Grouping pandas DataFrame by 10 minute intervals

With a little vectorised arithmetic, this should be possible (and be performant):

# Convert to datetime if not already.
# df['timestamp'] = pd.to_datetime(df['timestamp'], errors='coerce')    
u = (df.assign(timestamp=df['timestamp'].dt.floor('20min'))
       .groupby(pd.Grouper(key='timestamp',freq='10min'))
       .ngroup())

df['10min_period'] = np.char.add('period_', (pd.factorize(u)[0] + 1).astype(str))

Unfortunately, the drawback here is that, while this will produce the expected output for your sample data, there is no easy way to handle contiguous intervals of 10 minutes (pd.Grouper does not begin grouping from the first date in your column, so the dt.floor('20min') is necessary as an initial step—this will inadvertently lump in some—or most—rows from "period_{i+1}" under "period_{i}").

For reproduce your problem I do this:

index = pd.date_range(datetime.datetime.now().date() - datetime.timedelta(10), periods=100, freq='min')

This way I have this DataFrame:

a = pd.DataFrame(index)
a
                     0
0  2018-10-28 00:00:00
1  2018-10-28 00:01:00
2  2018-10-28 00:02:00
3  2018-10-28 00:03:00
4  2018-10-28 00:04:00
5  2018-10-28 00:05:00
6  2018-10-28 00:06:00
7  2018-10-28 00:07:00
8  2018-10-28 00:08:00
9  2018-10-28 00:09:00
10 2018-10-28 00:10:00
                   ...
[100 rows x 1 columns]

Then, I do this:

a['period'] = a.apply(lambda x: "perdio_%d"%(int(x[0].minute/10) + 1), axis=1)

And I have this solution:

                     0    period
0  2018-10-28 00:00:00  perdio_1
1  2018-10-28 00:01:00  perdio_1
2  2018-10-28 00:02:00  perdio_1
3  2018-10-28 00:03:00  perdio_1
4  2018-10-28 00:04:00  perdio_1
5  2018-10-28 00:05:00  perdio_1
6  2018-10-28 00:06:00  perdio_1
7  2018-10-28 00:07:00  perdio_1
8  2018-10-28 00:08:00  perdio_1
9  2018-10-28 00:09:00  perdio_1
10 2018-10-28 00:10:00  perdio_2
11 2018-10-28 00:11:00  perdio_2
12 2018-10-28 00:12:00  perdio_2
13 2018-10-28 00:13:00  perdio_2
14 2018-10-28 00:14:00  perdio_2
15 2018-10-28 00:15:00  perdio_2
                             ...

I hope it has been helpfull

df['timestamp'] = pd.to_datetime(df['timestamp'])
diffs = df['timestamp'] - df['timestamp'].shift()
laps = diffs > pd.Timedelta('10 min')
periods = laps.cumsum().apply(lambda x: 'period_{}'.format(x+1))
df['10min_period'] = periods

I have saved your dataframe in Notepad and called it timestamp.txt. It looks like this in Notepad:

And then I wrote this simple code:

import pandas as pd

timestamp = pd.read_csv("C:\\...path_of_your_file...\\timestamp.txt")  # read file
timestamp['10_Minute_Period'] = 0  # add column and initilize it to zero
numb_groups = int((timestamp.shape[0])/10)  # calculate number of groups
groups = 1  # initialize number of groups to one

while groups <= numb_groups+1:
   for idx, _ in timestamp.iterrows():  # iterate over row indexes
       # check if current row is below the group and the value is equal to 0
       if idx < groups*10 and timestamp.at[idx,'10_Minute_Period'] == 0:
           # in this case, write corresponding Period
           timestamp.loc[idx,'10_Minute_Period'] = ('Period' + str(groups))
   groups += 1  # increment groups and check while condition

print(timestamp)  # print the final modified timestamp

Hope it helps!