Pandas >= 0.16

pd.Series.mode is available!

Use groupby, GroupBy.agg, and apply the pd.Series.mode function to each group:

source.groupby(['Country','City'])['Short name'].agg(pd.Series.mode)

Country  City            
Russia   Sankt-Petersburg    Spb
USA      New-York             NY
Name: Short name, dtype: object

If this is needed as a DataFrame, use

source.groupby(['Country','City'])['Short name'].agg(pd.Series.mode).to_frame()

                         Short name
Country City                       
Russia  Sankt-Petersburg        Spb
USA     New-York                 NY

The useful thing about Series.mode is that it always returns a Series, making it very compatible with agg and apply, especially when reconstructing the groupby output. It is also faster.

# Accepted answer.
%timeit source.groupby(['Country','City']).agg(lambda x:x.value_counts().index[0])
# Proposed in this post.
%timeit source.groupby(['Country','City'])['Short name'].agg(pd.Series.mode)

5.56 ms ± 343 µs per loop (mean ± std. dev. of 7 runs, 100 loops each)
2.76 ms ± 387 µs per loop (mean ± std. dev. of 7 runs, 100 loops each)

Dealing with Multiple Modes

Series.mode also does a good job when there are multiple modes:

source2 = source.append(
    pd.Series({'Country': 'USA', 'City': 'New-York', 'Short name': 'New'}),
    ignore_index=True)

# Now `source2` has two modes for the 
# ("USA", "New-York") group, they are "NY" and "New".
source2

  Country              City Short name
0     USA          New-York         NY
1     USA          New-York        New
2  Russia  Sankt-Petersburg        Spb
3     USA          New-York         NY
4     USA          New-York        New

source2.groupby(['Country','City'])['Short name'].agg(pd.Series.mode)

Country  City            
Russia   Sankt-Petersburg          Spb
USA      New-York            [NY, New]
Name: Short name, dtype: object

Or, if you want a separate row for each mode, you can use GroupBy.apply:

source2.groupby(['Country','City'])['Short name'].apply(pd.Series.mode)

Country  City               
Russia   Sankt-Petersburg  0    Spb
USA      New-York          0     NY
                           1    New
Name: Short name, dtype: object

If you don't care which mode is returned as long as it's either one of them, then you will need a lambda that calls mode and extracts the first result.

source2.groupby(['Country','City'])['Short name'].agg(
    lambda x: pd.Series.mode(x)[0])

Country  City            
Russia   Sankt-Petersburg    Spb
USA      New-York             NY
Name: Short name, dtype: object

Alternatives to (not) consider

You can also use statistics.mode from python, but...

source.groupby(['Country','City'])['Short name'].apply(statistics.mode)

Country  City            
Russia   Sankt-Petersburg    Spb
USA      New-York             NY
Name: Short name, dtype: object

...it does not work well when having to deal with multiple modes; a StatisticsError is raised. This is mentioned in the docs:

If data is empty, or if there is not exactly one most common value, StatisticsError is raised.

But you can see for yourself...

statistics.mode([1, 2])
# ---------------------------------------------------------------------------
# StatisticsError                           Traceback (most recent call last)
# ...
# StatisticsError: no unique mode; found 2 equally common values

The issue with using x.mode()[0] is that pd.Series(['A', 'B', 'C']).mode() evaluates to ['A', 'B', 'C']. Meanwhile, pd.Series(['A', 'A', 'B']).mode() evaluates to ['A'].

Here is a function that will return the mode (if there is only one) and "x" if there is a tie (i.e., multiple modes).

import pandas as pd
lst = [["High", "A"], ["High", "A"], ["High", "B"],["Medium", "A"], ["Medium", "B"], ["Medium", "C"]]
df = pd.DataFrame(lst, columns=["Class", "Grade"])

def get_mode_or_x(series):
    mode = series.mode()
    if mode.size == 1:
        return mode[0]
    return "x"

df.loc[:, "majority_vote"] = df.groupby("Class")["Grade"].transform(get_mode_or_x)

This question is really a duplicate of this Stack Overflow post. The most intuitive solution is to use pd.Series.mode in groupby but the fastest solution is to drop duplicates of a value_counts result.

counts = df.value_counts(['season', 'league_name', 'home_winner'])
result = counts[~counts.droplevel('home_winner').index.duplicated()]

A toy example / output:

df = pd.DataFrame({
    'season': [1, 1, 1, 1, 1, 1], 
    'league_name': [*'AAABBB'], 
    'home_winner':[*'abaded']
})

counts = df.value_counts(['season', 'league_name', 'home_winner'])
result = counts[~counts.droplevel('home_winner').index.duplicated()]
print(result)

season  league_name  home_winner
1       A            a              2
        B            d              2