python - 填充 NAs : Min of values in group

Question

这是我的 DataFrame。

df = pd.DataFrame ( {'CNN': ['iphone 11 63 GB TMO','iphone 11 128 GB ATT','iphone 11 other carrier','iphone 12 256 GB TMO','iphone 12 64 GB TMO','iphone 12 other carrier'], 
                            'Family Name':['iphone 11', 'iphone 11', 'iphone 11', 'iphone 12', 'iphone 12', 'iphone 12'],
                            'Storage': [63, 128,np.nan, 256,64, np.nan]})

Output:

                       CNN Family Name  Storage
0      iphone 11 63 GB TMO   iphone 11     63.0
1     iphone 11 128 GB ATT   iphone 11    128.0
2  iphone 11 other carrier   iphone 11      NaN
3     iphone 12 256 GB TMO   iphone 12    256.0
4      iphone 12 64 GB TMO   iphone 12     64.0
5  iphone 12 other carrier   iphone 12      NaN

我想要实现的是找到 NA。标准是来自组(姓氏)的最小存储量。我尝试按 fillna(min()) 进行分组，但它似乎不起作用。

#Tried
df["Storage"] = df.groupby("Family Name").apply(lambda x: x.fillna(x.min()))

这是预期的最终输出。

Expected Output:

                       CNN Family Name  Storage
0      iphone 11 63 GB TMO   iphone 11     63.0
1     iphone 11 128 GB ATT   iphone 11    128.0
2  iphone 11 other carrier   iphone 11     63.0
3     iphone 12 256 GB TMO   iphone 12    256.0
4      iphone 12 64 GB TMO   iphone 12     64.0
5  iphone 12 other carrier   iphone 12     64.0

Answer 1

使用groupby.transform与 fillna :

df['Storage'] = df['Storage'].fillna(df.groupby('Family Name')['Storage'].transform('min'))

或者，可能更有效，groupby.min和 map与 boolean indexing取决于 DataFrame 的大小、NaN 的数量(很少)和组的数量:

s = df.groupby('Family Name')['Storage'].min()
m = df['Storage'].isna()
df.loc[m, 'Storage'] = df.loc[m, 'Family Name'].map(s)

输出:

                       CNN Family Name  Storage
0      iphone 11 63 GB TMO   iphone 11     63.0
1     iphone 11 128 GB ATT   iphone 11    128.0
2  iphone 11 other carrier   iphone 11     63.0
3     iphone 12 256 GB TMO   iphone 12    256.0
4      iphone 12 64 GB TMO   iphone 12     64.0
5  iphone 12 other carrier   iphone 12     64.0