I have a dataframe that can be simplified as:
date id
0 02/04/2015 02:34 1
1 06/04/2015 12:34 2
2 09/04/2015 23:03 3
3 12/04/2015 01:00 4
4 15/04/2015 07:12 5
5 21/04/2015 12:59 6
6 29/04/2015 17:33 7
7 04/05/2015 10:44 8
8 06/05/2015 11:12 9
9 10/05/2015 08:52 10
10 12/05/2015 14:19 11
11 19/05/2015 19:22 12
12 27/05/2015 22:31 13
13 01/06/2015 11:09 14
14 04/06/2015 12:57 15
15 10/06/2015 04:00 16
16 15/06/2015 03:23 17
17 19/06/2015 05:37 18
18 23/06/2015 13:41 19
19 27/06/2015 15:43 20
It can be created using:
tempDF = pd.DataFrame({ 'id': [1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20],
'date': ["02/04/2015 02:34","06/04/2015 12:34","09/04/2015 23:03","12/04/2015 01:00","15/04/2015 07:12","21/04/2015 12:59","29/04/2015 17:33","04/05/2015 10:44","06/05/2015 11:12","10/05/2015 08:52","12/05/2015 14:19","19/05/2015 19:22","27/05/2015 22:31","01/06/2015 11:09","04/06/2015 12:57","10/06/2015 04:00","15/06/2015 03:23","19/06/2015 05:37","23/06/2015 13:41","27/06/2015 15:43"]})
The data has the following types:
tempDF.dtypes
date object
id int64
dtype: object
I have set the 'date' variable to be Pandas datefime64 format (if that's the right way to describe it) using:
import numpy as np
import pandas as pd
tempDF['date'] = pd_to_datetime(tempDF['date'])
So now, the dtypes look like:
tempDF.dtypes
date datetime64[ns]
id int64
dtype: object
I want to change the hours of the original date data. I can use .normalize() to convert to midnight via the .dt accessor:
tempDF['date'] = tempDF['date'].dt.normalize()
And, I can get access to individual datetime components (e.g. year) using:
tempDF['date'].dt.year
This produces:
0 2015
1 2015
2 2015
3 2015
4 2015
5 2015
6 2015
7 2015
8 2015
9 2015
10 2015
11 2015
12 2015
13 2015
14 2015
15 2015
16 2015
17 2015
18 2015
19 2015
Name: date, dtype: int64
The question is, how can I change specific date and time components? For example, how could I change the midday (12:00) for all the dates? I've found that datetime.datetime has a .replace() function. However, having converted dates to Pandas format, it would make sense to keep in that format. Is there a way to do that without changing the format again?
EDIT :
A vectorized way to do this would be to normalize the series, and then add 12 hours to it using timedelta. Example -
tempDF['date'].dt.normalize() + datetime.timedelta(hours=12)
Demo -
In [59]: tempDF
Out[59]:
date id
0 2015-02-04 12:00:00 1
1 2015-06-04 12:00:00 2
2 2015-09-04 12:00:00 3
3 2015-12-04 12:00:00 4
4 2015-04-15 12:00:00 5
5 2015-04-21 12:00:00 6
6 2015-04-29 12:00:00 7
7 2015-04-05 12:00:00 8
8 2015-06-05 12:00:00 9
9 2015-10-05 12:00:00 10
10 2015-12-05 12:00:00 11
11 2015-05-19 12:00:00 12
12 2015-05-27 12:00:00 13
13 2015-01-06 12:00:00 14
14 2015-04-06 12:00:00 15
15 2015-10-06 12:00:00 16
16 2015-06-15 12:00:00 17
17 2015-06-19 12:00:00 18
18 2015-06-23 12:00:00 19
19 2015-06-27 12:00:00 20
In [60]: tempDF['date'].dt.normalize() + datetime.timedelta(hours=12)
Out[60]:
0 2015-02-04 12:00:00
1 2015-06-04 12:00:00
2 2015-09-04 12:00:00
