Question

我已经在数据集上编写了这两个groupby函数，第一个对数据进行了分组，并将数据的日期时间分隔为开始日期时间，结束日期时间。

这是数据集：

Blast Hole	East Coordinate	North Coordinate	Collar	Theoritical Depth	Tag Detector ID	Date and Time	Detection_Location	Detection Date & Time
64	16745.42	107390.32	2634.45	15.95	385656531	23-08-2018 2:39:34 PM	CV23	2018-09-08 14:18:17
61	16773.48	107382.6	2634.68	16.18	385760755	23-08-2018 2:38:32 PM	CV23	2018-09-08 14:24:19
63	16755.07	107387.68	2634.58	16.08	385262370	23-08-2018 2:39:30 PM	CV23	2018-09-08 14:12:42
105	16764.83	107347.67	2634.74	16.24	385742468	23-08-2018 2:41:29 PM	CV22	2018-09-06 20:02:46
100	16752.74	107360.32	2634.33	15.83	385112050	23-08-2018 2:41:08 PM	CV22	2018-09-06 20:15:42
99	16743.1	107362.96	2634.36	15.86	385087366	23-08-2018 2:41:05 PM	CV22	2018-09-06 20:49:21
35	16747.75	107417.68	2635.9	17.4	385453358	23-08-2018 2:36:09 PM	CV22	2018-09-23 05:47:44
5	16757.27	107452.4	2636	17.5	385662254	23-08-2018 2:35:03 PM	CV22	2018-09-23 05:01:12
19	16770.89	107420.83	2634.81	16.31	385826979	23-08-2018 2:35:50 PM	CV22	2018-09-23 05:52:54

第二部分帮助我将所有列保留在分组数据框中，就像之前用逗号分隔一样。

我在如何将这两个代码合并为一个代码并执行操作时遇到了问题：


df2 = (df1.groupby([pd.Grouper(key = 'Detection Date & Time', freq = 'H'),df.Detection_Location])
      ['Detection Date & Time'].agg(['first','last','size'])).reset_index()

df2 = df1.groupby("Detection date & Hour").agg({
        'Blast Hole': lambda x: ','.join([str(n) for n in x]),
        'East Coordinate': lambda x: ','.join([str(n) for n in x]),
        'North Coordinate': lambda x: ','.join([str(n) for n in x]),
        'Tag Detector ID': lambda x: ','.join([str(n) for n in x]),
        'Detection_Location': lambda x: min(x),
        'Detection Date & Time' : lambda x: len(x)}).reset_index().rename(columns = {'Detection Date & Time' : 'Tags'})

这是期望的结果：

Detection_Location_	first	last	size	Blast Hole	East Coordinate	North Coordinate	Tag Detector ID
CV22	2018-09-06 20:02:46	2018-09-06 20:49:21	3	105,100,99	16764.83,16752.74,16743.1	107347.67,107360.32,107362.96	385742468,385112050,385087366
CV23	2018-09-08 14:12:42	2018-09-08 14:24:19	3	64,61,63	16745.42,16773.48,16755.07	107390.32,107382.6,107387.68	385656531,385760755,385262370
CV22	2018-09-23 05:01:12	2018-09-23 05:52:54	3	35,5,19	16747.75,16757.27,16770.89	107417.68,107452.4,107420.83	385453358,385662254,385826979

谢谢

Answer 1

第一个想法是是否需要在groupby中使用不同的值-第一个df21与Grouper，第二个仅与Grouper：

df1['Date and Time'] = pd.to_datetime(df1['Date and Time'])
df1['Detection Date & Time'] = pd.to_datetime(df1['Detection Date & Time'])


df21 = (df1.groupby([pd.Grouper(key = 'Detection Date & Time', freq = 'H'),
                     df1.Detection_Location])
      ['Detection Date & Time'].agg(['first','last','size']))
#print (df21)


f = lambda x: ','.join(x.astype(str))
df22=(df1.groupby(pd.Grouper(key = 'Detection Date & Time', freq = 'H')).agg({
        'Blast Hole': f,
        'East Coordinate': f,
        'North Coordinate': f,
        'Tag Detector ID': f,
        'Detection_Location': 'min',
        'Detection Date & Time' : 'size'})
        .dropna()
        .rename(columns = {'Detection Date & Time' : 'Tags'})
        .set_index('Detection_Location', append=True))

#print (df22)

df = pd.merge(df21, df22, left_index=True, right_index=True).reset_index()
print (df)
  Detection Date & Time Detection_Location               first  \
0   2018-09-06 20:00:00               CV22 2018-09-06 20:02:46   
1   2018-09-08 14:00:00               CV23 2018-09-08 14:18:17   
2   2018-09-23 05:00:00               CV22 2018-09-23 05:47:44   

                 last  size  Blast Hole             East Coordinate  \
0 2018-09-06 20:49:21     3  105,100,99   16764.83,16752.74,16743.1   
1 2018-09-08 14:12:42     3    63,64,61  16755.07,16745.42,16773.48   
2 2018-09-23 05:52:54     3     5,35,19  16757.27,16747.75,16770.89   

