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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。

我有几个字段的数据：session_id, log_time, query,feature_i等。我想按顺序session_id将concat几行分组为一行。这样输出数据将以时间序列的方式表示用户的行为。log_time

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

我想要的是：

## 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 groupby 和 agg：

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

给出：

                      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使用代码进行排序：

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

给出：

                      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。有没有更好的办法groupby -> sort with in group -> aggregation？

Tips： 我们可以在SQL中使用STRING_AGGorGROUP_CONCAT来做组内排序。

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2 回答 2

3

使用DataFrame.sort_valuesbefore groupby，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]
于 2021-02-24T07:24:54.600 回答
0

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]
于 2021-02-25T23:15:35.223 回答