3 2015-12-04 12:00:00
4 2015-04-15 12:00:00
5 2015-04-21 12:00:00
6 2015-04-29 12:00:00
7 2015-04-05 12:00:00
8 2015-06-05 12:00:00
9 2015-10-05 12:00:00
10 2015-12-05 12:00:00
11 2015-05-19 12:00:00
12 2015-05-27 12:00:00
13 2015-01-06 12:00:00
14 2015-04-06 12:00:00
15 2015-10-06 12:00:00
16 2015-06-15 12:00:00
17 2015-06-19 12:00:00
18 2015-06-23 12:00:00
19 2015-06-27 12:00:00
dtype: datetime64[ns]
Timing information for both methods at bottom
One method would be to use Series.apply along with the .replace() method OP mentions in his post. Example -
tempDF['date'] = tempDF['date'].apply(lambda x:x.replace(hour=12,minute=0))
Demo -
In [12]: tempDF
Out[12]:
date id
0 2015-02-04 02:34:00 1
1 2015-06-04 12:34:00 2
2 2015-09-04 23:03:00 3
3 2015-12-04 01:00:00 4
4 2015-04-15 07:12:00 5
5 2015-04-21 12:59:00 6
6 2015-04-29 17:33:00 7
7 2015-04-05 10:44:00 8
8 2015-06-05 11:12:00 9
9 2015-10-05 08:52:00 10
10 2015-12-05 14:19:00 11
11 2015-05-19 19:22:00 12
12 2015-05-27 22:31:00 13
13 2015-01-06 11:09:00 14
14 2015-04-06 12:57:00 15
15 2015-10-06 04:00:00 16
16 2015-06-15 03:23:00 17
17 2015-06-19 05:37:00 18
18 2015-06-23 13:41:00 19
19 2015-06-27 15:43:00 20
In [13]: tempDF['date'] = tempDF['date'].apply(lambda x:x.replace(hour=12,minute=0))
In [14]: tempDF
Out[14]:
date id
0 2015-02-04 12:00:00 1
1 2015-06-04 12:00:00 2
2 2015-09-04 12:00:00 3
3 2015-12-04 12:00:00 4
4 2015-04-15 12:00:00 5
5 2015-04-21 12:00:00 6
6 2015-04-29 12:00:00 7
7 2015-04-05 12:00:00 8
8 2015-06-05 12:00:00 9
9 2015-10-05 12:00:00 10
10 2015-12-05 12:00:00 11
11 2015-05-19 12:00:00 12
12 2015-05-27 12:00:00 13
13 2015-01-06 12:00:00 14
14 2015-04-06 12:00:00 15
15 2015-10-06 12:00:00 16
16 2015-06-15 12:00:00 17
17 2015-06-19 12:00:00 18
18 2015-06-23 12:00:00 19
19 2015-06-27 12:00:00 20
Timing information
In [52]: df = pd.DataFrame([[datetime.datetime.now()] for _ in range(100000)],columns=['date'])
In [54]: %%timeit
....: df['date'].dt.normalize() + datetime.timedelta(hours=12)
....:
The slowest run took 12.53 times longer than the fastest. This could mean that an intermediate result is being cached
1 loops, best of 3: 32.3 ms per loop
In [57]: %%timeit
....: df['date'].apply(lambda x:x.replace(hour=12,minute=0))
....:
1 loops, best of 3: 1.09 s per loop
Here's the solution I used to replace the time component of the datetime values in a Pandas DataFrame. Not sure how efficient this solution is, but it fit my needs.
import pandas as pd
# Create a list of EOCY dates for a specified period
sDate = pd.Timestamp('2022-01-31 23:59:00')
eDate = pd.Timestamp('2060-01-31 23:59:00')
dtList = pd.date_range(sDate, eDate, freq='Y').to_pydatetime()
# Create a DataFrame with a single column called 'Date' and fill the rows with the list of EOCY dates.
df = pd.DataFrame({'Date': dtList})
# Loop through the DataFrame rows using the replace function to replace the hours and minutes of each date value.
for i in range(df.shape[0]):
df.iloc[i, 0]=df.iloc[i, 0].replace(hour=00, minute=00)
Not sure how efficient this solution is, but it fit my needs.
