perform operation opposite to pandas ffill

Question

Let's say I have the following DataFrame:

df = pd.DataFrame({'player': ['LBJ', 'LBJ', 'LBJ', 'Kyrie', 'Kyrie', 'LBJ', 'LBJ'],
                   'points': [25, 32, 26, 21, 29, 21, 35]})

How can I perform the operation opposite of ffill so I can get the following DataFrame:

df = pd.DataFrame({'player': ['LBJ', np.nan, np.nan, 'Kyrie', np.nan, 'LBJ', np.nan],
                   'points': [25, 32, 26, 21, 29, 21, 35]})

That is, I want to fill directly repeated values with NaN.

Here's what I have so far but I'm hoping there's a built-in pandas method or a better approach:

for i, (index, row) in enumerate(df.iterrows()):
    if i == 0:
        continue
    go_back = 1
    while True:
        past_player = df.ix[i-go_back, 'player']
        if pd.isnull(past_player):
            go_back += 1
            continue
        if row['player'] == past_player:
            df.set_value(index, 'player', value=np.nan)
        break

Answer 1

ffinv = lambda s: s.mask(s == s.shift())
df.assign(player=ffinv(df.player))

  player  points
0    LBJ      25
1    NaN      32
2    NaN      26
3  Kyrie      21
4    NaN      29
5    LBJ      21
6    NaN      35

Answer 2

Probably not the most efficient solution but working would be to use itertools.groupby and itertools.chain:

>>> df['player'] = list(itertools.chain.from_iterable([key] + [float('nan')]*(len(list(val))-1) 
                        for key, val in itertools.groupby(df['player'].tolist())))
>>> df
  player  points
0    LBJ      25
1    NaN      32
2    NaN      26
3  Kyrie      21
4    NaN      29
5    LBJ      21
6    NaN      35

More specifically this illustrates how it works:

for key, val in itertools.groupby(df['player']):
    print([key] + [float('nan')]*(len(list(val))-1))

giving:

['LBJ', nan, nan]
['Kyrie', nan]
['LBJ', nan]

which is then "chained" together.