pandas 熊猫 groupby 后缺少列
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missing column after pandas groupby
提问by user3439329
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 namecolumn. This column is mapped 1:1 with the desk_idcolumn. Is there anyway to include both in my final dataframe?
我有一个Pandas数据框df。我将其按 3 列分组，并计算结果。当我这样做时，我会丢失一些信息，特别是name列。此列与desk_id列按1:1 映射。无论如何都要将两者都包含在我的最终数据框中？
here is the dataframe:
这是数据框：
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
I group it like this:
我这样分组：
grouped = df.groupby(['desk_id', 'shift_id', 'shift_hour']).size()
grouped = grouped.reset_index()
And here is the result, missing the namecolumn.
这是结果，缺少name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
Also, anyway to rename the count column as 'count' instead of '0'?
另外，无论如何要将计数列重命名为“计数”而不是“0”？
采纳答案by CT Zhu
You need to include 'name'in groupbyby groups:
您需要包括'name'在groupby通过组：
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:
如果 name-to-id 关系是多对一类型，假设我们有一个 pete scott 用于同一组数据，结果将变为：
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
