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Generating list of 5 minute interval between two times

I have the following strings:
start = "07:00:00"
end = "17:00:00"
How can I generate a list of 5 minute interval between those times, ie
["07:00:00","07:05:00",...,"16:55:00","17:00:00"]

This works for me, I'm sure you can figure out how to put the results in the list instead of printing them out:
>>> import datetime
>>> start = "07:00:00"
>>> end = "17:00:00"
>>> delta = datetime.timedelta(minutes=5)
>>> start = datetime.datetime.strptime( start, '%H:%M:%S' )
>>> end = datetime.datetime.strptime( end, '%H:%M:%S' )
>>> t = start
>>> while t <= end :
... print datetime.datetime.strftime( t, '%H:%M:%S')
... t += delta
...
07:00:00
07:05:00
07:10:00
07:15:00
07:20:00
07:25:00
07:30:00
07:35:00
07:40:00
07:45:00
07:50:00
07:55:00
08:00:00
08:05:00
08:10:00
08:15:00
08:20:00
08:25:00
08:30:00
08:35:00
08:40:00
08:45:00
08:50:00
08:55:00
09:00:00
09:05:00
09:10:00
09:15:00
09:20:00
09:25:00
09:30:00
09:35:00
09:40:00
09:45:00
09:50:00
09:55:00
10:00:00
10:05:00
10:10:00
10:15:00
10:20:00
10:25:00
10:30:00
10:35:00
10:40:00
10:45:00
10:50:00
10:55:00
11:00:00
11:05:00
11:10:00
11:15:00
11:20:00
11:25:00
11:30:00
11:35:00
11:40:00
11:45:00
11:50:00
11:55:00
12:00:00
12:05:00
12:10:00
12:15:00
12:20:00
12:25:00
12:30:00
12:35:00
12:40:00
12:45:00
12:50:00
12:55:00
13:00:00
13:05:00
13:10:00
13:15:00
13:20:00
13:25:00
13:30:00
13:35:00
13:40:00
13:45:00
13:50:00
13:55:00
14:00:00
14:05:00
14:10:00
14:15:00
14:20:00
14:25:00
14:30:00
14:35:00
14:40:00
14:45:00
14:50:00
14:55:00
15:00:00
15:05:00
15:10:00
15:15:00
15:20:00
15:25:00
15:30:00
15:35:00
15:40:00
15:45:00
15:50:00
15:55:00
16:00:00
16:05:00
16:10:00
16:15:00
16:20:00
16:25:00
16:30:00
16:35:00
16:40:00
16:45:00
16:50:00
16:55:00
17:00:00
>>>

Try:
# import modules
from datetime import datetime, timedelta
# Create starting and end datetime object from string
start = datetime.strptime("07:00:00", "%H:%M:%S")
end = datetime.strptime("17:00:00", "%H:%M:%S")
# min_gap
min_gap = 5
# compute datetime interval
arr = [(start + timedelta(hours=min_gap*i/60)).strftime("%H:%M:%S")
for i in range(int((end-start).total_seconds() / 60.0 / min_gap))]
print(arr)
# ['07:00:00', '07:05:00', '07:10:00', '07:15:00', '07:20:00', '07:25:00', '07:30:00', ..., '16:55:00']
Explanations:
First, you need to convert string date to datetime object. The strptime does it!
Then, we will find the number of minutes between the starting date and the ending datetime. This discussion solved it! We can do it like this :
(end-start).total_seconds() / 60.0
However, in our case, we only want to iterate every n minutes. So, in our loop, we need to divide it by n.
Also, as we will iterate over this number of minutes, we need to convertit to int for the for loop. That results in:
int((end-start).total_seconds() / 60.0 / min_gap)
Then, on each element of our loop, we will add the number of minutes to the initial datetime. The tiemdelta function has been designed for. As parameter, we specify the number of hours we want to add : min_gap*i/60.
Finally, we convert this datetime object back to a string object using the strftime.

