pandas 在组内排序然后聚合

Question

I am doing query analysis of search engine. User may search different query one by one on google search engine at different time in one session.

I have data with several field: session_id , log_time , query , feature_i , etc. I want to group by session_id and then concat several rows into one by the order of log_time . So that output data will represent user's behaviors in a time series way.

dataset

Code:

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)

Output:

       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

What I want:

## 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]

My attempt

First we groupby and agg with code:

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

Gives:

                      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]

Then we sort with in each session with code:

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

Gives:

                      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]

My approach is quick slow. Is there any better way to do groupby -> sort with in group -> aggregation ?

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

Answer 1

Use DataFrame.sort_values before groupby , if need apply same function is possible use list of columns names:

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

try sorting by session_id and log_time before groupby

 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)

output

               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]