在两个熊猫数据帧之间分配值

Question

考虑两个数据框：

>> import pandas as pd
>> df1 = pd.DataFrame({"category": ["foo", "foo", "bar", "bar", "bar"], "quantity": [1,2,1,2,3]})
>> print(df1)

    category    quantity
0   foo         1
1   foo         2
2   bar         1
3   bar         2
4   bar         3

>> df2 = pd.DataFrame({
            "category": ["foo", "foo", "foo", "foo", "bar", "bar", "bar", "bar", "bar", "bar"], 
            "item": ["A", "B", "C", "D", "E", "F", "G", "H", "I", "J"]
        })
>> print(df2)
      category item
0      foo      A
1      foo      B
2      foo      C
3      foo      D
4      bar      E
5      bar      F
6      bar      G
7      bar      H
8      bar      I
9      bar      J

我如何在df1（名为df3的新数据框）中创建一个新列，它加入df1的category列并分配df2中的item列。因此，创建如下内容：

>> df3 = pd.DataFrame({
           "category": ["foo", "foo", "bar", "bar", "bar"], 
           "quantity": [1,2,1,2,3],
           "item": ["A", "B,C", "E", "F,G", "H,I,J"] 
})

     category  quantity   item
0      foo         1      A
1      foo         2      B,C
2      bar         1      E
3      bar         2      F,G
4      bar         3      H,I,J

Answer 1

您可以通过quantity列和Index.repeat与DataFrame.loc重复行来创建辅助DataFrame，将索引转换为列以避免丢失indices并在两个DataFrame中创建辅助列g以通过重复的categories合并GroupBy.cumcount，然后使用DataFrame.merge和聚合join：

df11 = (df1.loc[df1.index.repeat(df1['quantity'])].reset_index()
           .assign(g = lambda x: x.groupby('category').cumcount()))

df22 = df2.assign(g = df2.groupby('category').cumcount())

df = (df11.merge(df22, on=['g','category'], how='left')
          .groupby(['index','category','quantity'])['item']
          .agg(lambda x: ','.join(x.dropna()))
          .droplevel(0)
          .reset_index())
print (df)
  category  quantity   item
0      foo         1      A
1      foo         2    B,C
2      bar         1      E
3      bar         2    F,G
4      bar         3  H,I,J

Answer 2

您可以将迭代器与 itertools.islice 一起使用：

from itertools import islice

# aggregate the items as iterator
s = df2.groupby('category')['item'].agg(iter)

# for each category, allocate as many items as needed and join
df1['item'] = (df1.groupby('category', group_keys=False)['quantity']
                  .apply(lambda g:
                         g.map(lambda x: ','.join(list(islice(s[g.name], x)))))
               )

输出：

  category  quantity   item
0      foo         1      A
1      foo         2    B,C
2      bar         1      E
3      bar         2    F,G
4      bar         3  H,I,J

请注意，如果您没有足够的物品，这将只使用可用的物品。

使用在 F 之后截断的 df2 作为输入的示例：

  category  quantity item
0      foo         1    A
1      foo         2  B,C
2      bar         1    E
3      bar         2    F
4      bar         3