How to reset a date time column to be in increments of one minute in python

I have a dataframe that has a date time column called start time and it is set to a default of 12:00:00 AM. I would like to reset this column so that the first row is 00:01:00 and the second row is 00:02:00, that is one minute interval.
This is the original table.
ID State Time End Time
A001 12:00:00 12:00:00
A002 12:00:00 12:00:00
A003 12:00:00 12:00:00
A004 12:00:00 12:00:00
A005 12:00:00 12:00:00
A006 12:00:00 12:00:00
A007 12:00:00 12:00:00
I want to reset the start time column so that my output is this:
ID State Time End Time
A001 0:00:00 12:00:00
A002 0:00:01 12:00:00
A003 0:00:02 12:00:00
A004 0:00:03 12:00:00
A005 0:00:04 12:00:00
A006 0:00:05 12:00:00
A007 0:00:06 12:00:00
How do I go about this?

you could use pd.date_range:
df['Start Time'] = pd.date_range('00:00', periods=df['Start Time'].shape[0], freq='1min')
gives you
df
Out[23]:
Start Time
0 2019-09-30 00:00:00
1 2019-09-30 00:01:00
2 2019-09-30 00:02:00
3 2019-09-30 00:03:00
4 2019-09-30 00:04:00
5 2019-09-30 00:05:00
6 2019-09-30 00:06:00
7 2019-09-30 00:07:00
8 2019-09-30 00:08:00
9 2019-09-30 00:09:00
supply a full date/time string to get another starting date.

First we convert your State Time column to datetime type. Then we use pd.date_range and use the first time as starting point with a frequency of 1 minute.
df['State Time'] = pd.to_datetime(df['State Time'])
df['State Time'] = pd.date_range(start=df['State Time'].min(),
periods=len(df),
freq='min').time
Output
ID State Time End Time
0 A001 12:00:00 12:00:00
1 A002 12:01:00 12:00:00
2 A003 12:02:00 12:00:00
3 A004 12:03:00 12:00:00
4 A005 12:04:00 12:00:00
5 A006 12:05:00 12:00:00
6 A007 12:06:00 12:00:00

Convert datetime to the cloest time point

I have a dateset as below.
dummy
datetime
2015-10-25 06:00:00 1
2015-04-05 20:00:00 1
2015-11-24 00:00:00 1
2015-08-18 08:00:00 1
2015-10-21 12:00:00 1
I want to change the datetime to the cloest predefined time point, say 00:00:00 and 12:00:00
dummy
datetime
2015-10-25 00:00:00 1
2015-04-05 12:00:00 1
2015-11-24 00:00:00 1
2015-08-18 00:00:00 1
2015-10-21 12:00:00 1

Here is possible use DatetimeIndex.floor:
df.index = df.index.floor('12H')
print (df)
dummy
datetime
2015-10-25 00:00:00 1
2015-04-05 12:00:00 1
2015-11-24 00:00:00 1
2015-08-18 00:00:00 1
2015-10-21 12:00:00 1

Filtering pandas dataframe by difference of adjacent rows

I have a dataframe indexed by datetime. I want to filter out rows based on the difference between their index and the index of the previous row.
So, if my criteria is "remove all rows that are over one hour late than the previous row", the second row in the example below should be removed:
2005-07-15 17:00:00
2005-07-17 18:00:00
While in the following case, both rows stay:
2005-07-17 23:00:00
2005-07-18 00:00:00

It seems you need boolean indexing with diff for difference and compare with 1 hour Timedelta:
dates=['2005-07-15 17:00:00','2005-07-17 18:00:00', '2005-07-17 19:00:00',
'2005-07-17 23:00:00', '2005-07-18 00:00:00']
df = pd.DataFrame({'a':range(5)}, index=pd.to_datetime(dates))
print (df)
a
2005-07-15 17:00:00 0
2005-07-17 18:00:00 1
2005-07-17 19:00:00 2
2005-07-17 23:00:00 3
2005-07-18 00:00:00 4
diff = df.index.to_series().diff().fillna(0)
print (diff)
2005-07-15 17:00:00 0 days 00:00:00
2005-07-17 18:00:00 2 days 01:00:00
2005-07-17 19:00:00 0 days 01:00:00
2005-07-17 23:00:00 0 days 04:00:00
2005-07-18 00:00:00 0 days 01:00:00
dtype: timedelta64[ns]
mask = diff <= pd.Timedelta(1, unit='h')
print (mask)
2005-07-15 17:00:00 True
2005-07-17 18:00:00 False
2005-07-17 19:00:00 True
2005-07-17 23:00:00 False
2005-07-18 00:00:00 True
dtype: bool
df = df[mask]
print (df)
a
2005-07-15 17:00:00 0
2005-07-17 19:00:00 2
2005-07-18 00:00:00 4

How to resample a df with datetime index to exactly n equally sized periods?