Related
I want to extract the year from a datetime column into a new 'yyyy'-column AND I want the missing values (NaT) to be displayed as 'NaN', so the datetime-dtype of the new column should be changed I guess but there I'm stuck..
Initial df:
Date ID
0 2016-01-01 12
1 2015-01-01 96
2 NaT 20
3 2018-01-01 73
4 2017-01-01 84
5 NaT 26
6 2013-01-01 87
7 2016-01-01 64
8 2019-01-01 11
9 2014-01-01 34
Desired df:
Date ID yyyy
0 2016-01-01 12 2016
1 2015-01-01 96 2015
2 NaT 20 NaN
3 2018-01-01 73 2018
4 2017-01-01 84 2017
5 NaT 26 NaN
6 2013-01-01 87 2013
7 2016-01-01 64 2016
8 2019-01-01 11 2019
9 2014-01-01 34 2014
Code:
import pandas as pd
import numpy as np
# example df
df = pd.DataFrame({"ID": [12,96,20,73,84,26,87,64,11,34],
"Date": ['2016-01-01', '2015-01-01', np.nan, '2018-01-01', '2017-01-01', np.nan, '2013-01-01', '2016-01-01', '2019-01-01', '2014-01-01']})
df.ID = pd.to_numeric(df.ID)
df.Date = pd.to_datetime(df.Date)
print(df)
#extraction of year from date
df['yyyy'] = pd.to_datetime(df.Date).dt.strftime('%Y')
#Try to set NaT to NaN or datetime to numeric, PROBLEM: empty cells keep 'NaT'
df.loc[(df['yyyy'].isna()), 'yyyy'] = np.nan
#(try1)
df.yyyy = df.Date.astype(float)
#(try2)
df.yyyy = pd.to_numeric(df.Date)
#(try3)
print(df)
Use Series.dt.year with converting to integers with Int64:
df.Date = pd.to_datetime(df.Date)
df['yyyy'] = df.Date.dt.year.astype('Int64')
print (df)
ID Date yyyy
0 12 2016-01-01 2016
1 96 2015-01-01 2015
2 20 NaT <NA>
3 73 2018-01-01 2018
4 84 2017-01-01 2017
5 26 NaT <NA>
6 87 2013-01-01 2013
7 64 2016-01-01 2016
8 11 2019-01-01 2019
9 34 2014-01-01 2014
With no convert floats to integers:
df['yyyy'] = df.Date.dt.year
print (df)
ID Date yyyy
0 12 2016-01-01 2016.0
1 96 2015-01-01 2015.0
2 20 NaT NaN
3 73 2018-01-01 2018.0
4 84 2017-01-01 2017.0
5 26 NaT NaN
6 87 2013-01-01 2013.0
7 64 2016-01-01 2016.0
8 11 2019-01-01 2019.0
9 34 2014-01-01 2014.0
Your solution convert NaT to strings NaT, so is possible use replace.
Btw, in last versions of pandas replace is not necessary, it working correctly.
