熊猫在组内排序然后聚合

Question

我正在做搜索引擎的查询分析。用户可以在一个会话的不同时间在谷歌搜索引擎上一一搜索不同的查询。

我有多个字段的数据：session_id、log_time、query、feature_i 等。我想按 session_id 分组，然后将 concat 几行合并为一个 log_time 的顺序。这样输出数据将以时间序列的方式表示用户的行为。

数据集

代码：

toy_data = pd.DataFrame({'session_id':[1,2,1,2,3,3,],
             'log_time':[4,5,6,1,2,3],
             'query':['hi','dude','pandas','groupby','sort','agg'],
             'cate_feat_0':['apple','banana']*3,
             'num_feat_0':[1,2,3,4,5,6]})
print(toy_data)

输出：

       session_id  log_time query cate_feat_0  num_feat_0
0           1         4       hi       apple           1
1           2         5     dude      banana           2
2           1         6   pandas       apple           3
3           2         1  groupby      banana           4
4           3         2     sort       apple           5
5           3         3      agg      banana           6

我想要什么：

## note that all list are sorted by log time with each session_id group
session_id    query_list    log_time_list cate_feat_0_list    num_feat_0_list
    1         [hi, pandas]   [4,6]        [apple, apple]      [1,3]
    2         [groupby, dude] [1,5]       [banana, banana]    [4,2]  
    3         [sort,agg]      [2,3]       [apple, banana]     [5,6]

我的尝试

首先我们用代码进行分组和聚合：

toy_data_res = toy_data.groupby('session_id').agg({'query':list, 'log_time':list, 'cate_feat_0':list, 'num_feat_0':list})
toy_data_res

给予：

                      query log_time       cate_feat_0 num_feat_0
session_id                                                       
1              [hi, pandas]   [4, 6]    [apple, apple]     [1, 3]
2           [dude, groupby]   [5, 1]  [banana, banana]     [2, 4]
3               [sort, agg]   [2, 3]   [apple, banana]     [5, 6]

然后我们在每个会话中用代码排序：

for i in toy_data_res.index:
    sort_index = np.argsort(toy_data_res.loc[i,'log_time']) ##  get time order with in group
    for col in toy_data_res.columns.values:
        toy_data_res.loc[i,col] = [toy_data_res.loc[i,col][j] for j in sort_index] ## sort values in cols 
toy_data_res

给予：

                      query log_time       cate_feat_0 num_feat_0
session_id                                                       
1              [hi, pandas]   [4, 6]    [apple, apple]     [1, 3]
2           [groupby, dude]   [1, 5]  [banana, banana]     [4, 2]
3               [sort, agg]   [2, 3]   [apple, banana]     [5, 6]

我的方法是快慢。有没有更好的方法来做groupby -> sort with in group -> aggregation？

提示： We can use STRING_AGG or GROUP_CONCAT in SQL to do within group sorting.

Answer 1

在groupby 之前使用DataFrame.sort_values，如果需要应用相同的功能，可以使用列名列表：

df = (toy_data.sort_values(['session_id','log_time'])
              .groupby('session_id')[['query','log_time','cate_feat_0', 'num_feat_0']]
              .agg(list))

    
print (df)
                      query log_time       cate_feat_0 num_feat_0
session_id                                                       
1              [hi, pandas]   [4, 6]    [apple, apple]     [1, 3]
2           [groupby, dude]   [1, 5]  [banana, banana]     [4, 2]
3               [sort, agg]   [2, 3]   [apple, banana]     [5, 6]

Answer 2

尝试在 groupby 之前按 session_id 和 log_time 排序

 df = pd.DataFrame({'session_id':[1,2,1,2,3,3,],
         'log_time':[4,5,6,1,2,3],
         'query':['hi','dude','pandas','groupby','sort','agg'],
         'cate_feat_0':['apple','banana']*3,
         'num_feat_0':[1,2,3,4,5,6]})

 df=df.sort_values(by=['session_id','log_time'])

 grouped=df.groupby('session_id') 
 ['log_time','query','cate_feat_0','num_feat_0'].agg(list)
 print(grouped)

输出

               log_time    query            cate_feat_0       num_feat_0
  session_id                                                       
  1            [4, 6]      [hi, pandas]     [apple, apple]    [1, 3]
  2            [1, 5]      [groupby, dude]  [banana, banana]  [4, 2]
  3            [2, 3]      [sort, agg]      [apple, banana]   [5, 6]