I've got a large dataframe with a datetime index and need to resample data to exactly 10 equally sized periods.
So far, I've tried finding the first and last dates to determine the total number of days in the data, divide that by 10 to determine the size of each period, then resample using that number of days. eg:
first = df.reset_index().timesubmit.min()
last = df.reset_index().timesubmit.max()
periodsize = str((last-first).days/10) + 'D'
df.resample(periodsize,how='sum')
This doesn't guarantee exactly 10 periods in the df after resampling since the periodsize is a rounded down int. Using a float doesn't work in the resampling. Seems that either there's something simple that I'm missing here, or I'm attacking the problem all wrong.

import numpy as np
import pandas as pd
n = 10
nrows = 33
index = pd.date_range('2000-1-1', periods=nrows, freq='D')
df = pd.DataFrame(np.ones(nrows), index=index)
print(df)
# 0
# 2000-01-01 1
# 2000-01-02 1
# ...
# 2000-02-01 1
# 2000-02-02 1
first = df.index.min()
last = df.index.max() + pd.Timedelta('1D')
secs = int((last-first).total_seconds()//n)
periodsize = '{:d}S'.format(secs)
result = df.resample(periodsize, how='sum')
print('\n{}'.format(result))
assert len(result) == n
yields
0
2000-01-01 00:00:00 4
2000-01-04 07:12:00 3
2000-01-07 14:24:00 3
2000-01-10 21:36:00 4
2000-01-14 04:48:00 3
2000-01-17 12:00:00 3
2000-01-20 19:12:00 4
2000-01-24 02:24:00 3
2000-01-27 09:36:00 3
2000-01-30 16:48:00 3
The values in the 0-column indicate the number of rows that were aggregated, since the original DataFrame was filled with values of 1. The pattern of 4's and 3's is about as even as you can get since 33 rows can not be evenly grouped into 10 groups.
Explanation: Consider this simpler DataFrame:
n = 2
nrows = 5
index = pd.date_range('2000-1-1', periods=nrows, freq='D')
df = pd.DataFrame(np.ones(nrows), index=index)
# 0
# 2000-01-01 1
# 2000-01-02 1
# 2000-01-03 1
# 2000-01-04 1
# 2000-01-05 1
Using df.resample('2D', how='sum') gives the wrong number of groups
In [366]: df.resample('2D', how='sum')
Out[366]:
0
2000-01-01 2
2000-01-03 2
2000-01-05 1
Using df.resample('3D', how='sum') gives the right number of groups, but the
second group starts at 2000-01-04 which does not evenly divide the DataFrame
into two equally-spaced groups:
In [367]: df.resample('3D', how='sum')
Out[367]:
0
2000-01-01 3
2000-01-04 2
To do better, we need to work at a finer time resolution than in days. Since Timedeltas have a total_seconds method, let's work in seconds. So for the example above, the desired frequency string would be
In [374]: df.resample('216000S', how='sum')
Out[374]:
0
2000-01-01 00:00:00 3
2000-01-03 12:00:00 2
since there are 216000*2 seconds in 5 days:
In [373]: (pd.Timedelta(days=5) / pd.Timedelta('1S'))/2
Out[373]: 216000.0
Okay, so now all we need is a way to generalize this. We'll want the minimum and maximum dates in the index:
first = df.index.min()
last = df.index.max() + pd.Timedelta('1D')
We add an extra day because it makes the difference in days come out right. In
the example above, There are only 4 days between the Timestamps for 2000-01-05
and 2000-01-01,
In [377]: (pd.Timestamp('2000-01-05')-pd.Timestamp('2000-01-01')).days
Out[378]: 4
But as we can see in the worked example, the DataFrame has 5 rows representing 5
days. So it makes sense that we need to add an extra day.
Now we can compute the correct number of seconds in each equally-spaced group with:
secs = int((last-first).total_seconds()//n)

