如何用唯一的行（Python）表示每个用户？

Question

I have data like this:

UserId  Date    Part_of_day    Apps         Category   Frequency      Duration_ToT
1   2020-09-10  evening    Settings     System tool        1          3.436
1   2020-09-11  afternoon   Calendar    Calendar           5          9.965
1   2020-09-11  afternoon   Contacts    Phone_and_SMS      7          2.606
2   2020-09-11  afternoon   Facebook    Social             15         50.799
2   2020-09-11  afternoon   clock       System tool        2          5.223
3   2020-11-18  morning    Contacts    Phone_and_SMS       3          1.726
3   2020-11-18  morning     Google    Productivity         1          4.147
3   2020-11-18  morning    Instagram    Social             1          0.501
.......................................
67  2020-11-18  morning    Truecaller   Communication     1          1.246
67  2020-11-18  night      Instagram    Social            3          58.02

I'am trying to reduce the diemnsionnality of my dataframe to set the entries for k-means. I'd like to ask it's possible to represent each user by one row? what do you think to Embedding? How can i do please. I can't find any solution

Answer 1

This depends on how you want to aggregate the values. Here is a small example how to do it with groupby and agg .

First I create some sample data.

import pandas as pd
import random

df = pd.DataFrame({
   "id":   [int(i/3) for i in range(20)], 
   "val1": [random.random() for _ in range(20)], 
   "val2": [str(int(random.random()*100)) for _ in range(20)]
})
>>> df.head()
   id      val1 val2
0   0  0.174553   49
1   0  0.724547   95
2   0  0.369883    3
3   1  0.243191   64
4   1  0.575982   16
>>> df.dtypes
id        int64
val1    float64
val2     object
dtype: object

Then we group by the id and aggregate the values according to the functions you specify in the dictionary you pass to agg . In this example I sum up the float values and join the strings with an underscore separator. You could eg also pass the list function to store the values in a list.

>>> df.groupby("id").agg({"val1": sum, "val2": "__".join})
        val1        val2
id
0   1.268984   49__95__3
1   0.856992  64__16__54
2   2.186370  30__59__21
3   1.486925  29__47__77
4   1.523898  19__78__99
5   0.855413  59__74__73
6   0.201787      63__33

EDIT regarding the comment "But how can we make val2 contain the top 5 applications according to the duration of the application?":

The agg method is restricted in the sense that you cannot access other attributes while aggregating. To do that you should use the apply method. You pass it a function, that processes the whole group and returns a row as Series object.

In this example I still use the sum for val1, but for val2 I return the val2 of the row with the highest val1. This should make clear how to make the aggregation depend on other attributes.

def apply_func(group):
   return pd.Series({
      "id": group["id"].iat[0], 
      "val1": group["val1"].sum(), 
      "val2": group["val2"].iat[group["val1"].argmax()]
   })

>>> df.groupby("id").apply(apply_func)
    id      val1 val2
id
0    0  1.749955   95
1    1  0.344372   65
2    2  2.019035   70
3    3  2.444691   36
4    4  2.573576   92
5    5  1.453769   72
6    6  1.811516   94