                North Coordinate                Tag Detector ID  Tags  
0  107347.67,107360.32,107362.96  385742468,385112050,385087366     3  
1   107387.68,107390.32,107382.6  385262370,385656531,385760755     3  
2   107452.4,107417.68,107420.83  385662254,385453358,385826979     3

编辑：

如果需要按Grouper分组并一起列：

df1['Date and Time'] = pd.to_datetime(df1['Date and Time'])
df1['Detection Date & Time'] = pd.to_datetime(df1['Detection Date & Time'])


f = lambda x: ','.join(x.astype(str))
df2=(df1.groupby([pd.Grouper(key='Detection Date & Time',freq='H'),
                 df1.Detection_Location]).agg({
        'Blast Hole': f,
        'East Coordinate': f,
        'North Coordinate': f,
        'Tag Detector ID': f,
        'Detection Date & Time' : ['first','last','size']})
               .reset_index()
               .rename(columns = {'Detection Date & Time' : '', '<lambda>':''}))

df2.columns = df2.columns.map(''.join)
df2 = df2.rename(columns = {'' : 'Detection Date & Time'})

print (df2)
  Detection Date & Time Detection_Location  Blast Hole  \
0   2018-09-06 20:00:00               CV22  105,100,99   
1   2018-09-08 14:00:00               CV23    64,61,63   
2   2018-09-23 05:00:00               CV22     35,5,19   

              East Coordinate               North Coordinate  \
0   16764.83,16752.74,16743.1  107347.67,107360.32,107362.96   
1  16745.42,16773.48,16755.07   107390.32,107382.6,107387.68   
2  16747.75,16757.27,16770.89   107417.68,107452.4,107420.83   

                 Tag Detector ID               first                last  size  
0  385742468,385112050,385087366 2018-09-06 20:02:46 2018-09-06 20:49:21     3  
1  385656531,385760755,385262370 2018-09-08 14:18:17 2018-09-08 14:12:42     3  
2  385453358,385662254,385826979 2018-09-23 05:47:44 2018-09-23 05:52:54     3

Answer 2

这可能对您有用（我从您先前的问题中知道您的数据看起来如何）您可以只用agg(list)

将所有值聚合到一个列表中

df3=df.groupby([pd.Grouper(key = 'Detection_Date&Time', freq = 'H'),df.Detection_Location], sort=False).agg(list).reset_index()

然后，按如下所示合并另一个（从另一个问题到结果，此处为df2）

df2 = (df.groupby([pd.Grouper(key = 'Detection_Date&Time', freq = 'H'),df.Detection_Location], sort=False)['Detection_Date&Time']
   .agg(['first','last','size'])).reset_index()

df4 = pd.merge(df2, df3, on=['Detection_Date&Time','Detection_Location'])

获得的输出如下

Detection_Date&Time     Detection_Location  first   last    size    Blast_Hole  East_Coordinate     North_Coordinate    Collar  Theoritical_Depth   Tag_Detector_ID     Date_and_Time
0   2018-09-08 14:00:00     CV23    2018-09-08 14:18:00     2018-09-08 14:12:00     3   [64, 61, 63]    [16745.42, 16773.48, 16755.07]  [107390.32, 107382.6, 107387.68]    [2634.45, 2634.68, 2634.58]     [15.95, 16.18, 16.08]   [385656531, 385760755, 385262370]   [23-08-2018 2:39:34 PM, 23-08-2018 2:38:32 PM,...
1   2018-09-06 20:00:00     CV22    2018-09-06 20:02:00     2018-09-06 20:49:00     3   [105, 100, 99]  [16764.83, 16752.74, 16743.1]   [107347.67, 107360.32, 107362.96]   [2634.74, 2634.33, 2634.36]     [16.24, 15.83, 15.86]   [385742468, 385112050, 385087366]   [23-08-2018 2:41:29 PM, 23-08-2018 2:41:08 PM,...
2   2018-09-23 05:00:00     CV22    2018-09-23 05:47:00     2018-09-23 05:52:00     3   [35, 5, 19]     [16747.75, 16757.27, 16770.89]  [107417.68, 107452.4, 107420.83]    [2635.9, 2636.0, 2634.81]   [17.4, 17.5, 16.31]     [385453358, 385662254, 385826979]   [23-08-2018 2:36:09 PM, 23-08-2018 2:35:03 PM,...

结合Groupby功能代码（带有或不带有Grouper）

2 个答案: