Convert 'object' column to datetime - python

I currently have the following dataframe (with seven days, one day displayed below). Hours run from 01:00 to 24:00. How do I convert the HourEnding column to datetime format and combine it with the date_time column (which is already in datetime format)?
HourEnding LMP date_time
0 01:00 165.27 2021-02-20
1 02:00 155.89 2021-02-20
2 03:00 154.50 2021-02-20
3 04:00 153.44 2021-02-20
4 05:00 210.15 2021-02-20
5 06:00 298.90 2021-02-20
6 07:00 152.71 2021-02-20
7 08:00 204.61 2021-02-20
8 09:00 155.77 2021-02-20
9 10:00 90.64 2021-02-20
10 11:00 57.17 2021-02-20
11 12:00 43.74 2021-02-20
12 13:00 33.42 2021-02-20
13 14:00 5.05 2021-02-20
14 15:00 1.43 2021-02-20
15 16:00 0.99 2021-02-20
16 17:00 0.94 2021-02-20
17 18:00 12.13 2021-02-20
18 19:00 18.90 2021-02-20
19 20:00 19.04 2021-02-20
20 21:00 16.42 2021-02-20
21 22:00 14.47 2021-02-20
22 23:00 44.55 2021-02-20
23 24:00 40.51 2021-02-20
So far I've tried
df['time'] = pd.to_datetime(df['HourEnding'])
but that seems to fail because of the 24:00.
Similarly
df['time'] = pd.to_timedelta('HourEnding', 'h', errors = 'coerce')
yields a column of NaTs.


As you mentioned in the comments, hour 24 corresponds to midnight of the same day. I would simply start by replacing "24" by "00" :
df['HourEnding'] = df.HourEnding.str.replace('24:00', '00:00')
Then, convert date_time to string :
df['date_time'] = df.date_time.astype(str)
Create a new column that concatenates date_time and HourEnding :
df['date_and_hour'] = df.date_time + " " + df.HourEnding
df['date_and_hour'] = pd.to_datetime(df.date_and_hour)
Which gives you this :
>>> df
HourEnding LMP date_time date_and_hour
0 01:00 165.27 2021-02-20 2021-02-20 01:00:00
1 02:00 155.89 2021-02-20 2021-02-20 02:00:00
2 03:00 154.50 2021-02-20 2021-02-20 03:00:00
3 04:00 153.44 2021-02-20 2021-02-20 04:00:00
4 05:00 210.15 2021-02-20 2021-02-20 05:00:00
5 06:00 298.90 2021-02-20 2021-02-20 06:00:00
6 07:00 152.71 2021-02-20 2021-02-20 07:00:00
7 08:00 204.61 2021-02-20 2021-02-20 08:00:00
8 09:00 155.77 2021-02-20 2021-02-20 09:00:00
9 10:00 90.64 2021-02-20 2021-02-20 10:00:00
10 11:00 57.17 2021-02-20 2021-02-20 11:00:00
11 12:00 43.74 2021-02-20 2021-02-20 12:00:00
12 13:00 33.42 2021-02-20 2021-02-20 13:00:00
13 14:00 5.05 2021-02-20 2021-02-20 14:00:00
14 15:00 1.43 2021-02-20 2021-02-20 15:00:00
15 16:00 0.99 2021-02-20 2021-02-20 16:00:00
16 17:00 0.94 2021-02-20 2021-02-20 17:00:00
17 18:00 12.13 2021-02-20 2021-02-20 18:00:00
18 19:00 18.90 2021-02-20 2021-02-20 19:00:00
19 20:00 19.04 2021-02-20 2021-02-20 20:00:00
20 21:00 16.42 2021-02-20 2021-02-20 21:00:00
21 22:00 14.47 2021-02-20 2021-02-20 22:00:00
22 23:00 44.55 2021-02-20 2021-02-20 23:00:00
23 00:00 40.51 2021-02-20 2021-02-20 00:00:00
>>> df.dtypes
HourEnding object
LMP float64
date_time object
date_and_hour datetime64[ns]


