您可以使用drop_duplicates，然后使用set_axis重新索引输出，以强制在前几行上对齐索引：

s = df['B'].drop_duplicates()
#s = df.drop_duplicates()['B'] # alternative if you want to consider A+B

df['B'] = s.set_axis(df.index[:len(s)])

• 注意：此解决方案适用于df任何原始索引，而不仅仅适用于范围索引。*

输出：

A    B
0  1  1.0
1  1  2.0
2  2  3.0
3  3  4.0
4  3  NaN
5  4  NaN
6  4  NaN

你能做到

df['B'] = df['B'].drop_duplicates().reset_index(drop=True)
# or with DataFrame.drop_duplicates which can take a ignore_index parameter.
df['B'] = df[['B']].drop_duplicates(ignore_index=True)

print(df)

   A    B
0  1  1.0
1  1  2.0
2  2  3.0
3  3  4.0
4  3  NaN
5  4  NaN
6  4  NaN

如果默认索引从列表中重新创建列：

df['B'] = pd.Series(df['B'].drop_duplicates().tolist())
#alternative
#df['B'] = pd.Series(pd.unique(df['B']).tolist())
print (df)
   A    B
0  1  1.0
1  1  2.0
2  2  3.0
3  3  4.0
4  3  NaN
5  4  NaN
6  4  NaN

如果有索引也过滤，则首先按列表长度归纳：

L = df['B'].drop_duplicates().tolist()
#L = pd.unique(df['B']).tolist()
df['B'] = pd.Series(L, index=df.index[:len(L)])

或者：

a = pd.unique(df['B'])
df['B'] = np.hstack([a, np.full((len(df) - len(a), ), np.nan)])
print (df)
   A    B
0  1  1.0
1  1  2.0
2  2  3.0
3  3  4.0
4  3  NaN
5  4  NaN
6  4  NaN