Create regular time series from irregular interval with python - python

I wonder if is it possible to convert irregular time series interval to regular one without interpolating value from other column like this :
Index count
2018-01-05 00:00:00 1
2018-01-07 00:00:00 4
2018-01-08 00:00:00 15
2018-01-11 00:00:00 2
2018-01-14 00:00:00 5
2018-01-19 00:00:00 5
....
2018-12-26 00:00:00 6
2018-12-29 00:00:00 7
2018-12-30 00:00:00 8
And I expect the result to be something like this:
Index count
2018-01-01 00:00:00 0
2018-01-02 00:00:00 0
2018-01-03 00:00:00 0
2018-01-04 00:00:00 0
2018-01-05 00:00:00 1
2018-01-06 00:00:00 0
2018-01-07 00:00:00 4
2018-01-08 00:00:00 15
2018-01-09 00:00:00 0
2018-01-10 00:00:00 0
2018-01-11 00:00:00 2
2018-01-12 00:00:00 0
2018-01-13 00:00:00 0
2018-01-14 00:00:00 5
2018-01-15 00:00:00 0
2018-01-16 00:00:00 0
2018-01-17 00:00:00 0
2018-01-18 00:00:00 0
2018-01-19 00:00:00 5
....
2018-12-26 00:00:00 6
2018-12-27 00:00:00 0
2018-12-28 00:00:00 0
2018-12-29 00:00:00 7
2018-12-30 00:00:00 8
2018-12-31 00:00:00 0
So, far I just try resample from pandas but it only partially solved my problem.
Thanks in advance

Use DataFrame.reindex with date_range:
#if necessary
df.index = pd.to_datetime(df.index)
df = df.reindex(pd.date_range('2018-01-01','2018-12-31'), fill_value=0)
print (df)
count
2018-01-01 0
2018-01-02 0
2018-01-03 0
2018-01-04 0
2018-01-05 1
...
2018-12-27 0
2018-12-28 0
2018-12-29 7
2018-12-30 8
2018-12-31 0
[365 rows x 1 columns]

Related

Replacing NaNs with date and time format

I'm working with the following dataframes.
Date Light (umols) Time_difference
0 2018-01-12 07:16:52 2.5 NaT
1 2018-01-12 07:19:52 4.9 0 days 00:03:00
2 2018-01-12 07:22:52 4.9 0 days 00:03:00
3 2018-01-12 07:25:52 7.4 0 days 00:03:00
4 2018-01-12 07:28:50 9.9 0 days 00:02:58
... ... ... ...
6252 2018-12-18 17:54:24 12.2 0 days 00:03:00
6253 2018-12-18 17:57:24 7.6 0 days 00:03:00
6254 2018-12-18 18:00:24 4.9 0 days 00:03:00
6255 2018-12-18 18:03:24 2.5 0 days 00:03:00
6256 2018-12-18 18:06:24 0.2 0 days 00:03:00
Date Light (umols) Time_difference
0 2019-01-10 00:00:00 500.4 NaT
1 2019-01-10 00:00:01 451.2 0 days 00:00:01
2 2019-01-10 00:00:02 343.7 0 days 00:00:01
3 2019-01-10 00:00:03 354.5 0 days 00:00:01
4 2019-01-10 00:00:04 176.4 0 days 00:00:00
... ... ... ...
81264 2021-02-22 23:59:55 937.7 0 days 00:00:00
81265 2021-02-22 23:59:56 634.4 0 days 00:00:00
81266 2021-02-22 23:59:57 574.3 0 days 00:00:00
81267 2021-02-22 23:59:58 598.9 0 days 00:00:00
81268 2021-02-22 23:59:59 676.9 0 days 00:00:00
I'm wanting to calculate where there are gaps, how long they are and how many there are. The idea is to have a consistent timeline every 3 minutes in a day tops, and anything above that needs to flagged up, the idea would be to merge the two dataframes together afterwards. There are some pesky NaTs in both their first rows, and I want to replace each one with something like '0 days 00:00:00'. I tried writing the following code with little success:
better = clean['Date'] == '2018-01-12 07:16:52'
clean.loc[better, 'Time_difference'] = clean.loc[clean, 'Time_difference'].replace('NaT', '0 days 00:00:00')
Any suggestions?

compare dates within a dataframe and assign a value to another variable

I have two dataframes (df and df1) like as shown below
df = pd.DataFrame({'person_id': [101,101,101,101,202,202,202],
'start_date':['5/7/2013 09:27:00 AM','09/08/2013 11:21:00 AM','06/06/2014 08:00:00 AM', '06/06/2014 05:00:00 AM','12/11/2011 10:00:00 AM','13/10/2012 12:00:00 AM','13/12/2012 11:45:00 AM']})
df.start_date = pd.to_datetime(df.start_date)
df['end_date'] = df.start_date + timedelta(days=5)
df['enc_id'] = ['ABC1','ABC2','ABC3','ABC4','DEF1','DEF2','DEF3']
df1 = pd.DataFrame({'person_id': [101,101,101,101,101,101,101,202,202,202,202,202,202,202,202],'date_1':['07/07/2013 11:20:00 AM','05/07/2013 02:30:00 PM','06/07/2013 02:40:00 PM','08/06/2014 12:00:00 AM','11/06/2014 12:00:00 AM','02/03/2013 12:30:00 PM','13/06/2014 12:00:00 AM','12/11/2011 12:00:00 AM','13/10/2012 07:00:00 AM','13/12/2015 12:00:00 AM','13/12/2012 12:00:00 AM','13/12/2012 06:30:00 PM','13/07/2011 10:00:00 AM','18/12/2012 10:00:00 AM', '19/12/2013 11:00:00 AM']})
df1['date_1'] = pd.to_datetime(df1['date_1'])
df1['within_id'] = ['ABC','ABC','ABC','ABC','ABC','ABC','ABC','DEF','DEF','DEF','DEF','DEF','DEF','DEF',np.nan]
What I would like to do is
a) Pick each person from df1 who doesnt have NA in 'within_id' column and check whether their date_1 is between (df.start_date - 1) and (df.end_date + 1) of the same person in df and for the same within_idor enc_id
ex: for subject = 101 and within_id = ABC, we have date_1 is 7/7/2013, you check whether they are between 4/7/2013 (df.start_date - 1) and 11/7/2013 (df.end_date + 1).
As the first-row comparison itself gave us the result, we don't have to compare our date_1 with rest of the records in df for subject 101. If not, we need to find/scan until we find the interval within which date_1 falls.
b) If date interval found, then assign the corresponding enc_id from df to the within_id in df1
c) If not then assign, "Out of Range"
I tried the below
t1 = df.groupby('person_id').apply(pd.DataFrame.sort_values, 'start_date')
t2 = df1.groupby('person_id').apply(pd.DataFrame.sort_values, 'date_1')
t3= pd.concat([t1, t2], axis=1)
t3['within_id'] = np.where((t3['date_1'] >= t3['start_date'] && t3['person_id'] == t3['person_id_x'] && t3['date_2'] >= t3['end_date']),enc_id]
I expect my output (also see 14th row at the bottom of my screenshot) to be as shown below. As I intend to apply the solution on big data (4/5 million records and there might be 5000-6000 unique person_ids), any efficient and elegant solution is helpful
14 202 2012-12-13 11:00:00 NA
Let's do:
d = df1.merge(df.assign(within_id=df['enc_id'].str[:3]),
on=['person_id', 'within_id'], how='left', indicator=True)
m = d['date_1'].between(d['start_date'] - pd.Timedelta(days=1),
d['end_date'] + pd.Timedelta(days=1))
d = df1.merge(d[m | d['_merge'].ne('both')], on=['person_id', 'date_1'], how='left')
d['within_id'] = d['enc_id'].fillna('out of range').mask(d['_merge'].eq('left_only'))
d = d[df1.columns]
Details:
Left merge the dataframe df1 with df on person_id and within_id:
print(d)
person_id date_1 within_id start_date end_date enc_id _merge
0 101 2013-07-07 11:20:00 ABC 2013-05-07 09:27:00 2013-05-12 09:27:00 ABC1 both
1 101 2013-07-07 11:20:00 ABC 2013-09-08 11:21:00 2013-09-13 11:21:00 ABC2 both
2 101 2013-07-07 11:20:00 ABC 2014-06-06 08:00:00 2014-06-11 08:00:00 ABC3 both
3 101 2013-07-07 11:20:00 ABC 2014-06-06 05:00:00 2014-06-11 10:00:00 DEF1 both
....
47 202 2012-12-18 10:00:00 DEF 2012-10-13 00:00:00 2012-10-18 00:00:00 DEF2 both
48 202 2012-12-18 10:00:00 DEF 2012-12-13 11:45:00 2012-12-18 11:45:00 DEF3 both
49 202 2013-12-19 11:00:00 NaN NaT NaT NaN left_only
Create a boolean mask m to represent the condition where date_1 is between df.start_date - 1 days and df.end_date + 1 days:
print(m)
0 False
1 False
2 False
3 False
...
47 False
48 True
49 False
dtype: bool
Again left merge the dataframe df1 with the dataframe filtered using mask m on columns person_id and date_1:
print(d)
person_id date_1 within_id_x within_id_y start_date end_date enc_id _merge
0 101 2013-07-07 11:20:00 ABC NaN NaT NaT NaN NaN
1 101 2013-05-07 14:30:00 ABC ABC 2013-05-07 09:27:00 2013-05-12 09:27:00 ABC1 both
2 101 2013-06-07 14:40:00 ABC NaN NaT NaT NaN NaN
3 101 2014-08-06 00:00:00 ABC NaN NaT NaT NaN NaN
4 101 2014-11-06 00:00:00 ABC NaN NaT NaT NaN NaN
5 101 2013-02-03 12:30:00 ABC NaN NaT NaT NaN NaN
6 101 2014-06-13 00:00:00 ABC NaN NaT NaT NaN NaN
7 202 2011-12-11 00:00:00 DEF DEF 2011-12-11 10:00:00 2011-12-16 10:00:00 DEF1 both
8 202 2012-10-13 07:00:00 DEF DEF 2012-10-13 00:00:00 2012-10-18 00:00:00 DEF2 both
9 202 2015-12-13 00:00:00 DEF NaN NaT NaT NaN NaN
10 202 2012-12-13 00:00:00 DEF DEF 2012-12-13 11:45:00 2012-12-18 11:45:00 DEF3 both
11 202 2012-12-13 18:30:00 DEF DEF 2012-12-13 11:45:00 2012-12-18 11:45:00 DEF3 both
12 202 2011-07-13 10:00:00 DEF NaN NaT NaT NaN NaN
13 202 2012-12-18 10:00:00 DEF DEF 2012-12-13 11:45:00 2012-12-18 11:45:00 DEF3 both
14 202 2013-12-19 11:00:00 NaN NaN NaT NaT NaN left_only
Populate the values in within_id column from enc_id and using Series.fillna fill the NaN excluding the ones that doesn't match from df with out of range, finally filter the columns to get the result:
print(d)
person_id date_1 within_id
0 101 2013-07-07 11:20:00 out of range
1 101 2013-05-07 14:30:00 ABC1
2 101 2013-06-07 14:40:00 out of range
3 101 2014-08-06 00:00:00 out of range
4 101 2014-11-06 00:00:00 out of range
5 101 2013-02-03 12:30:00 out of range
6 101 2014-06-13 00:00:00 out of range
7 202 2011-12-11 00:00:00 DEF1
8 202 2012-10-13 07:00:00 DEF2
9 202 2015-12-13 00:00:00 out of range
10 202 2012-12-13 00:00:00 DEF3
11 202 2012-12-13 18:30:00 DEF3
12 202 2011-07-13 10:00:00 out of range
13 202 2012-12-18 10:00:00 DEF3
14 202 2013-12-19 11:00:00 NaN
I used df and df1 as provided above.
The basic approach is to iterate over df1 and extract the matching values of enc_id.
I added a 'rule' column, to show how each value got populated.
Unfortunately, I was not able to reproduce the expected results. Perhaps the general approach will be useful.
df1['rule'] = 0
for t in df1.itertuples():
person = (t.person_id == df.person_id)
b = (t.date_1 >= df.start_date) & (t.date_2 <= df.end_date)
c = (t.date_1 >= df.start_date) & (t.date_2 >= df.end_date)
d = (t.date_1 <= df.start_date) & (t.date_2 <= df.end_date)
e = (t.date_1 <= df.start_date) & (t.date_2 <= df.start_date) # start_date at BOTH ends
if (m := person & b).any():
df1.at[t.Index, 'within_id'] = df.loc[m, 'enc_id'].values[0]
df1.at[t.Index, 'rule'] += 1
elif (m := person & c).any():
df1.at[t.Index, 'within_id'] = df.loc[m, 'enc_id'].values[0]
df1.at[t.Index, 'rule'] += 10
elif (m := person & d).any():
df1.at[t.Index, 'within_id'] = df.loc[m, 'enc_id'].values[0]
df1.at[t.Index, 'rule'] += 100
elif (m := person & e).any():
df1.at[t.Index, 'within_id'] = 'out of range'
df1.at[t.Index, 'rule'] += 1_000
else:
df1.at[t.Index, 'within_id'] = 'impossible!'
df1.at[t.Index, 'rule'] += 10_000
df1['within_id'] = df1['within_id'].astype('Int64')
The results are:
print(df1)
person_id date_1 date_2 within_id rule
0 11 1961-12-30 00:00:00 1962-01-01 00:00:00 11345678901 1
1 11 1962-01-30 00:00:00 1962-02-01 00:00:00 11345678902 1
2 12 1962-02-28 00:00:00 1962-03-02 00:00:00 34567892101 100
3 12 1989-07-29 00:00:00 1989-07-31 00:00:00 34567892101 1
4 12 1989-09-03 00:00:00 1989-09-05 00:00:00 34567892101 10
5 12 1989-10-02 00:00:00 1989-10-04 00:00:00 34567892103 1
6 12 1989-10-01 00:00:00 1989-10-03 00:00:00 34567892103 1
7 13 1999-03-29 00:00:00 1999-03-31 00:00:00 56432718901 1
8 13 1999-04-20 00:00:00 1999-04-22 00:00:00 56432718901 10
9 13 1999-06-02 00:00:00 1999-06-04 00:00:00 56432718904 1
10 13 1999-06-03 00:00:00 1999-06-05 00:00:00 56432718904 1
11 13 1999-07-29 00:00:00 1999-07-31 00:00:00 56432718905 1
12 14 2002-02-03 10:00:00 2002-02-05 10:00:00 24680135791 1
13 14 2002-02-03 10:00:00 2002-02-05 10:00:00 24680135791 1

