大 pandas groupby后失踪列 | groupby后失踪列

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问题描述

我有一个熊猫数据框

您需要在<$ c中包含'name'

 在[43]中：
 
分组= df.groupby（['desk_id'，'shift_id'，'shift_hour'，'name']）。size（）
 grouped = grouped.reset_index（）
 grouped.columns = np。其中（grouped.columns == 0，'count'，grouped.columns）＃取消默认0到'count'
打印分组
 desk_id shift_id shift_hour名字数
 0 15557987 37423064 0 AdamScott 1 
 1 15557987 37423064 2 Adam Scott 3 
 15557987 37423064 3 Adam Scott 1

如果name-to-id关系是多对一的类型，比如说我们有一个pete scott来处理同一组数据，那么结果就会变成：

  desk_id shift_id shift_hour name count 
 0 15557987 37423064 0 Adam Scott 1 
 1 15557987 37423064 0 Pete Scott 1 
 2 15557987 37423064 2 Adam Scott 3 
 3 15557987 37423064 2 Pete Scott 3 
 4 15557987 37423064 3 Adam Scott 1 
 15557987 37423064 3 Pete Scott 1

I've got a pandas dataframe df. I group it by 3 columns, and count the results. When I do this I lose some information, specifically, the name column. This column is mapped 1:1 with the desk_id column. Is there anyway to include both in my final dataframe?

here is the dataframe:

   shift_id    shift_start_time      shift_end_time        name                   end_time       desk_id  shift_hour
0  37423064 2014-01-17 08:00:00 2014-01-17 12:00:00  Adam Scott 2014-01-17 10:16:41.040000  15557987           2
1  37423064 2014-01-17 08:00:00 2014-01-17 12:00:00  Adam Scott 2014-01-17 10:16:41.096000  15557987           2
2  37423064 2014-01-17 08:00:00 2014-01-17 12:00:00  Adam Scott 2014-01-17 10:52:17.402000  15557987           2
3  37423064 2014-01-17 08:00:00 2014-01-17 12:00:00  Adam Scott 2014-01-17 11:06:59.083000  15557987           3
4  37423064 2014-01-17 08:00:00 2014-01-17 12:00:00  Adam Scott 2014-01-17 08:27:57.998000  15557987           0

I group it like this:

grouped = df.groupby(['desk_id', 'shift_id', 'shift_hour']).size()
grouped = grouped.reset_index()

And here is the result, missing the name column.

    desk_id  shift_id  shift_hour  0
0  14468690  37729081           0  7
1  14468690  37729081           1  3
2  14468690  37729081           2  6
3  14468690  37729081           3  5
4  14468690  37729082           0  5

Also, anyway to rename the count column as 'count' instead of '0'?

解决方案

You need to include 'name' in groupby by groups:

In [43]:

grouped = df.groupby(['desk_id', 'shift_id', 'shift_hour', 'name']).size()
grouped = grouped.reset_index()
grouped.columns=np.where(grouped.columns==0, 'count', grouped.columns) #replace the default 0 to 'count'
print grouped
    desk_id  shift_id  shift_hour        name  count
0  15557987  37423064           0  Adam Scott      1
1  15557987  37423064           2  Adam Scott      3
2  15557987  37423064           3  Adam Scott      1

If the name-to-id relationship is a many-to-one type, say we have a pete scott for the same set of data, the result will become:

    desk_id  shift_id  shift_hour        name  count
0  15557987  37423064           0  Adam Scott      1
1  15557987  37423064           0  Pete Scott      1
2  15557987  37423064           2  Adam Scott      3
3  15557987  37423064           2  Pete Scott      3
4  15557987  37423064           3  Adam Scott      1
5  15557987  37423064           3  Pete Scott      1

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