python - 用n-1替换缺失值

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例如：我有，

df = pd.DataFrame({0: [420, np.nan, 455, np.nan, np.nan, np.nan]})

df

       0
0  420.0
1    NaN
2  455.0
3    NaN
4    NaN
5    NaN

然后使用：

df[0].isnull().astype(int)

0    0
1    1
2    0
3    1
4    1
5    1
Name: 0, dtype: int64

我明白了

df[0].fillna(method='ffill') - df[0].isnull().astype(int)

0    420.0
1    419.0
2    455.0
3    454.0
4    454.0
5    454.0
Name: 0, dtype: float64

我想得到0,1,0,1,2,3，最后：
df[0]=420419455；454453452

最佳答案

groupby，cumcount

df[0].ffill() - df.groupby(df[0].notna().cumsum()).cumcount()

0    420.0
1    419.0
2    455.0
3    454.0
4    453.0
5    452.0
dtype: float64

详情
定义组

df[0].notna().cumsum()

0    1
1    1
2    2
3    2
4    2
5    2
Name: 0, dtype: int64

与groupby一起在cumcount中使用

df.groupby(df[0].notna().cumsum()).cumcount()

0    0
1    1
2    0
3    1
4    2
5    3
dtype: int64