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

pandas groupby time series by 10 min and also keep some columns

i have this information; where "opid" is categorical
datetime id nut opid user amount
2018-01-01 07:01:00 1531 3hrnd 1 mherrera 1
2018-01-01 07:05:00 9510 sd45f 1 svasqu 1
2018-01-01 07:06:00 8125 5s8fr 15 urubi 1
2018-01-01 07:08:15 6324 sd5d6 1 jgonza 1
2018-01-01 07:12:01 0198 tgfg5 1 julmaf 1
2018-01-01 07:13:50 6589 mbkg4 15 jdjiep 1
2018-01-01 07:16:10 9501 wurf4 15 polga 1
the result i'm looking for is something like this
datetime opid amount
2018-01-01 07:00:00 1 3
2018-01-01 07:00:00 15 1
2018-01-01 07:10:00 1 1
2018-01-01 07:10:00 15 2
so... basically i need to know how many of each "opid" are done every 10 min
P.D "amount" is always 1, "opid" is from 1 - 15
Using grouper:
df.set_index('datetime').groupby(['opid', pd.Grouper(freq='10min')]).amount.sum()
opid datetime
1 2018-01-01 07:00:00 3
2018-01-01 07:10:00 1
15 2018-01-01 07:00:00 1
2018-01-01 07:10:00 2
Name: amount, dtype: int64

pandas datetime: groupy hourly and every monday

I'm new to pandas / python:
I have a dataframe (events.number) indexed by a datetime object.
I'm trying to extract an event count hourly, on every Monday (or other particular weekday). I wrote:
hour_tally_monday = events.number.groupby(lambda x: (x.hour & x.weekday==0) ).count()
but this does not work correctly.
I can drop the "& x.weekday==1" and it works but presumably uses all the days in the frame. What's the right (simplest) syntax to just average over Mondays?
I think you need first filter dataframe with boolean indexing and then use groupby with size:
import pandas as pd
start = pd.to_datetime('2016-02-01')
end = pd.to_datetime('2016-02-25')
rng = pd.date_range(start, end, freq='12H')
events = pd.DataFrame({'number': [1] * 20 + [2] * 15 + [3] * 14}, index=rng)
print events
number
2016-02-01 00:00:00 1
2016-02-01 12:00:00 1
2016-02-02 00:00:00 1
2016-02-02 12:00:00 1
2016-02-03 00:00:00 1
2016-02-03 12:00:00 1
2016-02-04 00:00:00 1
2016-02-04 12:00:00 1
2016-02-05 00:00:00 1
2016-02-05 12:00:00 1
2016-02-06 00:00:00 1
2016-02-06 12:00:00 1
2016-02-07 00:00:00 1
...
...
filtered = events[events.index.weekday == 0]
print filtered
number
2016-02-01 00:00:00 1
2016-02-01 12:00:00 1
2016-02-08 00:00:00 1
2016-02-08 12:00:00 1
2016-02-15 00:00:00 2
2016-02-15 12:00:00 2
2016-02-22 00:00:00 3
2016-02-22 12:00:00 3
In version 0.18.1 you can use new method DatetimeIndex.weekday_name:
filtered = events[events.index.weekday_name == 'Monday']
print filtered
number
2016-02-01 00:00:00 1
2016-02-01 12:00:00 1
2016-02-08 00:00:00 1
2016-02-08 12:00:00 1
2016-02-15 00:00:00 2
2016-02-15 12:00:00 2
2016-02-22 00:00:00 3
2016-02-22 12:00:00 3
print filtered.groupby(filtered.index.hour).size()
0 4
12 4
dtype: int64

Categories

Resources