Convert both columns to strings, then join them into a new 'datetime' column, and finally convert the 'datetime' column to datetime.
EDIT: To deal with the 1-24 hour problem, build a function to split the string and subtract 1 from each of the hours and then join:
def subtract_hour(t):
t = t.split(':')
t[0] = str(int(t[0]) - 1)
if len(t[0]) < 2:
t[0] = '0' + t[0]
return ':'.join(t)
Then you can apply this to your hour column (e.g., df['hour'] = df['hour'].apply(subtract_hour)) and proceed with joining columns and then parsing using pd.to_datetime.
EDIT 2: You just want to change '24' to '00', my bad.
def mod_midnight(t):
t = t.split(':')
if t[0] == '24':
t[0] = '00'
return ':'.join(t)

Related

Splitting Dataframe time into morning and evening

I have a df that looks like this (shortened):
DateTime Value Date Time
0 2022-09-18 06:00:00 5.4 18/09/2022 06:00
1 2022-09-18 07:00:00 6.0 18/09/2022 07:00
2 2022-09-18 08:00:00 6.5 18/09/2022 08:00
3 2022-09-18 09:00:00 6.7 18/09/2022 09:00
8 2022-09-18 14:00:00 7.9 18/09/2022 14:00
9 2022-09-18 15:00:00 7.8 18/09/2022 15:00
10 2022-09-18 16:00:00 7.6 18/09/2022 16:00
11 2022-09-18 17:00:00 6.8 18/09/2022 17:00
12 2022-09-18 18:00:00 6.4 18/09/2022 18:00
13 2022-09-18 19:00:00 5.7 18/09/2022 19:00
14 2022-09-18 20:00:00 4.8 18/09/2022 20:00
15 2022-09-18 21:00:00 5.4 18/09/2022 21:00
16 2022-09-18 22:00:00 4.7 18/09/2022 22:00
17 2022-09-18 23:00:00 4.3 18/09/2022 23:00
18 2022-09-19 00:00:00 4.1 19/09/2022 00:00
19 2022-09-19 01:00:00 4.4 19/09/2022 01:00
22 2022-09-19 04:00:00 3.5 19/09/2022 04:00
23 2022-09-19 05:00:00 2.8 19/09/2022 05:00
24 2022-09-19 06:00:00 3.8 19/09/2022 06:00
I want to create a new column where i split the between day and night like this:
00:00 - 05:00 night ,
06:00 - 18:00 day ,
19:00 - 23:00 night
But apparently one can't use same label? How can I solve this problem? Here is my code
df['period'] = pd.cut(pd.to_datetime(df.DateTime).dt.hour,
bins=[0, 5, 17, 23],
labels=['night', 'morning', 'night'],
include_lowest=True)
It's returning
ValueError: labels must be unique if ordered=True; pass ordered=False for duplicate labels

if i understood correctly, if time is between 00:00 - 05:00 or 19:00 - 23:00, you want your new column to say 'night', else 'day', well here's that code:
df['day/night'] = df['Time'].apply(lambda x: 'night' if '00:00' <= x <= '05:00' or '19:00' <= x <= '23:00' else 'day')
or you can add ordered = false parameter using your method
input ->
df = pd.DataFrame(columns=['DateTime', 'Value', 'Date', 'Time'], data=[
['2022-09-18 06:00:00', 5.4, '18/09/2022', '06:00'],
['2022-09-18 07:00:00', 6.0, '18/09/2022', '07:00'],
['2022-09-18 08:00:00', 6.5, '18/09/2022', '08:00'],
['2022-09-18 09:00:00', 6.7, '18/09/2022', '09:00'],
['2022-09-18 14:00:00', 7.9, '18/09/2022', '14:00'],
['2022-09-18 15:00:00', 7.8, '18/09/2022', '15:00'],
['2022-09-18 16:00:00', 7.6, '18/09/2022', '16:00'],
['2022-09-18 17:00:00', 6.8, '18/09/2022', '17:00'],
['2022-09-18 18:00:00', 6.4, '18/09/2022', '18:00'],
['2022-09-18 19:00:00', 5.7, '18/09/2022', '19:00'],
['2022-09-18 20:00:00', 4.8, '18/09/2022', '20:00'],
['2022-09-18 21:00:00', 5.4, '18/09/2022', '21:00'],
['2022-09-18 22:00:00', 4.7, '18/09/2022', '22:00'],
['2022-09-18 23:00:00', 4.3, '18/09/2022', '23:00'],
['2022-09-19 00:00:00', 4.1, '19/09/2022', '00:00'],
['2022-09-19 01:00:00', 4.4, '19/09/2022', '01:00'],
['2022-09-19 04:00:00', 3.5, '19/09/2022', '04:00'],
['2022-09-19 05:00:00', 2.8, '19/09/2022', '05:00'],
['2022-09-19 06:00:00', 3.8, '19/09/2022', '06:00']])
output ->
DateTime Value Date Time is_0600_0900
0 2022-09-18 06:00:00 5.4 18/09/2022 06:00 day
1 2022-09-18 07:00:00 6.0 18/09/2022 07:00 day
2 2022-09-18 08:00:00 6.5 18/09/2022 08:00 day
3 2022-09-18 09:00:00 6.7 18/09/2022 09:00 day
4 2022-09-18 14:00:00 7.9 18/09/2022 14:00 day
5 2022-09-18 15:00:00 7.8 18/09/2022 15:00 day
6 2022-09-18 16:00:00 7.6 18/09/2022 16:00 day
7 2022-09-18 17:00:00 6.8 18/09/2022 17:00 day
8 2022-09-18 18:00:00 6.4 18/09/2022 18:00 day
9 2022-09-18 19:00:00 5.7 18/09/2022 19:00 night
10 2022-09-18 20:00:00 4.8 18/09/2022 20:00 night
11 2022-09-18 21:00:00 5.4 18/09/2022 21:00 night
12 2022-09-18 22:00:00 4.7 18/09/2022 22:00 night
13 2022-09-18 23:00:00 4.3 18/09/2022 23:00 night
14 2022-09-19 00:00:00 4.1 19/09/2022 00:00 night
15 2022-09-19 01:00:00 4.4 19/09/2022 01:00 night
16 2022-09-19 04:00:00 3.5 19/09/2022 04:00 night
17 2022-09-19 05:00:00 2.8 19/09/2022 05:00 night
18 2022-09-19 06:00:00 3.8 19/09/2022 06:00 day

