Python 组合Groupby函数代码，带和不带grouper
我已经在我的数据集上编写了这两个groupby函数，第一个将我的数据分组，并将数据的日期时间分隔为开始日期时间和结束日期时间 这是数据集：Python 组合Groupby函数代码，带和不带grouper,python,pandas,group-by,Python,Pandas,Group By,我已经在我的数据集上编写了这两个groupby函数，第一个将我的数据分组，并将数据的日期时间分隔为开始日期时间和结束日期时间 这是数据集： 炮孔东坐标北坐标轴环理论深度标签探测器ID日期和时间检测\u位置检测日期和时间 64 16745.42 107390.32 2634.45 15.95 385656531 23-08-2018 2:39:34下午CV23 2018-09-08 14:18:17 61 16773.48 107382.6 2634.68 16.18 385760755 23-
炮孔东坐标北坐标轴环理论深度标签探测器ID日期和时间检测\u位置检测日期和时间
64 16745.42 107390.32 2634.45 15.95 385656531 23-08-2018 2:39:34下午CV23 2018-09-08 14:18:17
61 16773.48 107382.6 2634.68 16.18 385760755 23-08-2018 2:38:32下午CV23 2018-09-08 14:24:19
63 16755.07 107387.68 2634.58 16.08 385262370 23-08-2018 2:39:30下午CV23 2018-09-08 14:12:42
105 16764.83 107347.67 2634.74 16.24 385742468 23-08-2018 2:41:29下午CV22 2018-09-06 20:02:46
100 16752.74 107360.32 2634.33 15.83 385112050 23-08-2018 2:41:08下午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下午CV22 2018-09-23 05:47:44
5 16757.27 107452.4 2636 17.5 385662254 23-08-2018 2:35:03下午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第一个想法是如果需要在groupby
中使用不同的值-第一个df21
使用石斑鱼，第二个仅使用石斑鱼：
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)
编辑：
如果需要按Grouper
和列进行分组：
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'})
这可能对你有用（我从你前面的问题中知道你的数据是什么样子的）
您可以使用agg（list）
然后，将另一个问题（从另一个问题到结果，这里是df2）合并如下
获得的输出如下所示
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,...
Hi@jezrael-这给了我这个错误：列重叠但没有指定后缀：Index（['Detection\u Location']，dtype='object'）我认为重叠的列很少，我需要剪切几行代码。我会努力的。Thanks@ShrutiGaur-Trydf2=df21.join（df22，lsuffix=''）
@jexrael-df22在加入df2Perfect后显示为空！！谢谢你，汤姆！在数据帧中，“检测日期和时间”和“检测位置”是相同的。它对我有用。请看前面的问题。在那里，有一些照片是人们所期待的。这就是为什么我提到我知道这些数据@莫哈尼-谢谢。虽然当我们创建第一个数据帧df3时，列检测位置已经存在，并且当我们将（list）传递给聚合函数时，它给出了以下错误：无法插入检测位置，已经存在exists@jezrael-是，此问题的数据和预期结果仅在此问题中。你是right@mohanys-您显示的输出非常完美，我将再次尝试代码。谢谢
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
df3=df.groupby([pd.Grouper(key = 'Detection_Date&Time', freq = 'H'),df.Detection_Location], sort=False).agg(list).reset_index()
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'])
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,...