Here is one way to ensure equal-size sub-periods by using np.linspace() on pd.Timedelta and then classifying each obs into different bins using pd.cut.
import pandas as pd
import numpy as np
# generate artificial data
np.random.seed(0)
df = pd.DataFrame(np.random.randn(100, 2), columns=['A', 'B'], index=pd.date_range('2015-01-01 00:00:00', periods=100, freq='8H'))
Out[87]:
A B
2015-01-01 00:00:00 1.7641 0.4002
2015-01-01 08:00:00 0.9787 2.2409
2015-01-01 16:00:00 1.8676 -0.9773
2015-01-02 00:00:00 0.9501 -0.1514
2015-01-02 08:00:00 -0.1032 0.4106
2015-01-02 16:00:00 0.1440 1.4543
2015-01-03 00:00:00 0.7610 0.1217
2015-01-03 08:00:00 0.4439 0.3337
2015-01-03 16:00:00 1.4941 -0.2052
2015-01-04 00:00:00 0.3131 -0.8541
2015-01-04 08:00:00 -2.5530 0.6536
2015-01-04 16:00:00 0.8644 -0.7422
2015-01-05 00:00:00 2.2698 -1.4544
2015-01-05 08:00:00 0.0458 -0.1872
2015-01-05 16:00:00 1.5328 1.4694
... ... ...
2015-01-29 08:00:00 0.9209 0.3187
2015-01-29 16:00:00 0.8568 -0.6510
2015-01-30 00:00:00 -1.0342 0.6816
2015-01-30 08:00:00 -0.8034 -0.6895
2015-01-30 16:00:00 -0.4555 0.0175
2015-01-31 00:00:00 -0.3540 -1.3750
2015-01-31 08:00:00 -0.6436 -2.2234
2015-01-31 16:00:00 0.6252 -1.6021
2015-02-01 00:00:00 -1.1044 0.0522
2015-02-01 08:00:00 -0.7396 1.5430
2015-02-01 16:00:00 -1.2929 0.2671
2015-02-02 00:00:00 -0.0393 -1.1681
2015-02-02 08:00:00 0.5233 -0.1715
2015-02-02 16:00:00 0.7718 0.8235
2015-02-03 00:00:00 2.1632 1.3365
[100 rows x 2 columns]
# cutoff points, 10 equal-size group requires 11 points
# measured by timedelta 1 hour
time_delta_in_hours = (df.index - df.index[0]) / pd.Timedelta('1h')
n = 10
ts_cutoff = np.linspace(0, time_delta_in_hours[-1], n+1)
# labels, time index
time_index = df.index[0] + np.array([pd.Timedelta(str(time_delta)+'h') for time_delta in ts_cutoff])
# create a categorical reference variables
df['start_time_index'] = pd.cut(time_delta_in_hours, bins=10, labels=time_index[:-1])
# for clarity, reassign labels using end-period index
df['end_time_index'] = pd.cut(time_delta_in_hours, bins=10, labels=time_index[1:])
Out[89]:
A B start_time_index end_time_index
2015-01-01 00:00:00 1.7641 0.4002 2015-01-01 00:00:00 2015-01-04 07:12:00
2015-01-01 08:00:00 0.9787 2.2409 2015-01-01 00:00:00 2015-01-04 07:12:00
2015-01-01 16:00:00 1.8676 -0.9773 2015-01-01 00:00:00 2015-01-04 07:12:00
2015-01-02 00:00:00 0.9501 -0.1514 2015-01-01 00:00:00 2015-01-04 07:12:00
2015-01-02 08:00:00 -0.1032 0.4106 2015-01-01 00:00:00 2015-01-04 07:12:00
2015-01-02 16:00:00 0.1440 1.4543 2015-01-01 00:00:00 2015-01-04 07:12:00
2015-01-03 00:00:00 0.7610 0.1217 2015-01-01 00:00:00 2015-01-04 07:12:00
2015-01-03 08:00:00 0.4439 0.3337 2015-01-01 00:00:00 2015-01-04 07:12:00
2015-01-03 16:00:00 1.4941 -0.2052 2015-01-01 00:00:00 2015-01-04 07:12:00
2015-01-04 00:00:00 0.3131 -0.8541 2015-01-01 00:00:00 2015-01-04 07:12:00
2015-01-04 08:00:00 -2.5530 0.6536 2015-01-04 07:12:00 2015-01-07 14:24:00
