Python 熊猫在组内排序，然后聚集

我正在做搜索引擎的查询分析。用户可以在一个会话中的不同时间在google搜索引擎上逐个搜索不同的查询

我有几个字段的数据：

session\u id

，

log\u time

，

query

，

feature\u I

，等等。我想按

session\u id

分组，然后按

log\u time

的顺序将几行

压缩成一行。因此，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：

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，我们使用代码编写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

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           [dude, groupby]   [5, 1]  [banana, banana]     [2, 4]
3               [sort, agg]   [2, 3]   [apple, banana]     [5, 6]

                      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]

然后，我们在每个会话中使用代码进行排序：

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

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           [dude, groupby]   [5, 1]  [banana, banana]     [2, 4]
3               [sort, agg]   [2, 3]   [apple, banana]     [5, 6]

                      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执行
groupby->sort with in group->aggregation

小贴士:
在
groupby
之前使用，如果需要可以应用相同的功能，请使用列名称列表：

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]

在groupby之前，尝试按会话id和日志时间进行排序

 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]

谢谢你，兄弟！我以前心碎了，没有考虑过预排序方法。记住，如果它解决了您的问题，您可以接受答案：）