df['yyyy'] = pd.to_datetime(df.Date).dt.strftime('%Y').replace('NaT', np.nan)
Isn't it:
df['yyyy'] = df.Date.dt.year
Output:
Date ID yyyy
0 2016-01-01 12 2016.0
1 2015-01-01 96 2015.0
2 NaT 20 NaN
3 2018-01-01 73 2018.0
4 2017-01-01 84 2017.0
5 NaT 26 NaN
6 2013-01-01 87 2013.0
7 2016-01-01 64 2016.0
8 2019-01-01 11 2019.0
9 2014-01-01 34 2014.0
For pandas 0.24.2+, you can use Int64 data type for nullable integers:
df['yyyy'] = df.Date.dt.year.astype('Int64')
which gives:
Date ID yyyy
0 2016-01-01 12 2016
1 2015-01-01 96 2015
2 NaT 20 <NA>
3 2018-01-01 73 2018
4 2017-01-01 84 2017
5 NaT 26 <NA>
6 2013-01-01 87 2013
7 2016-01-01 64 2016
8 2019-01-01 11 2019
9 2014-01-01 34 2014
I am trying to develop a more efficient loop to complete a problem. At the moment, the code below applies a string if it aligns with a specific value. However, the values are in identical order so a loop could make this process more efficient.
Using the df below as an example, using integers to represent time periods, each integer increase equates to a 15 min period. So 1 == 8:00:00 and 2 == 8:15:00 etc. At the moment I would repeat this process until the last time period. If this gets up to 80 it could become very inefficient. Could a loop be incorporated here?
import pandas as pd
d = ({
'Time' : [1,1,1,2,2,2,3,3,3,4,4,4,5,5,5,6,6,6],
})
df = pd.DataFrame(data = d)
def time_period(row) :
if row['Time'] == 1 :
return '8:00:00'
if row['Time'] == 2 :
return '8:15:00'
if row['Time'] == 3 :
return '8:30:00'
if row['Time'] == 4 :
return '8:45:00'
if row['Time'] == 5 :
return '9:00:00'
if row['Time'] == 6 :
return '9:15:00'
.....
if row['Time'] == 80 :
return '4:00:00'
df['24Hr Time'] = df.apply(lambda row: time_period(row), axis=1)
print(df)
Out:
Time 24Hr Time
0 1 8:00:00
1 1 8:00:00
2 1 8:00:00
3 2 8:15:00
4 2 8:15:00
5 2 8:15:00
6 3 8:30:00
7 3 8:30:00
8 3 8:30:00
9 4 8:45:00
10 4 8:45:00
11 4 8:45:00
12 5 9:00:00
13 5 9:00:00
14 5 9:00:00
15 6 9:15:00
16 6 9:15:00
17 6 9:15:00
This is possible with some simple timdelta arithmetic:
df['24Hr Time'] = (
pd.to_timedelta((df['Time'] - 1) * 15, unit='m') + pd.Timedelta(hours=8))
df.head()
Time 24Hr Time
0 1 08:00:00
1 1 08:00:00
2 1 08:00:00
3 2 08:15:00
4 2 08:15:00
df.dtypes
Time int64
24Hr Time timedelta64[ns]
dtype: object
If you need a string, use pd.to_datetime with unit and origin:
df['24Hr Time'] = (
pd.to_datetime((df['Time']-1) * 15, unit='m', origin='8:00:00')
.dt.strftime('%H:%M:%S'))
df.head()
Time 24Hr Time
0 1 08:00:00
1 1 08:00:00
2 1 08:00:00
3 2 08:15:00
4 2 08:15:00
df.dtypes
Time int64
24Hr Time object
dtype: object
In general, you want to make a dictionary and apply
my_dict = {'old_val1': 'new_val1',...}
df['24Hr Time'] = df['Time'].map(my_dict)
But, in this case, you can do with time delta:
df['24Hr Time'] = pd.to_timedelta(df['Time']*15, unit='T') + pd.to_timedelta('7:45:00')
Output (note that the new column is of type timedelta, not string)
Time 24Hr Time
0 1 08:00:00
1 1 08:00:00
2 1 08:00:00
3 2 08:15:00
4 2 08:15:00
5 2 08:15:00
6 3 08:30:00
7 3 08:30:00
8 3 08:30:00
9 4 08:45:00
10 4 08:45:00
11 4 08:45:00
12 5 09:00:00