You have two options.
Either you don't care about the order and you can set ordered=False as parameter of cut:
df['period'] = pd.cut(pd.to_datetime(df.DateTime).dt.hour,
bins=[0, 5, 17, 23],
labels=['night', 'morning', 'night'],
ordered=False,
include_lowest=True)
Or you care to have night and morning ordered, in which case you can further convert to ordered Categorical:
df['period'] = pd.Categorical(df['period'], categories=['night', 'morning'], ordered=True)
output:
DateTime Value Date Time period
0 2022-09-18 06:00:00 5.4 18/09/2022 06:00 morning
1 2022-09-18 07:00:00 6.0 18/09/2022 07:00 morning
2 2022-09-18 08:00:00 6.5 18/09/2022 08:00 morning
3 2022-09-18 09:00:00 6.7 18/09/2022 09:00 morning
8 2022-09-18 14:00:00 7.9 18/09/2022 14:00 morning
9 2022-09-18 15:00:00 7.8 18/09/2022 15:00 morning
10 2022-09-18 16:00:00 7.6 18/09/2022 16:00 morning
11 2022-09-18 17:00:00 6.8 18/09/2022 17:00 morning
12 2022-09-18 18:00:00 6.4 18/09/2022 18:00 night
13 2022-09-18 19:00:00 5.7 18/09/2022 19:00 night
14 2022-09-18 20:00:00 4.8 18/09/2022 20:00 night
15 2022-09-18 21:00:00 5.4 18/09/2022 21:00 night
16 2022-09-18 22:00:00 4.7 18/09/2022 22:00 night
17 2022-09-18 23:00:00 4.3 18/09/2022 23:00 night
18 2022-09-19 00:00:00 4.1 19/09/2022 00:00 night
19 2022-09-19 01:00:00 4.4 19/09/2022 01:00 night
22 2022-09-19 04:00:00 3.5 19/09/2022 04:00 night
23 2022-09-19 05:00:00 2.8 19/09/2022 05:00 night
24 2022-09-19 06:00:00 3.8 19/09/2022 06:00 morning
column:
df['period']
0 morning
1 morning
2 morning
...
23 night
24 morning
Name: period, dtype: category
Categories (2, object): ['morning', 'night']

Find maximum of value in a column of one dataframe given the value constraint of two columns in another dataframe

I have a dataframe, df1 with two columns representing the start and end time of task. I have another dataframe, df2 with two columns representing time and the stock available at that time. I want to create another column in df1 named as max_stock which has maximum value of stock values for a time range given by ST and ET of df1. For instance, first task has start time 7/11/2021 1:00 and end time 7/11/2021 2:00 so for this value of max_stock is maximum of values in stock column of df2 which is maximum of 10, 26, and 48 at time 7/11/2021 1:00, and 7/11/2021 1:30, and 7/11/2021 2:00, respectively.
df1
ST ET
7/11/2021 1:00 7/11/2021 2:00
7/11/2021 2:00 7/11/2021 3:00
7/11/2021 3:00 7/11/2021 4:00
7/11/2021 4:00 7/11/2021 5:00
7/11/2021 5:00 7/11/2021 6:00
7/11/2021 6:00 7/11/2021 7:00
7/11/2021 7:00 7/11/2021 8:00
7/11/2021 8:00 7/11/2021 9:00
7/11/2021 9:00 7/11/2021 10:00
df2
Time stock
7/11/2021 1:00 10
7/11/2021 1:30 26
7/11/2021 2:00 48
7/11/2021 2:30 35
7/11/2021 3:00 32
7/11/2021 3:30 80
7/11/2021 4:00 31
7/11/2021 4:30 81
7/11/2021 5:00 65
7/11/2021 5:30 83
7/11/2021 6:00 40
7/11/2021 6:30 84
7/11/2021 7:00 41
7/11/2021 7:30 15
7/11/2021 8:00 65
7/11/2021 8:30 18
7/11/2021 9:00 80
7/11/2021 9:30 12
7/11/2021 10:00 5
Required df
ST ET max_stock
7/11/2021 1:00 7/11/2021 2:00 48.00
7/11/2021 2:00 7/11/2021 3:00 48.00
7/11/2021 3:00 7/11/2021 4:00 80.00
7/11/2021 4:00 7/11/2021 5:00 81.00
7/11/2021 5:00 7/11/2021 6:00 83.00
7/11/2021 6:00 7/11/2021 7:00 84.00
7/11/2021 7:00 7/11/2021 8:00 65.00
7/11/2021 8:00 7/11/2021 9:00 80.00
7/11/2021 9:00 7/11/2021 10:00 80.00

One option is via conditional_join from pyjanitor to simulate greater than and less than conditions, before grouping and aggregating:
# pip install pyjanitor
import pandas as pd
import janitor
(df1.conditional_join(
df2,
('ST', 'Time', '<='),
('ET', 'Time', '>='))
.groupby(['ST', 'ET'], as_index = False)
.stock
.max()
)
ST ET stock
0 2021-07-11 01:00:00 2021-07-11 02:00:00 48
1 2021-07-11 02:00:00 2021-07-11 03:00:00 48
2 2021-07-11 03:00:00 2021-07-11 04:00:00 80
3 2021-07-11 04:00:00 2021-07-11 05:00:00 81
4 2021-07-11 05:00:00 2021-07-11 06:00:00 83
5 2021-07-11 06:00:00 2021-07-11 07:00:00 84
6 2021-07-11 07:00:00 2021-07-11 08:00:00 65
7 2021-07-11 08:00:00 2021-07-11 09:00:00 80
8 2021-07-11 09:00:00 2021-07-11 10:00:00 80
You can use a cartesian join and filter afterwards (for large dataframes, this might be memory inefficient):
(df1.merge(df2, how='cross')
.query('ST <=Time <= ET')
.groupby(['ST', 'ET'], as_index = False)
.stock
.max()
)
Out[113]:
ST ET stock
0 2021-07-11 01:00:00 2021-07-11 02:00:00 48
1 2021-07-11 02:00:00 2021-07-11 03:00:00 48
2 2021-07-11 03:00:00 2021-07-11 04:00:00 80
3 2021-07-11 04:00:00 2021-07-11 05:00:00 81
4 2021-07-11 05:00:00 2021-07-11 06:00:00 83
5 2021-07-11 06:00:00 2021-07-11 07:00:00 84
6 2021-07-11 07:00:00 2021-07-11 08:00:00 65
7 2021-07-11 08:00:00 2021-07-11 09:00:00 80
8 2021-07-11 09:00:00 2021-07-11 10:00:00 80
Another option is with interval index (a longer process here, since the resulting interval has overlapping values):
box = pd.IntervalIndex.from_arrays(df1.ST, df1.ET, closed='both')
df1.index = box
# create temporary Series
temp = (df2.Time
.apply(lambda x: box[box.get_loc(x)])
.explode(ignore_index = False)
)
temp.name = 'interval'
# lump back to main dataframe (df2)
temp = pd.concat([df2, temp], axis = 1)
# aggregate:
temp = temp.groupby('interval').stock.max()
# join back to df1 to get final output
df1.join(temp).reset_index(drop=True)
ST ET stock
0 2021-07-11 01:00:00 2021-07-11 02:00:00 48
1 2021-07-11 02:00:00 2021-07-11 03:00:00 48
2 2021-07-11 03:00:00 2021-07-11 04:00:00 80
3 2021-07-11 04:00:00 2021-07-11 05:00:00 81
4 2021-07-11 05:00:00 2021-07-11 06:00:00 83
5 2021-07-11 06:00:00 2021-07-11 07:00:00 84
6 2021-07-11 07:00:00 2021-07-11 08:00:00 65
7 2021-07-11 08:00:00 2021-07-11 09:00:00 80
8 2021-07-11 09:00:00 2021-07-11 10:00:00 80

How to crate new column based on time interval?

I want to create a new column, based on time interval of 6hours from datetime column how can I do that?
C/A UNIT SCP STATION LINENAME DIVISION DATE TIME DESC ENTRIES EXITS
0 A002 R051 02-00-00 59 ST NQR456W BMT 05/29/2021 00:00:00 REGULAR 7578734 2590325
1 A002 R051 02-00-00 59 ST NQR456W BMT 05/29/2021 04:00:00 REGULAR 7578740 2590327
2 A002 R051 02-00-00 59 ST NQR456W BMT 05/29/2021 08:00:00 REGULAR 7578749 2590340
3 A002 R051 02-00-00 59 ST NQR456W BMT 05/29/2021 12:00:00 REGULAR 7578789 2590386
4 A002 R051 02-00-00 59 ST NQR456W BMT 05/29/2021 16:00:00 REGULAR 7578897 259041

pandas has floor function for time
df['DATETIME'].dt.floor('6H')
this column needs to be datetime type
0 1
0 2021-06-06 00:00:00 2021-06-06 00:00:00
1 2021-06-06 01:00:00 2021-06-06 00:00:00
2 2021-06-06 02:00:00 2021-06-06 00:00:00
3 2021-06-06 03:00:00 2021-06-06 00:00:00
4 2021-06-06 04:00:00 2021-06-06 00:00:00
5 2021-06-06 05:00:00 2021-06-06 00:00:00
6 2021-06-06 06:00:00 2021-06-06 06:00:00
7 2021-06-06 07:00:00 2021-06-06 06:00:00
8 2021-06-06 08:00:00 2021-06-06 06:00:00
9 2021-06-06 09:00:00 2021-06-06 06:00:00

If you want to get a new column with date/time being 6 hours offset from DATETIME column, you can use pd.DateOffset, as follows:
df['New_DATETIME'] = pd.to_datetime(df['DATETIME']) + pd.DateOffset(hours=6)

Fixing inconsistent formatting of 24-hour to 12-hour

I need to essentially measure how much each employee gets paid during each hour of work. There was some data cleaning to do and so I'm trying to make the formatting consistent.
It is a homework problem and its proving tough. I am new to python so please feel free to compress the code. I'm trying to use the pandas database.
csv file in pandas
break_notes end_time pay_rate start_time
0 15-18 23:00 10.0 10:00
1 18.30-19.00 23:00 12.0 18:00
2 4PM-5PM 22:30 14.0 12:00
3 3-4 18:00 10.0 09:00
4 4-4.10PM 23:00 20.0 09:00
5 15 - 17 23:00 10.0 11:00
6 11 - 13 16:00 10.0 10:00
'''
import pandas as pd
import datetime
import numpy as np
work_shifts = pd.read_csv('work_shifts.csv')
break_shifts = work_shifts['break_notes'].str.extract('(?P<start>[\d\.]+)?\D*(?P<end>[\d\.]+)?')
print(work_shifts)
for i in range(len(break_shifts['start'])):
if '.' not in break_shifts['start'][i]:
break_shifts['start'][i] = break_shifts['start'][i] + ':00'
else:
break_shifts['start'][i] = break_shifts['start'][i].replace('.',':')
for i in range(len(break_shifts['end'])):
if '.' in str(break_shifts['end'][i]):
break_shifts['end'][i] = break_shifts['end'][i].replace('.',':')
elif '.' not in str(break_shifts['end'][i]):
break_shifts['end'][i] = break_shifts['end'][i] + ':00'
for i in range(len(break_shifts['end'])):
break_shifts['end'][i] = datetime.datetime.strptime(break_shifts['end'][i], '%H:%M').time()
break_shifts['start'][i] = datetime.datetime.strptime(break_shifts['start'][i], '%H:%M').time()
work_shifts[['start_break','end_break']] = break_shifts[['start', 'end']]
for i in range(len(work_shifts['end_time'])):
work_shifts['end_time'][i] = datetime.datetime.strptime(work_shifts['end_time'][i], '%H:%M').time()
for i in range(len(work_shifts['start_time'])):
work_shifts['start_time'][i] = datetime.datetime.strptime(work_shifts['start_time'][i], '%H:%M').time()
print(work_shifts)
this is the result
break_notes end_time pay_rate start_time start_break end_break
0 15-18 23:00:00 10.0 10:00:00 15:00:00 18:00:00
1 18.30-19.00 23:00:00 12.0 18:00:00 18:30:00 19:00:00
2 4PM-5PM 22:30:00 14.0 12:00:00 04:00:00 05:00:00
3 3-4 18:00:00 10.0 09:00:00 03:00:00 04:00:00
4 4-4.10PM 23:00:00 20.0 09:00:00 04:00:00 04:10:00
5 15 - 17 23:00:00 10.0 11:00:00 15:00:00 17:00:00
6 11 - 13 16:00:00 10.0 10:00:00 11:00:00 13:00:00
I tried adding the times but they are inconsistent types. If theres a different approach then please provide guidance. I need to calculate how many employees are working at what time and then calculate how much pay is given to the employees per hour.
My approach was to convert the formatting of the break notes into time then convert the 12-hour to 12 provided both end_break and start_break was before datetime.datetime(12,0,0).
I'm not sure how to calculate the money per hour. Maybe using if statements?

Pandas : merge on date and hour from datetime index

I have two data frames like following, data frame A has datetime even with minutes, data frame B only has hour.
df:A
dataDate original
2018-09-30 11:20:00 3
2018-10-01 12:40:00 10
2018-10-02 07:00:00 5
2018-10-27 12:50:00 5
2018-11-28 19:45:00 7
df:B
dataDate count
2018-09-30 10:00:00 300
2018-10-01 12:00:00 50
2018-10-02 07:00:00 120
2018-10-27 12:00:00 234
2018-11-28 19:05:00 714
I like to merge the two on the basis of hour date and hour, so that now in dataframe A should have all the rows filled on the basis of merge on date and hour
I can try to do it via
A['date'] = A.dataDate.date
B['date'] = B.dataDate.date
A['hour'] = A.dataDate.hour
B['hour'] = B.dataDate.hour
and then merge
merge_df = pd.merge(A,B, how='left', left_on=['date', 'hour'],
right_on=['date', 'hour'])
but its a very long process, Is their an efficient way to perform the same operation with the help of pandas time series or date functionality?

Use map if need append only one column from B to A with floor for set minutes and seconds if exist to 0:
d = dict(zip(B.dataDate.dt.floor('H'), B['count']))
A['count'] = A.dataDate.dt.floor('H').map(d)
print (A)
dataDate original count
0 2018-09-30 11:20:00 3 NaN
1 2018-10-01 12:40:00 10 50.0
2 2018-10-02 07:00:00 5 120.0
3 2018-10-27 12:50:00 5 234.0
4 2018-11-28 19:45:00 7 714.0
For general solution use DataFrame.join:
A.index = A.dataDate.dt.floor('H')
B.index = B.dataDate.dt.floor('H')
A = A.join(B, lsuffix='_left')
print (A)
dataDate_left original dataDate count
dataDate
2018-09-30 11:00:00 2018-09-30 11:20:00 3 NaT NaN
2018-10-01 12:00:00 2018-10-01 12:40:00 10 2018-10-01 12:00:00 50.0
2018-10-02 07:00:00 2018-10-02 07:00:00 5 2018-10-02 07:00:00 120.0
2018-10-27 12:00:00 2018-10-27 12:50:00 5 2018-10-27 12:00:00 234.0
2018-11-28 19:00:00 2018-11-28 19:45:00 7 2018-11-28 19:05:00 714.0

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