熊猫DataFrame高级索引

Question

I am looking for some help with pandas DataFrame sorting. I have a Data frame of 8 columns that go like;

 ['Date' , 'S ID', 'Se ID', 'S #', 'File Size (Mb)', 'HD name', 'Start Time', 'End time'] 

I've then done a:

 DataFile.groupby(['HD Name','Date','Se ID','S ID'])['File Size (Mb)'].agg({'Sequence #':'count','File Size (Mb)':'sum'}).reset_index().rename(columns={'Sequence #':'# of Files'}) 

which takes my data and groups it by the matching grouby() parameters and sums the file sizes. I would like to add two columns on to this which holds the first 'Start Time' and last 'End Time' how would I go about doing that?
I'm thinking my only option may be to loop over the data or create a duplicate dataframe to et start and end times of grouped data.
Any ideas would be appreciated!

Example dataframe:

'Hard Drive Name' : [H5 , H5 , H5 , H5 , H5]
'S ID' : [LA , LA , LA , SD , SD]
'Se ID' : [1200, 1200, 1200, 30, 30]
'Date' : ['10/01/2018' , '10/01/2018' , '10/01/2018' , '09/03/2018' , '09/03/2018']
'#' : [1 , 2 , 3 , 1 , 2]
'Start Time' : [[08:09:54] , [08:58:31] , [09:39:38] , [05:04:13] , [05:41:13] ]
'End Time' : [[08:28:54] , [09:17:31] , [09:58:38] , [05:23:12] , [06:00:12]]

 {'Date': {34: '10/01/2018',
35: '10/01/2018',
36: '10/01/2018',
37: '10/01/2018',
38: '10/01/2018',
39: '10/01/2018',
40: '10/01/2018',
41: '10/01/2018',
42: '10/01/2018',
661: '09/03/2018'},  

'End Time': {34: ['08:28:54'],
35: ['09:17:31'],
36: ['09:58:38'],
37: ['10:37:41'],
38: ['11:21:32'],
39: ['12:04:42'],
40: ['12:45:31'],
41: ['13:25:23'],
42: ['14:04:03'],
661: ['05:53:36']},  

'File Size (Mb)': {34: 1074.256284,
35: 1074.842244,
36: 1074.759444,
37: 1074.836956,
38: 1074.516156,
39: 1074.547044,
40: 1074.8363,
41: 1074.891492,
42: 1074.792068,
661: 1074.428204},  

'Hard Drive Name': {34: 'H5',
35: 'H5',
36: 'H5',
37: 'H5',
38: 'H5',
39: 'H5',
40: 'H5',
41: 'H5',
42: 'H5',
661: 'H5'},  

'Sensor ID': {34: '1207',
35: '1207',
36: '1207',
37: '1207',
38: '1207',
39: '1207',
40: '1207',
41: '1207',
42: '1207',
661: '1207'},  

'Sequence #': {34: 's005',
35: 's006',
36: 's007',
37: 's008',
38: 's009',
39: 's010',
40: 's011',
41: 's012',
42: 's013',
661: 's000'},  

'Site ID': {34: 'SD',
35: 'SD',
36: 'SD',
37: 'SD',
38: 'SD',
39: 'SD',
40: 'SD',
41: 'SD',
42: 'SD',
661: 'SDO'},  

'Start Time': {34: ['08:09:54'],
35: ['08:58:31'],
36: ['09:39:38'],
37: ['10:18:41'],
38: ['11:02:32'],
39: ['11:45:42'],
40: ['12:26:31'],
41: ['13:06:23'],
42: ['13:45:03'],
661: ['05:34:37']}}

Answer 1

Okay, you need to use pd.to_timedelta with .str accessor:

Where d equals to your df.head(10).to_dict() output:

df = pd.DataFrame(d)

df['Start Time'] = pd.to_timedelta(df['Start Time'].str[0])
df['End Time']  = pd.to_timedelta(df['End Time'].str[0])

df_out = df.groupby(['Hard Drive Name','Date','Sensor ID','Site ID'])['Sequence #',
                                                                      'File Size (Mb)',
                                                                      'Start Time',
                                                                      'End Time']\
           .agg({'Sequence #':'count',
                 'File Size (Mb)':'sum',
                 'Start Time':'min',
                 'End Time':'max'})\
           .reset_index()\
           .rename(columns={'Sequence #':'# of Files'})

Output:

  Hard Drive Name        Date Sensor ID Site ID  # of Files  File Size (Mb) Start Time End Time
0              H5  09/03/2018      1207     SDO           1     1074.428204   05:34:37 05:53:36
1              H5  10/01/2018      1207      SD           9     9672.277988   08:09:54 14:04:03