13 5 09:00:00
14 5 09:00:00
15 6 09:15:00
16 6 09:15:00
17 6 09:15:00
I end up using this
pd.to_datetime((df.Time-1)*15*60+8*60*60,unit='s').dt.time
0 08:00:00
1 08:00:00
2 08:00:00
3 08:15:00
4 08:15:00
5 08:15:00
6 08:30:00
7 08:30:00
8 08:30:00
9 08:45:00
10 08:45:00
11 08:45:00
12 09:00:00
13 09:00:00
14 09:00:00
15 09:15:00
16 09:15:00
17 09:15:00
Name: Time, dtype: object
A fun way is using pd.timedelta_range and index.repeat
n = df.Time.nunique()
c = df.groupby('Time').size()
df['24_hr'] = pd.timedelta_range(start='8 hours', periods=n, freq='15T').repeat(c)
Out[380]:
Time 24_hr
0 1 08:00:00
1 1 08:00:00
2 1 08:00:00
3 2 08:15:00
4 2 08:15:00
5 2 08:15:00
6 3 08:30:00
7 3 08:30:00
8 3 08:30:00
9 4 08:45:00
10 4 08:45:00
11 4 08:45:00
12 5 09:00:00
13 5 09:00:00
14 5 09:00:00
15 6 09:15:00
16 6 09:15:00
17 6 09:15:00
I have a question of comparing data of datetime64[ns] and date like '2017-01-01'.
here is the code:
df.loc[(df['Date'] >= datetime.date(2017.1.1), 'TimeRange'] = '2017.1'
but , an error has been showed and said descriptor 'date' requires a 'datetime.datetime' object but received a 'int'.
how can i compare a datetime64 to data (2017-01-01 or 2-17-6-1 and likes)
Thanks
Demo:
Source DF:
In [83]: df = pd.DataFrame({'tm':pd.date_range('2000-01-01', freq='9999T', periods=20)})
In [84]: df
Out[84]:
tm
0 2000-01-01 00:00:00
1 2000-01-07 22:39:00
2 2000-01-14 21:18:00
3 2000-01-21 19:57:00
4 2000-01-28 18:36:00
5 2000-02-04 17:15:00
6 2000-02-11 15:54:00
7 2000-02-18 14:33:00
8 2000-02-25 13:12:00
9 2000-03-03 11:51:00
10 2000-03-10 10:30:00
11 2000-03-17 09:09:00
12 2000-03-24 07:48:00
13 2000-03-31 06:27:00
14 2000-04-07 05:06:00
15 2000-04-14 03:45:00
16 2000-04-21 02:24:00
17 2000-04-28 01:03:00
18 2000-05-04 23:42:00
19 2000-05-11 22:21:00
Filtering:
In [85]: df.loc[df.tm > '2000-03-01']
Out[85]:
tm
9 2000-03-03 11:51:00
10 2000-03-10 10:30:00
11 2000-03-17 09:09:00
12 2000-03-24 07:48:00
13 2000-03-31 06:27:00
14 2000-04-07 05:06:00
15 2000-04-14 03:45:00
16 2000-04-21 02:24:00
17 2000-04-28 01:03:00
18 2000-05-04 23:42:00
19 2000-05-11 22:21:00
In [86]: df.loc[df.tm > '2000-3-1']
Out[86]:
tm
9 2000-03-03 11:51:00
10 2000-03-10 10:30:00
11 2000-03-17 09:09:00
12 2000-03-24 07:48:00
13 2000-03-31 06:27:00
14 2000-04-07 05:06:00
15 2000-04-14 03:45:00
16 2000-04-21 02:24:00
17 2000-04-28 01:03:00
18 2000-05-04 23:42:00
19 2000-05-11 22:21:00
not standard date format:
In [87]: df.loc[df.tm > pd.to_datetime('03/01/2000')]
Out[87]:
tm
9 2000-03-03 11:51:00
10 2000-03-10 10:30:00
11 2000-03-17 09:09:00
12 2000-03-24 07:48:00
13 2000-03-31 06:27:00
14 2000-04-07 05:06:00
15 2000-04-14 03:45:00
16 2000-04-21 02:24:00
17 2000-04-28 01:03:00
18 2000-05-04 23:42:00
19 2000-05-11 22:21:00
You need to ensure that the data you're comparing it with is also in the same format. Assuming that you have two datetime objects, you can do it like this:
import datetime
print(df.loc[(df['Date'] >= datetime.date(2017, 1, 1), 'TimeRange'])
This will create a datetime object and list out the filtered results. You can also assign the results an updated value as you have mentioned above.
