假设我有以下DataFrame:
df = pd.DataFrame({'player': ['LBJ', 'LBJ', 'LBJ', 'Kyrie', 'Kyrie', 'LBJ', 'LBJ'],
'points': [25, 32, 26, 21, 29, 21, 35]})
如何执行ffill相反的操作,以便获得以下DataFrame:
df = pd.DataFrame({'player': ['LBJ', np.nan, np.nan, 'Kyrie', np.nan, 'LBJ', np.nan],
'points': [25, 32, 26, 21, 29, 21, 35]})
也就是说,我想用NaN直接填充重复的值。
到目前为止,这是我想要的,但是我希望有一个内置的pandas方法或更好的方法:
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
最佳答案
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
关于python - 执行与 Pandas 填充相反的操作,我们在Stack Overflow上找到一个类似的问题:https://stackoverflow.com/questions/46479437/