Selecting Data between Specific hours in a pandas dataframe

Question

My Pandas Dataframe frame looks something like this

 1. 2013-10-09 09:00:05
 2. 2013-10-09 09:05:00
 3. 2013-10-09 10:00:00
 4.  ............
 5.   ............
 6.   ............
 7. 2013-10-10 09:00:05
 8. 2013-10-10 09:05:00 
 9. 2013-10-10 10:00:00

I want the data lying in between hours 9 and 10 ...if anyone has worked on something like this ,it would be really helpful.

Answer 1

 In [7]: index = date_range('20131009 08:30','20131010 10:05',freq='5T')

In [8]: df = DataFrame(randn(len(index),2),columns=list('AB'),index=index)

In [9]: df
Out[9]: 
<class 'pandas.core.frame.DataFrame'>
DatetimeIndex: 308 entries, 2013-10-09 08:30:00 to 2013-10-10 10:05:00
Freq: 5T
Data columns (total 2 columns):
A    308  non-null values
B    308  non-null values
dtypes: float64(2)

In [10]: df.between_time('9:00','10:00')
Out[10]: 
                            A         B
2013-10-09 09:00:00 -0.664639  1.597453
2013-10-09 09:05:00  1.197290 -0.500621
2013-10-09 09:10:00  1.470186 -0.963553
2013-10-09 09:15:00  0.181314 -0.242415
2013-10-09 09:20:00  0.969427 -1.156609
2013-10-09 09:25:00  0.261473  0.413926
2013-10-09 09:30:00 -0.003698  0.054953
2013-10-09 09:35:00  0.418147 -0.417291
2013-10-09 09:40:00  0.413565 -1.096234
2013-10-09 09:45:00  0.460293  1.200277
2013-10-09 09:50:00 -0.702444 -0.041597
2013-10-09 09:55:00  0.548385 -0.832382
2013-10-09 10:00:00 -0.526582  0.758378
2013-10-10 09:00:00  0.926738  0.178204
2013-10-10 09:05:00 -1.178534  0.184205
2013-10-10 09:10:00  1.408258  0.948526
2013-10-10 09:15:00  0.523318  0.327390
2013-10-10 09:20:00 -0.193174  0.863294
2013-10-10 09:25:00  1.355610 -2.160864
2013-10-10 09:30:00  1.930622  0.174683
2013-10-10 09:35:00  0.273551  0.870682
2013-10-10 09:40:00  0.974756 -0.327763
2013-10-10 09:45:00  1.808285  0.080267
2013-10-10 09:50:00  0.842119  0.368689
2013-10-10 09:55:00  1.065585  0.802003
2013-10-10 10:00:00 -0.324894  0.781885

Answer 2

Make a new column for the time after splitting your original column . Use the below code to split your time for hours, minutes, and seconds:-

df[['h','m','s']] = df['Time'].astype(str).str.split(':', expand=True).astype(int)

Once you are done with that, you have to select the data by filtering it out:-

df9to10 =df[df['h'].between(9, 10, inclusive=True)]

And, it's dynamic, if you want to take another period between apart from 9 and 10.

Answer 3

Another method that uses query. Tested with Python 3.9.

from Pandas import Timestamp
from datetime import time
df = pd.DataFrame({"timestamp": 
[Timestamp("2017-01-03 09:30:00.049"), Timestamp("2017-01-03 09:30:00.049"),
 Timestamp("2017-12-29 16:12:34.214"), Timestamp("2017-12-29 16:17:19.006")]})
df["time"] = df.timestamp.dt.time
start_time = time(9,20,0)
end_time = time(10,0,0)
df_times = df.query("time >= @start_time and time <= @end_time")

In:

              timestamp
2017-01-03 09:30:00.049
2017-01-03 09:30:00.049
2017-12-29 16:12:34.214
2017-12-29 16:17:19.006

Out:

              timestamp             time
2017-01-03 09:30:00.049  09:30:00.049000
2017-01-03 09:30:00.049  09:30:00.049000

As a bonus, arbitrarily complex expressions can be used within a query, e.g. selecting everything within two separate time ranges (this is impossible with between_time).