I have a pandas series with two columns and lots of rows like:
r =
1-10-2010 3.4
1-11-2010 4.5
1-12-2010 3.7
... ...
What I'd like to do is to remove days of the week not in a custom week. So to remove Fridays and Saturdays, do something like this:
r = amazingfunction(r, ('Sun', 'Mon', 'Tue', 'Wed', Thu'))
r =
1-10-2010 3.4
1-11-2010 4.5
1-12-2010 3.7
1-13-2010 3.4
1-14-2010 4.1
1-17-2010 4.5
1-18-2010 3.7
... ...
How can I go about this?
You can use dt.dayofweek and isin to filter the df, here Friday and Saturday are 4,5 respectively, we negate the boolean mask using ~:
In [12]:
df = pd.DataFrame({'dates':pd.date_range(dt.datetime(2015,1,1), dt.datetime(2015,2,1))})
df['dayofweek'] = df['dates'].dt.dayofweek
df
Out[12]:
dates dayofweek
0 2015-01-01 3
1 2015-01-02 4
2 2015-01-03 5
3 2015-01-04 6
4 2015-01-05 0
5 2015-01-06 1
6 2015-01-07 2
7 2015-01-08 3
8 2015-01-09 4
9 2015-01-10 5
10 2015-01-11 6
11 2015-01-12 0
12 2015-01-13 1
13 2015-01-14 2
14 2015-01-15 3
15 2015-01-16 4
16 2015-01-17 5
17 2015-01-18 6
18 2015-01-19 0
19 2015-01-20 1
20 2015-01-21 2
21 2015-01-22 3
22 2015-01-23 4
23 2015-01-24 5
24 2015-01-25 6
25 2015-01-26 0
26 2015-01-27 1
27 2015-01-28 2
28 2015-01-29 3
29 2015-01-30 4
30 2015-01-31 5
31 2015-02-01 6
In [13]:
df[~df['dates'].dt.dayofweek.isin([4,5])]
Out[13]:
dates dayofweek
0 2015-01-01 3
3 2015-01-04 6
4 2015-01-05 0
5 2015-01-06 1
6 2015-01-07 2
7 2015-01-08 3
10 2015-01-11 6
11 2015-01-12 0
12 2015-01-13 1
13 2015-01-14 2
14 2015-01-15 3
17 2015-01-18 6
18 2015-01-19 0
19 2015-01-20 1
20 2015-01-21 2
21 2015-01-22 3
24 2015-01-25 6
25 2015-01-26 0
26 2015-01-27 1
27 2015-01-28 2
28 2015-01-29 3
31 2015-02-01 6
EDIT
As your data is a Series your dates are your index so the following should work:
r[~r.index.dayofweek.isin([4,5])]
I have this dataframe df:
U,Datetime
01,2015-01-01 20:00:00
01,2015-02-01 20:05:00
01,2015-04-01 21:00:00
01,2015-05-01 22:00:00
01,2015-07-01 22:05:00
02,2015-08-01 20:00:00
02,2015-09-01 21:00:00
02,2014-01-01 23:00:00
02,2014-02-01 22:05:00
02,2015-01-01 20:00:00
02,2014-03-01 21:00:00
03,2015-10-01 20:00:00
03,2015-11-01 21:00:00
03,2015-12-01 23:00:00
03,2015-01-01 22:05:00
03,2015-02-01 20:00:00
03,2015-05-01 21:00:00
03,2014-01-01 20:00:00
03,2014-02-01 21:00:00
made by U and a Datetime object. What I would like to do is to filter U values having at least three consecutive occurrences in months/year. So far I have grouped by by U, year and month as:
m = df.groupby(['U',df.index.year,df.index.month]).size()
obtaining:
U
1 2015 1 1
2 1
4 1
5 1
7 1
2 2014 1 1
2 1
3 1
2015 1 1
8 1
9 1
3 2014 1 1
2 1
2015 1 1
2 1
5 1
10 1
11 1
12 1
The third column is related to the occurrences in different months/year. In this case only U values of 02 and 03 contain at least three consecutive values in months/year. Now I can't figured out how can I select those users and getting them out in a list, for instance, or just keeping them in the original dataframe df and discard the others. I tried also:
g = m.groupby(level=[0,1]).diff()
But I can't get any useful information.