2015-01-04 16:00:00 0.8644 -0.7422 2015-01-04 07:12:00 2015-01-07 14:24:00
2015-01-05 00:00:00 2.2698 -1.4544 2015-01-04 07:12:00 2015-01-07 14:24:00
2015-01-05 08:00:00 0.0458 -0.1872 2015-01-04 07:12:00 2015-01-07 14:24:00
2015-01-05 16:00:00 1.5328 1.4694 2015-01-04 07:12:00 2015-01-07 14:24:00
... ... ... ... ...
2015-01-29 08:00:00 0.9209 0.3187 2015-01-27 09:36:00 2015-01-30 16:48:00
2015-01-29 16:00:00 0.8568 -0.6510 2015-01-27 09:36:00 2015-01-30 16:48:00
2015-01-30 00:00:00 -1.0342 0.6816 2015-01-27 09:36:00 2015-01-30 16:48:00
2015-01-30 08:00:00 -0.8034 -0.6895 2015-01-27 09:36:00 2015-01-30 16:48:00
2015-01-30 16:00:00 -0.4555 0.0175 2015-01-27 09:36:00 2015-01-30 16:48:00
2015-01-31 00:00:00 -0.3540 -1.3750 2015-01-30 16:48:00 2015-02-03 00:00:00
2015-01-31 08:00:00 -0.6436 -2.2234 2015-01-30 16:48:00 2015-02-03 00:00:00
2015-01-31 16:00:00 0.6252 -1.6021 2015-01-30 16:48:00 2015-02-03 00:00:00
2015-02-01 00:00:00 -1.1044 0.0522 2015-01-30 16:48:00 2015-02-03 00:00:00
2015-02-01 08:00:00 -0.7396 1.5430 2015-01-30 16:48:00 2015-02-03 00:00:00
2015-02-01 16:00:00 -1.2929 0.2671 2015-01-30 16:48:00 2015-02-03 00:00:00
2015-02-02 00:00:00 -0.0393 -1.1681 2015-01-30 16:48:00 2015-02-03 00:00:00
2015-02-02 08:00:00 0.5233 -0.1715 2015-01-30 16:48:00 2015-02-03 00:00:00
2015-02-02 16:00:00 0.7718 0.8235 2015-01-30 16:48:00 2015-02-03 00:00:00
2015-02-03 00:00:00 2.1632 1.3365 2015-01-30 16:48:00 2015-02-03 00:00:00
[100 rows x 4 columns]
df.groupby('start_time_index').agg('sum')
Out[90]:
A B
start_time_index
2015-01-01 00:00:00 8.6133 2.7734
2015-01-04 07:12:00 1.9220 -0.8069
2015-01-07 14:24:00 -8.1334 0.2318
2015-01-10 21:36:00 -2.7572 -4.2862
2015-01-14 04:48:00 1.1957 7.2285
2015-01-17 12:00:00 3.2485 6.6841
2015-01-20 19:12:00 -0.8903 2.2802
2015-01-24 02:24:00 -2.1025 1.3800
2015-01-27 09:36:00 -1.1017 1.3108
2015-01-30 16:48:00 -0.0902 -2.5178
Another potential shorter way to do this is to specify your sampling freq as the time delta. But the problem, as shown in below, is that it delivers 11 sub-samples instead of 10. I believe the reason is that the resample implements a left-inclusive/right-exclusive (or left-exclusive/right-inclusive) sub-sampling scheme so that the very last obs at '2015-02-03 00:00:00' is considered as a separate group. If we use pd.cut to do it ourself, we can specify include_lowest=True so that it gives us exactly 10 sub-samples rather than 11.
n = 10
time_delta_str = str((df.index[-1] - df.index[0]) / (pd.Timedelta('1s') * n)) + 's'
df.resample(pd.Timedelta(time_delta_str), how='sum')
Out[114]:
A B
2015-01-01 00:00:00 8.6133 2.7734
2015-01-04 07:12:00 1.9220 -0.8069
2015-01-07 14:24:00 -8.1334 0.2318
2015-01-10 21:36:00 -2.7572 -4.2862
2015-01-14 04:48:00 1.1957 7.2285
2015-01-17 12:00:00 3.2485 6.6841
2015-01-20 19:12:00 -0.8903 2.2802
2015-01-24 02:24:00 -2.1025 1.3800
2015-01-27 09:36:00 -1.1017 1.3108
2015-01-30 16:48:00 -2.2534 -3.8543
2015-02-03 00:00:00 2.1632 1.3365