          3             49

In [22]: res = df.groupby([df["Date/Time"].dt.year, df["Date/Time"].dt.hour]).mean()

In [23]: res.index.names = ["year", "hour"]

In [24]: res
Out[24]:
           Value
year hour
2010 0        60
     1        50
     2        52
     3        49

If it's a datetime64 index you can do:

In [31]: df1.groupby([df1.index.year, df1.index.hour]).mean()
Out[31]:
        Value
2010 0     60
     1     50
     2     52
     3     49

---

Old answer (will be slower):

Assuming Date/Time was the index* you can use a mapping function in the groupby:

In [11]: year_hour_means = df1.groupby(lambda x: (x.year, x.hour)).mean()

In [12]: year_hour_means
Out[12]:
           Value
(2010, 0)     60
(2010, 1)     50
(2010, 2)     52
(2010, 3)     49

For a more useful index, you could then create a MultiIndex from the tuples:

In [13]: year_hour_means.index = pd.MultiIndex.from_tuples(year_hour_means.index,
                                                           names=['year', 'hour'])

In [14]: year_hour_means
Out[14]:
           Value
year hour
2010 0        60
     1        50
     2        52
     3        49

* if not, then first use set_index:

df1 = df.set_index('Date/Time')

If your date/time column were in the datetime format (see dateutil.parser for automatic parsing options), you can use pandas resample as below:

year_hour_means = df.resample('H',how = 'mean')

which will keep your data in the datetime format. This may help you with whatever you are going to be doing with your data down the line.