Question

我的previous question的扩展名。我有一个源数据框，其中包含三列：客户，日期和项目。我想添加一个包含“项目历史记录”的新列，该列是该客户在早期（由日期定义）行中所有项目的数组。 如果客户在同一日期进行了多次购买，则该行的任何一项都不应在另一项的历史记录中列出。

因此，鉴于此示例数据：

df = pd.DataFrame({'Customer':['Bert', 'Bert', 'Bert', 'Bert', 'Bert', 'Ernie', 'Ernie', 'Ernie', 'Ernie', 'Steven', 'Steven'], 'Date':['01/01/2019', '15/01/2019', '20/01/2019', '20/01/2019', '22/01/2019', '01/01/2019', '15/01/2019', '20/01/2019', '22/01/2019', '01/01/2019' ,'15/01/2019'], 'Item':['Bread', 'Cheese', 'Apples', 'Pears', 'Toothbrush', 'Toys', 'Shellfish', 'Dog', 'Yoghurt', 'Toilet', 'Dominos']})

Customer    Date    Item
Bert    01/01/2019  Bread
Bert    15/01/2019  Cheese
Bert    20/01/2019  Apples
Bert    20/01/2019  Pears
Bert    22/01/2019  Toothbrush
Ernie   01/01/2019  Toys
Ernie   15/01/2019  Shellfish
Ernie   20/01/2019  Dog
Ernie   22/01/2019  Yoghurt
Steven  01/01/2019  Toilet
Steven  15/01/2019  Dominos

我想看到的输出是：

Customer    Date    Item        Item History
Bert    01/01/2019  Bread       NaN
Bert    15/01/2019  Cheese      [Bread]
Bert    20/01/2019  Apples      [Bread, Cheese]
Bert    20/01/2019  Pears       [Bread, Cheese]
Bert    22/01/2019  Toothbrush  [Bread, Cheese, Apples, Pears]
Ernie   01/01/2019  Toys        NaN
Ernie   15/01/2019  Shellfish   [Toys]
Ernie   20/01/2019  Dog         [Toys, Shellfish]
Ernie   22/01/2019  Yoghurt     [Toys, Shellfish, Dog]
Steven  01/01/2019  Toilet      NaN
Steven  15/01/2019  Dominos     [Toilet]

请注意，对于Bert在20/01/2019的购买，``历史记录''列均未包含对方的商品。对于他在2019年1月22日购买的商品，都包括了2019年1月20日起的两个商品。

上一个问题的答案是列表理解的精妙之处，形式为：

df['Item History'] = [x.Item[:i].tolist() for j, x in df.groupby('Customer') 
                                          for i in range(len(x))]

df.loc[~df['Item History'].astype(bool), 'Item History']= np.nan

但是显然x.Item[:i]中的“ i”需要计算出日期与当前行不同的最后一行。任何关于实现该目标的建议都将受到赞赏。

Answer 1

使用apply和np.cumsum()的另一种方式：

#aggregates Item as list per 'Customer'& 'Date'
m=df.groupby(['Customer','Date'])['Item'].apply(lambda x: 
                               [*itertools.chain.from_iterable([x])])
#groups each Customer and cumsum the list with shift
n=m.groupby(level=0).apply(lambda x:np.cumsum(x).shift())
df.set_index(['Customer','Date']).assign(Item=n).reset_index() #assign back

   Customer        Date                            Item
0      Bert  01/01/2019                             NaN
1      Bert  15/01/2019                         [Bread]
2      Bert  20/01/2019                 [Bread, Cheese]
3      Bert  20/01/2019                 [Bread, Cheese]
4      Bert  22/01/2019  [Bread, Cheese, Apples, Pears]
5     Ernie  01/01/2019                             NaN
6     Ernie  15/01/2019                          [Toys]
7     Ernie  20/01/2019               [Toys, Shellfish]
8     Ernie  22/01/2019          [Toys, Shellfish, Dog]
9    Steven  01/01/2019                             NaN
10   Steven  15/01/2019                        [Toilet]

Answer 2

想法是通过DataFrame.duplicated来区分每组重复的值，然后将这些值替换为NaN并用正向填充缺失值。

每组的第一个值始终为空字符串，因此不必按组替换：

df['Item History'] = [x.Item[:i].tolist() for j, x in df.groupby('Customer') 
                                          for i in range(len(x))]

df['Item History'] = df['Item History'].mask(df.duplicated(['Customer','Date'])).ffill()

df.loc[~df['Item History'].astype(bool), 'Item History']= np.nan
print (df)
   Customer        Date        Item                    Item History
0      Bert  01/01/2019       Bread                             NaN
1      Bert  15/01/2019      Cheese                         [Bread]
2      Bert  20/01/2019      Apples                 [Bread, Cheese]
3      Bert  20/01/2019       Pears                 [Bread, Cheese]
4      Bert  22/01/2019  Toothbrush  [Bread, Cheese, Apples, Pears]
5     Ernie  01/01/2019        Toys                             NaN
6     Ernie  15/01/2019   Shellfish                          [Toys]
7     Ernie  20/01/2019         Dog               [Toys, Shellfish]
8     Ernie  22/01/2019     Yoghurt          [Toys, Shellfish, Dog]
9    Steven  01/01/2019      Toilet                             NaN
10   Steven  15/01/2019     Dominos                        [Toilet]

Answer 3

仅使用apply一个可能更简单的答案-这可能比其他方法慢：

df['item history'] = df.apply(lambda x: 
            [i for i in list(df.loc[(df.Date<x.Date)&(df.Customer==x.Customer),'Item'])], axis=1)

结果：

   Customer               ...                                  item history
0      Bert               ...                                            []
1      Bert               ...                                       [Bread]
2      Bert               ...                               [Bread, Cheese]
3      Bert               ...                               [Bread, Cheese]
4      Bert               ...                [Bread, Cheese, Apples, Pears]
5     Ernie               ...                                            []
6     Ernie               ...                                        [Toys]
7     Ernie               ...                             [Toys, Shellfish]
8     Ernie               ...                        [Toys, Shellfish, Dog]
9    Steven               ...                                            []
10   Steven               ...                                      [Toilet]

如果要列出唯一项，则可能要在结果中添加list(set())。

在熊猫数据框中串联来自具有不同日期的较早行的值

3 个答案: