Let's say I have the below Dataframe. How would I do to get an extra column 'flag' with 1's where a day has a age bigger than 90 and only if it happens in 2 consecutive days (48h in this case)? The output should contain 1' on 2 or more days, depending on how many days the condition is met The dataset is much bigger, but I put here just a small portion so you get an idea.
Age
Dates
2019-01-01 00:00:00 29
2019-01-01 01:00:00 56
2019-01-01 02:00:00 82
2019-01-01 03:00:00 13
2019-01-01 04:00:00 35
2019-01-01 05:00:00 53
2019-01-01 06:00:00 25
2019-01-01 07:00:00 23
2019-01-01 08:00:00 21
2019-01-01 09:00:00 12
2019-01-01 10:00:00 15
2019-01-01 11:00:00 9
2019-01-01 12:00:00 13
2019-01-01 13:00:00 87
2019-01-01 14:00:00 9
2019-01-01 15:00:00 63
2019-01-01 16:00:00 62
2019-01-01 17:00:00 52
2019-01-01 18:00:00 43
2019-01-01 19:00:00 77
2019-01-01 20:00:00 95
2019-01-01 21:00:00 79
2019-01-01 22:00:00 77
2019-01-01 23:00:00 5
2019-01-02 00:00:00 78
2019-01-02 01:00:00 41
2019-01-02 02:00:00 10
2019-01-02 03:00:00 10
2019-01-02 04:00:00 88
2019-01-02 05:00:00 19
This would be the desired output:
Dates Age flag
0 2019-01-01 00:00:00 29 1
1 2019-01-01 01:00:00 56 1
2 2019-01-01 02:00:00 82 1
3 2019-01-01 03:00:00 13 1
4 2019-01-01 04:00:00 35 1
5 2019-01-01 05:00:00 53 1
6 2019-01-01 06:00:00 25 1
7 2019-01-01 07:00:00 23 1
8 2019-01-01 08:00:00 21 1
9 2019-01-01 09:00:00 12 1
10 2019-01-01 10:00:00 15 1
11 2019-01-01 11:00:00 9 1
12 2019-01-01 12:00:00 13 1
13 2019-01-01 13:00:00 87 1
14 2019-01-01 14:00:00 9 1
15 2019-01-01 15:00:00 63 1
16 2019-01-01 16:00:00 62 1
17 2019-01-01 17:00:00 52 1
18 2019-01-01 18:00:00 43 1
19 2019-01-01 19:00:00 77 1
20 2019-01-01 20:00:00 95 1
21 2019-01-01 21:00:00 79 1
22 2019-01-01 22:00:00 77 1
23 2019-01-01 23:00:00 5 1
24 2019-01-02 00:00:00 78 0
25 2019-01-02 01:00:00 41 0
26 2019-01-02 02:00:00 10 0
27 2019-01-02 03:00:00 10 0
28 2019-01-02 04:00:00 88 0
29 2019-01-02 05:00:00 19 0
The dates is the index of the dataframe and is incremented by 1h.
thanks
You can first compare column by Series.gt, then grouping by DatetimeIndex.date and ccheck if at least one True per groups by GroupBy.transform with GroupBy.any, last cast mask to integers for True/False to 1/0 mapping, then combinae it with previous answer:
df = pd.DataFrame({'Age': 10}, index=pd.date_range('2019-01-01', freq='5H', periods=24))
#for test 1H timestamp use
#df = pd.DataFrame({'Age': 10}, index=pd.date_range('2019-01-01', freq='H', periods=24 * 5))
df.loc[pd.Timestamp('2019-01-02 01:00:00'), 'Age'] = 95
df.loc[pd.Timestamp('2019-01-03 02:00:00'), 'Age'] = 95
df.loc[pd.Timestamp('2019-01-05 19:00:00'), 'Age'] = 95
#print (df)
#for test 48 consecutive values change N = 48
N = 10
s = df['Age'].gt(90)
s1 = (s.groupby(df.index.date).transform('any'))
g1 = s1.ne(s1.shift()).cumsum()
df['flag'] = (s.groupby(g1).transform('size').ge(N) & s1).astype(int)
print (df)
Age flag
2019-01-01 00:00:00 10 0
2019-01-01 05:00:00 10 0
2019-01-01 10:00:00 10 0
2019-01-01 15:00:00 10 0
2019-01-01 20:00:00 10 0
2019-01-02 01:00:00 95 1
2019-01-02 06:00:00 10 1
2019-01-02 11:00:00 10 1
2019-01-02 16:00:00 10 1
2019-01-02 21:00:00 10 1
2019-01-03 02:00:00 95 1
2019-01-03 07:00:00 10 1
2019-01-03 12:00:00 10 1
2019-01-03 17:00:00 10 1
2019-01-03 22:00:00 10 1
2019-01-04 03:00:00 10 0
2019-01-04 08:00:00 10 0
2019-01-04 13:00:00 10 0
2019-01-04 18:00:00 10 0
2019-01-04 23:00:00 10 0
2019-01-05 04:00:00 10 0
2019-01-05 09:00:00 10 0
2019-01-05 14:00:00 10 0
2019-01-05 19:00:00 95 0
Apparently, this could be a solution to the first version of the question: how to add a column whose row values are 1 if at least one of the rows with the same date (y-m-d) has an Age value greater than 90.
import pandas as pd
df = pd.DataFrame({
'Dates':['2019-01-01 00:00:00',
'2019-01-01 01:00:00',
'2019-01-01 02:00:00',
'2019-01-02 00:00:00',
'2019-01-02 01:00:00',
'2019-01-03 02:00:00',
'2019-01-03 03:00:00',],
'Age':[29, 56, 92, 13, 1, 2, 93],})
df.set_index('Dates', inplace=True)
df.index = pd.to_datetime(df.index)
df['flag'] = pd.DatetimeIndex(df.index).day
df['flag'] = df.flag.isin(df['flag'][df['Age']>90]).astype(int)
It returns:
Age flag
Dates
2019-01-01 00:00:00 29 1
2019-01-01 01:00:00 56 1
2019-01-01 02:00:00 92 1
2019-01-02 00:00:00 13 0
2019-01-02 01:00:00 1 0
2019-01-03 02:00:00 2 1
2019-01-03 03:00:00 93 1
Related
I am trying to create my own custom day of the week mapping using python. I have used a few different methods such as dayofweek and isoweekday. Yet all of these dont provide me with what I need.
0 2026-01-01 00:00:00
1 2026-01-01 01:00:00
2 2026-01-01 02:00:00
3 2026-01-01 03:00:00
4 2026-01-01 04:00:00
5 2026-01-01 05:00:00
6 2026-01-01 06:00:00
7 2026-01-01 07:00:00
8 2026-01-01 08:00:00
9 2026-01-01 09:00:00
10 2026-01-01 10:00:00
11 2026-01-01 11:00:00
12 2026-01-01 12:00:00
13 2026-01-01 13:00:00
14 2026-01-01 14:00:00
15 2026-01-01 15:00:00
16 2026-01-01 16:00:00
17 2026-01-01 17:00:00
18 2026-01-01 18:00:00
19 2026-01-01 19:00:00
Name: Date, dtype: datetime64[ns]
It contiunes to
0 2026-01-01 00:00:00
1 2026-01-01 01:00:00
2 2026-01-01 02:00:00
3 2026-01-01 03:00:00
4 2026-01-01 04:00:00
95 2026-01-04 23:00:00
96 2026-01-05 00:00:00
97 2026-01-05 01:00:00
98 2026-01-05 02:00:00
99 2026-01-05 03:00:00
example of my code
power_data['Day of Week'] = power_data['Date'].dt.dayofweek+1
power_data['Day of Weekv2'] = power_data['Date'].dt.isoweekday
Above is a example portion of my dataframe, the data formatting I would like to follow is Sunday be 1, monday be 2...etc and Saturday be equal to 7. Please let me know if I can do this how its currently presented
As per pandas documentation for weekday, it mentions that:
The day of the week with Monday=0, Sunday=6.
Which means, you just need to add 2 to the weekday value as you want to shift Monday from 0 to 2, then take modulo by 8
# df is your dataframe, and date is the column name consisting pandas Timestamp
>>> (df['date'].dt.weekday+2)%8
#output:
0 5
1 5
2 5
3 5
4 5
5 5
6 5
7 5
8 5
9 5
10 5
11 5
12 5
13 5
14 5
15 5
16 5
17 5
18 5
Name: date, dtype: int64
I have the following two dataframes:
print(df_diff)
print(df_census_occupation)
pacients
2019-01-01 00:10:00 1
2019-01-01 00:20:00 1
2019-01-01 00:30:00 -1
2019-01-02 10:00:00 1
2019-01-02 11:30:00 1
2019-01-03 00:00:00 -1
2019-01-03 15:00:00 -1
2019-01-03 23:30:00 -1
2019-01-04 00:00:00 1
2019-01-04 00:00:00 1
2019-01-04 10:00:00 -1
2019-01-04 10:00:00 -1
pacients_census
2019-01-01 10
2019-01-02 20
2019-01-03 30
2019-01-04 10
And I need to transform them into:
pacients
2019-01-01 00:00:00 10
2019-01-01 00:10:00 11
2019-01-01 00:20:00 12
2019-01-01 00:30:00 11
2019-01-02 00:00:00 20
2019-01-02 10:00:00 21
2019-01-02 11:30:00 22
2019-01-03 00:00:00 30
2019-01-03 00:00:00 29
2019-01-03 15:00:00 28
2019-01-03 23:30:00 27
2019-01-04 00:00:00 10
2019-01-04 00:00:00 11
2019-01-04 00:00:00 12
2019-01-04 10:00:00 11
2019-01-04 10:00:00 10
It's like a cumsum by day, where each day starts over again based on another dataframe (df_census_occupation). Attention must be taken to consider repeated hours, there may be days where we have exactly the same hour in df_diff, and such hours may also coincide with the start of the day in df_census_occupation. This is what happens in 2019-01-04 00:00:00 for example.
I tried using cumsum with masks and shifts, and also some groupby operations, but the code was becoming difficult to understand and it was not considering the repeated hours issue.
Auxiliary code to generate the two dataframes:
import datetime
df_diff_index = [
"2019-01-01 00:10:00",
"2019-01-01 00:20:00",
"2019-01-01 00:30:00",
"2019-01-02 10:00:00",
"2019-01-02 11:30:00",
"2019-01-03 00:00:00",
"2019-01-03 15:00:00",
"2019-01-03 23:30:00",
"2019-01-04 00:00:00",
"2019-01-04 00:00:00",
"2019-01-04 10:00:00",
"2019-01-04 10:00:00",
]
df_diff_index = [datetime.datetime.strptime(date, "%Y-%m-%d %H:%M:%S") for date in df_diff_index]
df_census_occupation_index = [
"2019-01-01",
"2019-01-02",
"2019-01-03",
"2019-01-04",
]
df_census_occupation_index = [datetime.datetime.strptime(date, "%Y-%m-%d") for date in df_census_occupation_index]
df_diff = pd.DataFrame({"pacients": [1, 1, -1, 1, 1, -1, -1, -1, 1, 1, -1, -1]}, index=df_diff_index)
df_census_occupation = pd.DataFrame({"pacients_census": [10, 20, 30, 10]}, index=df_census_occupation_index)
Concatenate to data, sort by index, then groupby on the day and cumsum:
out = pd.concat([df_census_occupation.rename(columns={'pacients_census':'pacients'}), df_diff]
).sort_index().groupby(pd.Grouper(freq='D')).cumsum()
Output:
pacients
2019-01-01 00:00:00 10
2019-01-01 00:10:00 11
2019-01-01 00:20:00 12
2019-01-01 00:30:00 11
2019-01-02 00:00:00 20
2019-01-02 10:00:00 21
2019-01-02 11:30:00 22
2019-01-03 00:00:00 30
2019-01-03 00:00:00 29
2019-01-03 15:00:00 28
2019-01-03 23:30:00 27
2019-01-04 00:00:00 10
2019-01-04 00:00:00 11
2019-01-04 00:00:00 12
2019-01-04 10:00:00 11
2019-01-04 10:00:00 10
note you may want to pass kind='mergesort' to sort_index so the sort is stable, i.e. concensus goes before the data.
I'm trying to use pd.cut to divide 24 hours into the following interval:
[6,11),[11,14),[14,17),[17,22),[22,6)
How could I achieve the last bin [22,6)?
Assuming some form of datetime column, try offsetting the datetime by 6 hours so that the lower bound becomes midnight. Then cutting based on those hours instead, with the custom labels:
import pandas as pd
# sample data
df = pd.DataFrame({
'datetime': pd.date_range('2021-01-01', periods=24, freq='H')
})
df['bins'] = pd.cut((df['datetime'] - pd.Timedelta(hours=6)).dt.hour,
bins=[0, 5, 8, 11, 16, 24],
labels=['[6,11)', '[11,14)', '[14,17)',
'[17,22)', '[22,6)'],
right=False)
df:
datetime bins
0 2021-01-01 00:00:00 [22,6)
1 2021-01-01 01:00:00 [22,6)
2 2021-01-01 02:00:00 [22,6)
3 2021-01-01 03:00:00 [22,6)
4 2021-01-01 04:00:00 [22,6)
5 2021-01-01 05:00:00 [22,6)
6 2021-01-01 06:00:00 [6,11)
7 2021-01-01 07:00:00 [6,11)
8 2021-01-01 08:00:00 [6,11)
9 2021-01-01 09:00:00 [6,11)
10 2021-01-01 10:00:00 [6,11)
11 2021-01-01 11:00:00 [11,14)
12 2021-01-01 12:00:00 [11,14)
13 2021-01-01 13:00:00 [11,14)
14 2021-01-01 14:00:00 [14,17)
15 2021-01-01 15:00:00 [14,17)
16 2021-01-01 16:00:00 [14,17)
17 2021-01-01 17:00:00 [17,22)
18 2021-01-01 18:00:00 [17,22)
19 2021-01-01 19:00:00 [17,22)
20 2021-01-01 20:00:00 [17,22)
21 2021-01-01 21:00:00 [17,22)
22 2021-01-01 22:00:00 [22,6)
23 2021-01-01 23:00:00 [22,6)
I have a dataframe df that contains datetimes for every hour of a day between 2003-02-12 to 2017-06-30 and I want to delete all datetimes between 24th Dec and 1st Jan of EVERY year.
An extract of my data frame is:
...
7505,2003-12-23 17:00:00
7506,2003-12-23 18:00:00
7507,2003-12-23 19:00:00
7508,2003-12-23 20:00:00
7509,2003-12-23 21:00:00
7510,2003-12-23 22:00:00
7511,2003-12-23 23:00:00
7512,2003-12-24 00:00:00
7513,2003-12-24 01:00:00
7514,2003-12-24 02:00:00
7515,2003-12-24 03:00:00
7516,2003-12-24 04:00:00
7517,2003-12-24 05:00:00
7518,2003-12-24 06:00:00
...
7723,2004-01-01 19:00:00
7724,2004-01-01 20:00:00
7725,2004-01-01 21:00:00
7726,2004-01-01 22:00:00
7727,2004-01-01 23:00:00
7728,2004-01-02 00:00:00
7729,2004-01-02 01:00:00
7730,2004-01-02 02:00:00
7731,2004-01-02 03:00:00
7732,2004-01-02 04:00:00
7733,2004-01-02 05:00:00
7734,2004-01-02 06:00:00
7735,2004-01-02 07:00:00
...
and my expected output is:
...
7505,2003-12-23 17:00:00
7506,2003-12-23 18:00:00
7507,2003-12-23 19:00:00
7508,2003-12-23 20:00:00
7509,2003-12-23 21:00:00
7510,2003-12-23 22:00:00
7511,2003-12-23 23:00:00
...
7728,2004-01-02 00:00:00
7729,2004-01-02 01:00:00
7730,2004-01-02 02:00:00
7731,2004-01-02 03:00:00
7732,2004-01-02 04:00:00
7733,2004-01-02 05:00:00
7734,2004-01-02 06:00:00
7735,2004-01-02 07:00:00
...
Sample dataframe:
dates
0 2003-12-23 23:00:00
1 2003-12-24 05:00:00
2 2004-12-27 05:00:00
3 2003-12-13 23:00:00
4 2002-12-23 23:00:00
5 2004-01-01 05:00:00
6 2014-12-24 05:00:00
Solution:
If you want it for every year between the following dates excluded, then extract the month and dates first:
df['month'] = df['dates'].dt.month
df['day'] = df['dates'].dt.day
And now put the condition check:
dec_days = [24, 25, 26, 27, 28, 29, 30, 31]
## if the month is dec, then check for these dates
## if the month is jan, then just check for the day to be 1 like below
df = df[~(((df.month == 12) & (df.day.isin(dec_days))) | ((df.month == 1) & (df.day == 1)))]
Sample output:
dates month day
0 2003-12-23 23:00:00 12 23
3 2003-12-13 23:00:00 12 13
4 2002-12-23 23:00:00 12 23
This takes advantage of the fact that datetime-strings in the form mm-dd are sortable. Read everything in from the CSV file then filter for the dates you want:
df = pd.read_csv('...', parse_dates=['DateTime'])
s = df['DateTime'].dt.strftime('%m-%d')
excluded = (s == '01-01') | (s >= '12-24') # Jan 1 or >= Dec 24
df[~excluded]
You can try dropping on conditionals. Maybe with a pattern match to the date string or parsing the date as a number (like in Java) and conditionally removing.
datesIdontLike = df[df['colname'] == <stringPattern>].index
newDF = df.drop(datesIdontLike, inplace=True)
Check this out: https://thispointer.com/python-pandas-how-to-drop-rows-in-dataframe-by-conditions-on-column-values/
(If you have issues, let me know.)
You can use pandas and boolean filtering with strftime
# version 0.23.4
import pandas as pd
# make df
df = pd.DataFrame(pd.date_range('20181223', '20190103', freq='H'), columns=['date'])
# string format the date to only include the month and day
# then set it strictly less than '12-24' AND greater than or equal to `01-02`
df = df.loc[
(df.date.dt.strftime('%m-%d') < '12-24') &
(df.date.dt.strftime('%m-%d') >= '01-02')
].copy()
print(df)
date
0 2018-12-23 00:00:00
1 2018-12-23 01:00:00
2 2018-12-23 02:00:00
3 2018-12-23 03:00:00
4 2018-12-23 04:00:00
5 2018-12-23 05:00:00
6 2018-12-23 06:00:00
7 2018-12-23 07:00:00
8 2018-12-23 08:00:00
9 2018-12-23 09:00:00
10 2018-12-23 10:00:00
11 2018-12-23 11:00:00
12 2018-12-23 12:00:00
13 2018-12-23 13:00:00
14 2018-12-23 14:00:00
15 2018-12-23 15:00:00
16 2018-12-23 16:00:00
17 2018-12-23 17:00:00
18 2018-12-23 18:00:00
19 2018-12-23 19:00:00
20 2018-12-23 20:00:00
21 2018-12-23 21:00:00
22 2018-12-23 22:00:00
23 2018-12-23 23:00:00
240 2019-01-02 00:00:00
241 2019-01-02 01:00:00
242 2019-01-02 02:00:00
243 2019-01-02 03:00:00
244 2019-01-02 04:00:00
245 2019-01-02 05:00:00
246 2019-01-02 06:00:00
247 2019-01-02 07:00:00
248 2019-01-02 08:00:00
249 2019-01-02 09:00:00
250 2019-01-02 10:00:00
251 2019-01-02 11:00:00
252 2019-01-02 12:00:00
253 2019-01-02 13:00:00
254 2019-01-02 14:00:00
255 2019-01-02 15:00:00
256 2019-01-02 16:00:00
257 2019-01-02 17:00:00
258 2019-01-02 18:00:00
259 2019-01-02 19:00:00
260 2019-01-02 20:00:00
261 2019-01-02 21:00:00
262 2019-01-02 22:00:00
263 2019-01-02 23:00:00
264 2019-01-03 00:00:00
This will work with multiple years because we are only filtering on the month and day.
# change range to include 2017
df = pd.DataFrame(pd.date_range('20171223', '20190103', freq='H'), columns=['date'])
df = df.loc[
(df.date.dt.strftime('%m-%d') < '12-24') &
(df.date.dt.strftime('%m-%d') >= '01-02')
].copy()
print(df)
date
0 2017-12-23 00:00:00
1 2017-12-23 01:00:00
2 2017-12-23 02:00:00
3 2017-12-23 03:00:00
4 2017-12-23 04:00:00
5 2017-12-23 05:00:00
6 2017-12-23 06:00:00
7 2017-12-23 07:00:00
8 2017-12-23 08:00:00
9 2017-12-23 09:00:00
10 2017-12-23 10:00:00
11 2017-12-23 11:00:00
12 2017-12-23 12:00:00
13 2017-12-23 13:00:00
14 2017-12-23 14:00:00
15 2017-12-23 15:00:00
16 2017-12-23 16:00:00
17 2017-12-23 17:00:00
18 2017-12-23 18:00:00
19 2017-12-23 19:00:00
20 2017-12-23 20:00:00
21 2017-12-23 21:00:00
22 2017-12-23 22:00:00
23 2017-12-23 23:00:00
240 2018-01-02 00:00:00
241 2018-01-02 01:00:00
242 2018-01-02 02:00:00
243 2018-01-02 03:00:00
244 2018-01-02 04:00:00
245 2018-01-02 05:00:00
... ...
8779 2018-12-23 19:00:00
8780 2018-12-23 20:00:00
8781 2018-12-23 21:00:00
8782 2018-12-23 22:00:00
8783 2018-12-23 23:00:00
9000 2019-01-02 00:00:00
9001 2019-01-02 01:00:00
9002 2019-01-02 02:00:00
9003 2019-01-02 03:00:00
9004 2019-01-02 04:00:00
9005 2019-01-02 05:00:00
9006 2019-01-02 06:00:00
9007 2019-01-02 07:00:00
9008 2019-01-02 08:00:00
9009 2019-01-02 09:00:00
9010 2019-01-02 10:00:00
9011 2019-01-02 11:00:00
9012 2019-01-02 12:00:00
9013 2019-01-02 13:00:00
9014 2019-01-02 14:00:00
9015 2019-01-02 15:00:00
9016 2019-01-02 16:00:00
9017 2019-01-02 17:00:00
9018 2019-01-02 18:00:00
9019 2019-01-02 19:00:00
9020 2019-01-02 20:00:00
9021 2019-01-02 21:00:00
9022 2019-01-02 22:00:00
9023 2019-01-02 23:00:00
9024 2019-01-03 00:00:00
Since you want this to happen for every year, we can first define a series that where we replace the year by a static value (2000 for example). Let date be the column that stores the date, we can generate such column as:
dt = pd.to_datetime({'year': 2000, 'month': df['date'].dt.month, 'day': df['date'].dt.day})
For the given sample data, we get:
>>> dt
0 2000-12-23
1 2000-12-23
2 2000-12-23
3 2000-12-23
4 2000-12-23
5 2000-12-23
6 2000-12-23
7 2000-12-24
8 2000-12-24
9 2000-12-24
10 2000-12-24
11 2000-12-24
12 2000-12-24
13 2000-12-24
14 2000-01-01
15 2000-01-01
16 2000-01-01
17 2000-01-01
18 2000-01-01
19 2000-01-02
20 2000-01-02
21 2000-01-02
22 2000-01-02
23 2000-01-02
24 2000-01-02
25 2000-01-02
26 2000-01-02
dtype: datetime64[ns]
Next we can filter the rows, like:
from datetime import date
df[(dt >= date(2000,1,2)) & (dt < date(2000,12,24))]
This gives us the following data for your sample data:
>>> df[(dt >= date(2000,1,2)) & (dt < date(2000,12,24))]
id dt
0 7505 2003-12-23 17:00:00
1 7506 2003-12-23 18:00:00
2 7507 2003-12-23 19:00:00
3 7508 2003-12-23 20:00:00
4 7509 2003-12-23 21:00:00
5 7510 2003-12-23 22:00:00
6 7511 2003-12-23 23:00:00
19 7728 2004-01-02 00:00:00
20 7729 2004-01-02 01:00:00
21 7730 2004-01-02 02:00:00
22 7731 2004-01-02 03:00:00
23 7732 2004-01-02 04:00:00
24 7733 2004-01-02 05:00:00
25 7734 2004-01-02 06:00:00
26 7735 2004-01-02 07:00:00
So regardless what the year is, we will only consider dates between the 2nd of January and the 23rd of December (both inclusive).
I'm not able to create a Pandas Series of every hour (as datetime objects) of a given year without iterating and adding one hour to the last, and that's slow. Is there any way to do that paralelly.
My input would be a year and the output should be a Pandas Series of every hour of that year.
You can use pd.date_range with freq='H' which is hourly frequency:
Edit with 23:00:00 after comment by #ALollz
year = 2019
pd.Series(pd.date_range(start=f'{year}-01-01', end=f'{year}-12-31 23:00:00', freq='H'))
0 2019-01-01 00:00:00
1 2019-01-01 01:00:00
2 2019-01-01 02:00:00
3 2019-01-01 03:00:00
4 2019-01-01 04:00:00
5 2019-01-01 05:00:00
6 2019-01-01 06:00:00
7 2019-01-01 07:00:00
8 2019-01-01 08:00:00
9 2019-01-01 09:00:00
10 2019-01-01 10:00:00
11 2019-01-01 11:00:00
12 2019-01-01 12:00:00
13 2019-01-01 13:00:00
14 2019-01-01 14:00:00
15 2019-01-01 15:00:00
16 2019-01-01 16:00:00
17 2019-01-01 17:00:00
18 2019-01-01 18:00:00
19 2019-01-01 19:00:00
20 2019-01-01 20:00:00
21 2019-01-01 21:00:00
22 2019-01-01 22:00:00
23 2019-01-01 23:00:00
24 2019-01-02 00:00:00
25 2019-01-02 01:00:00
26 2019-01-02 02:00:00
27 2019-01-02 03:00:00
28 2019-01-02 04:00:00
29 2019-01-02 05:00:00
30 2019-01-02 06:00:00
31 2019-01-02 07:00:00
32 2019-01-02 08:00:00
33 2019-01-02 09:00:00
34 2019-01-02 10:00:00
35 2019-01-02 11:00:00
36 2019-01-02 12:00:00
37 2019-01-02 13:00:00
38 2019-01-02 14:00:00
39 2019-01-02 15:00:00
40 2019-01-02 16:00:00
41 2019-01-02 17:00:00
42 2019-01-02 18:00:00
43 2019-01-02 19:00:00
44 2019-01-02 20:00:00
45 2019-01-02 21:00:00
46 2019-01-02 22:00:00
47 2019-01-02 23:00:00
48 2019-01-03 00:00:00
49 2019-01-03 01:00:00
...
8711 2019-12-29 23:00:00
8712 2019-12-30 00:00:00
8713 2019-12-30 01:00:00
8714 2019-12-30 02:00:00
8715 2019-12-30 03:00:00
8716 2019-12-30 04:00:00
8717 2019-12-30 05:00:00
8718 2019-12-30 06:00:00
8719 2019-12-30 07:00:00
8720 2019-12-30 08:00:00
8721 2019-12-30 09:00:00
8722 2019-12-30 10:00:00
8723 2019-12-30 11:00:00
8724 2019-12-30 12:00:00
8725 2019-12-30 13:00:00
8726 2019-12-30 14:00:00
8727 2019-12-30 15:00:00
8728 2019-12-30 16:00:00
8729 2019-12-30 17:00:00
8730 2019-12-30 18:00:00
8731 2019-12-30 19:00:00
8732 2019-12-30 20:00:00
8733 2019-12-30 21:00:00
8734 2019-12-30 22:00:00
8735 2019-12-30 23:00:00
8736 2019-12-31 00:00:00
8737 2019-12-31 01:00:00
8738 2019-12-31 02:00:00
8739 2019-12-31 03:00:00
8740 2019-12-31 04:00:00
8741 2019-12-31 05:00:00
8742 2019-12-31 06:00:00
8743 2019-12-31 07:00:00
8744 2019-12-31 08:00:00
8745 2019-12-31 09:00:00
8746 2019-12-31 10:00:00
8747 2019-12-31 11:00:00
8748 2019-12-31 12:00:00
8749 2019-12-31 13:00:00
8750 2019-12-31 14:00:00
8751 2019-12-31 15:00:00
8752 2019-12-31 16:00:00
8753 2019-12-31 17:00:00
8754 2019-12-31 18:00:00
8755 2019-12-31 19:00:00
8756 2019-12-31 20:00:00
8757 2019-12-31 21:00:00
8758 2019-12-31 22:00:00
8759 2019-12-31 23:00:00
8760 2020-01-01 00:00:00
Length: 8761, dtype: datetime64[ns]
Note if your Python version is lower than 3.6 use .format for string formatting:
year = 2019
pd.Series(pd.date_range(start='{}-01-01'.format(year), end='{}-01-01 23:00:00'.format(year), freq='H'))