python - 根据行条件替换 nan 值

Question

这是我的原始数据框 df_:

 index_label,id_label,morning,evening,night
 a,x,nan,eating,sleep
 b,x,shower,eating,nan
 c,x,nan,nan,nan
 d,y,work,reading,travel
 e,y,nan,reading,nan
 f,y,work,nan,nan
 g,z,shower,nan,travel
 h,z,shower,eating,nan

我尝试用基于相同 id_labels 从相同数据帧 df 中获取的非值替换 nan 值。每列“早上”、“晚上”都需要从 nan 中清除。 “夜晚”列应保持不变。

例如，我为“早上”栏目写这个

crit_nan_ = pd.isna(df_[['morning']])
df_nan_ = df_.loc[crit_nan_]
df_clean_ = df_.loc[~crit_nan_]

但是我如何获得结果数据框:

 index_label,id_label,morning,evening,night
 a,x,shower,eating,sleep
 b,x,shower,eating,nan
 c,x,shower,eating,nan
 d,y,work,reading,travel
 e,y,work,reading,nan
 f,y,work,reading,nan
 g,z,shower,eating,travel
 h,z,shower,eating,nan

Answer 1

结果数据框可以使用df.groupby获得& df.fillna :

def fill_na(x):
    return x.fillna(method="ffill").fillna(method="bfill")

for col in ("morning", "evening", ):
    d[col] = d.groupby("id_label")[col].transform(fill_na)