大 pandas groupby后失踪列 [英] missing column after pandas groupby
问题描述
我有一个熊猫数据框 df 。我把它分成3列，并计算结果。当我这样做时，我会丢失一些信息，特别是 name 列。该列与 desk_id 列进行1：1映射。无论如何都要在我的最终数据框中包含这两个选项？
这里是数据框：
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
我这样分组：
<$ （$'code> groupped = df.groupby（['desk_id'，'shift_id'，'shift_hour']）size（）
grouped = grouped.reset_index（）
这里是结果，缺少名称列。
desk_id shift_id shift_hour 0
0 14468690 37729081 0 7
1 14468690 37729081 1 3
2 14468690 37729081 2 6
3 14468690 37729081 3 5
4 14468690 37729082 0 5
另外，无论如何要将count列重命名为'count'而不是'0' ？
您需要在<$ 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'?
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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