将一行分布在共享相同键的其他行上

Question

我有一个如下所示的数据框：

+------+------------+-------+--------------+
| name |    date    | value | replacement |
+------+------------+-------+--------------+
| A    | 20/11/2016 |    10 | NaN          |
| C    | 20/11/2016 |    8  | [A,B]        |
| B    | 20/11/2016 |    12 | NaN          |
| E    | 25/12/2016 |    16 | NaN          |
| F    | 25/12/2016 |    18 | NaN          |
| D    | 25/12/2016 |    11 | [E,F]        |
+------+------------+-------+--------------+

我想做什么：
对于在 'replacement' 列中有名称列表的每一行，我希望分发它的 'value'在包含相同日期的那些替换项的行上均等。
对于前面的示例，输出将如下所示：

+------+------------+-------+------------------+
| name |    date    | value | additional value |
+------+------------+-------+------------------+
| A    | 20/11/2016 |    10 |                4 |
| B    | 20/11/2016 |    12 |                4 |
| A    | 25/12/2016 |    16 |              5.5 |
| B    | 25/12/2016 |    18 |              5.5 |
+------+------------+-------+------------------+

我设法通过拆分这些行并按名称 + 日期分组来找到一种无需创建新列即可直接执行分发的方法，但是 1/ 它太慢了 + 2/ 我确实需要创建该附加列并且找不到这样做的方法。

Answer 1

想法是按 replacement 列表的长度创建新列，其中 Series.str.len and then DataFrame.explode (pandas 0.25+) them to scalars. Divide columns value by new and merge 按原始列名添加原始列：

df1 = df.assign(new=df['replacement'].str.len()).explode('replacement')
df1['new'] = df1['value'].div(df1['new'])

df1 = df1[['name','date','value']].merge(df1[['replacement','date','new']],
                                    left_on=['name','date'],
                                    right_on=['replacement','date'])
df1['replacement'] = df1.pop('new')
print (df1)
  name        date  value  replacement
0    A  20/11/2016     10          4.0
1    B  20/11/2016     12          4.0
2    A  25/12/2016     16          5.5
3    B  25/12/2016     18          5.5

类似的解决方案，使用删除而不是选择：

df1 = df.assign(new=df['replacement'].str.len()).explode('replacement')
df1['new'] = df1['value'].div(df1['new'])

df1 = df1.drop(['replacement','new'],1).merge(df1.drop(['name','value'],1),
                                        left_on=['name','date'],
                                        right_on=['replacement','date'])
df1['replacement'] = df1.pop('new')
print (df1)
  name        date  value  replacement
0    A  20/11/2016     10          4.0
1    B  20/11/2016     12          4.0
2    A  25/12/2016     16          5.5
3    B  25/12/2016     18          5.5

Answer 2

这是另一种使用 explode（需要 pandas 0.25+）和 groupby 的方法：

m = df[[isinstance(i,list) for i in df.replacement]] #df which has lists in replacement col

g = m.explode('replacement').groupby('date') #explode and groupby by date
#drop indices of m and assign the divided value
final = df.drop(m.index).set_index('date').assign(
      replacement=(g['value'].mean()/g.size())).reset_index()

         date name  value  replacement
0  20/11/2016    A   10.0          4.0
1  20/11/2016    B   12.0          4.0
2  25/12/2016    A   16.0          5.5
3  25/12/2016    B   18.0          5.5

将一行分布在共享相同键的其他行上

Distribute a row over other rows sharing same key

python

pandas

dataframe

data-processing