Finally I could come up with the solution :) .
to give you an idea of how custom function works , simply it subtracts the value of the month from it's preceding value , the result should be one of course , and this should happen twice , for example if you have a list of numbers [5 , 6 , 7] , so 7 - 6 = 1 and 6 - 5 = 1 , 1 here appeared twice so the condition has been fulfilled
In [80]:
df.reset_index(inplace=True)
In [281]:
df['month'] = df.Datetime.dt.month
df['year'] = df.Datetime.dt.year
df
Out[281]:
Datetime U month year
0 2015-01-01 20:00:00 1 1 2015
1 2015-02-01 20:05:00 1 2 2015
2 2015-04-01 21:00:00 1 4 2015
3 2015-05-01 22:00:00 1 5 2015
4 2015-07-01 22:05:00 1 7 2015
5 2015-08-01 20:00:00 2 8 2015
6 2015-09-01 21:00:00 2 9 2015
7 2014-01-01 23:00:00 2 1 2014
8 2014-02-01 22:05:00 2 2 2014
9 2015-01-01 20:00:00 2 1 2015
10 2014-03-01 21:00:00 2 3 2014
11 2015-10-01 20:00:00 3 10 2015
12 2015-11-01 21:00:00 3 11 2015
13 2015-12-01 23:00:00 3 12 2015
14 2015-01-01 22:05:00 3 1 2015
15 2015-02-01 20:00:00 3 2 2015
16 2015-05-01 21:00:00 3 5 2015
17 2014-01-01 20:00:00 3 1 2014
18 2014-02-01 21:00:00 3 2 2014
In [284]:
g = df.groupby([df['U'] , df.year])
In [86]:
res = g.filter(lambda x : is_at_least_three_consec(x['month'].diff().values.tolist()))
res
Out[86]:
Datetime U month year
7 2014-01-01 23:00:00 2 1 2014
8 2014-02-01 22:05:00 2 2 2014
10 2014-03-01 21:00:00 2 3 2014
11 2015-10-01 20:00:00 3 10 2015
12 2015-11-01 21:00:00 3 11 2015
13 2015-12-01 23:00:00 3 12 2015
14 2015-01-01 22:05:00 3 1 2015
15 2015-02-01 20:00:00 3 2 2015
16 2015-05-01 21:00:00 3 5 2015
if you want to see the result of the custom function
In [84]:
res = g['month'].agg(lambda x : is_at_least_three_consec(x.diff().values.tolist()))
res
Out[84]:
U year
1 2015 False
2 2014 True
2015 False
3 2014 False
2015 True
Name: month, dtype: bool
this is how custom function implemented
In [53]:
def is_at_least_three_consec(month_diff):
consec_count = 0
#print(month_diff)
for index , val in enumerate(month_diff):
if index != 0 and val == 1:
consec_count += 1
if consec_count == 2:
return True
else:
consec_count = 0
return False