这是我一直在努力解决的Python熊猫问题。假设我有一个简单的数据帧df,其中df[‘a']=[1,2,3,1,4,6]和df[‘b']=[10,20,30,40,60]。我想创建第三列"c",其中如果df[‘a’]==1,df[‘c’]=df[‘b’]的值。如果这是false,则df['c']=df['c']的前一个值。我尝试过使用np.where来实现这一点,但结果并不是我所期望的。有什么建议吗?
df = pd.DataFrame()
df['a'] = [1,2,3,1,4,6]
df['b'] = [10,20,30,40,50,60]
df['c'] = np.nan
df['c'] = np.where(df['a'] == 1, df['b'], df['c'].shift(1))
结果是:
a b c
0 1 10 10.0
1 2 20 NaN
2 3 30 NaN
3 1 40 40.0
4 4 50 NaN
5 6 60 NaN
而我本以为:
a b c
0 1 10 10.0
1 2 20 10.0
2 3 30 10.0
3 1 40 40.0
4 4 50 40.0
5 6 60 40.0
试试这个:
df.c.ffill(inplace=True)
输出:
a b c
0 1 10 10.0
1 2 20 10.0
2 3 30 10.0
3 1 40 40.0
4 4 50 40.0
5 6 60 40.0