I have a Dataframe with a timestamp column:
tij_pd.datetime[0:5]
Out[29]:
0 2016-01-09 05:27:00
1 2016-01-09 06:49:00
2 2016-01-09 08:05:00
3 2016-01-09 12:09:00
4 2016-01-09 14:54:00
Name: datetime, dtype: datetime64[ns]
I need to select times between '00:00' and '04:00' and add 1Hour.
In[31]: tij_pd.set_index('datetime').between_time('00:00','04:00').reset_index().datetime
Out[31]:
0 2016-03-09 01:01:00
1 2016-10-09 00:31:00
...
16 2016-03-09 01:40:00
17 2016-09-23 00:46:00
Name: datetime, dtype: datetime64[ns]
How can I add 1Hour to the datetime column of this subset?
tij_pd['datetime'] = tij_pd['datetime']+pd.to_timedelta(1,'d')
IIUC:
tij_pd.loc[(tij_pd.datetime.dt.hour >= 0) &
(tij_pd.datetime.dt.hour <= 4),
'datetime'] += \
pd.to_timedelta('1H')
Related
I have multiple Pandas Series of datetime64 values that I want to bin into groups using arbitrary bin sizes.
I've found the Series.to_period() function which does exactly what I want except that I need more control over the chosen bin size. to_period allows me to bin by full years, months, days, etc. but I also want to bin by 5 years, 6 hours or 15 minutes. Using a syntax like 5Y, 6H or 15min works in other corners of Pandas but apparently not here.
s = pd.Series(["2020-02-01", "2020-02-02", "2020-02-03", "2020-02-04"], dtype="datetime64[ns]")
# Output as expected
s.dt.to_period("M").value_counts()
2020-02 4
Freq: M, dtype: int64
# Output as expected
s.dt.to_period("W").value_counts()
2020-01-27/2020-02-02 2
2020-02-03/2020-02-09 2
Freq: W-SUN, dtype: int64
# Output as expected
s.dt.to_period("D").value_counts()
2020-02-01 1
2020-02-02 1
2020-02-03 1
2020-02-04 1
Freq: D, dtype: int64
# Output unexpected (and wrong?)
s.dt.to_period("2D").value_counts()
2020-02-01 1
2020-02-02 1
2020-02-03 1
2020-02-04 1
Freq: 2D, dtype: int64
I believe that pd.Grouper is what you're looking for.
https://pandas.pydata.org/docs/reference/api/pandas.Grouper.html
It provides the flexibility of having multiple frequencies in addition to the standard ones: https://pandas.pydata.org/pandas-docs/stable/user_guide/timeseries.html#offset-aliases
From the documentation:
>>> start, end = '2000-10-01 23:30:00', '2000-10-02 00:30:00'
>>> rng = pd.date_range(start, end, freq='7min')
>>> ts = pd.Series(np.arange(len(rng)) * 3, index=rng)
>>> ts
2000-10-01 23:30:00 0
2000-10-01 23:37:00 3
2000-10-01 23:44:00 6
2000-10-01 23:51:00 9
2000-10-01 23:58:00 12
2000-10-02 00:05:00 15
2000-10-02 00:12:00 18
2000-10-02 00:19:00 21
2000-10-02 00:26:00 24
Freq: 7T, dtype: int64
>>> ts.groupby(pd.Grouper(freq='17min')).sum()
2000-10-01 23:14:00 0
2000-10-01 23:31:00 9
2000-10-01 23:48:00 21
2000-10-02 00:05:00 54
2000-10-02 00:22:00 24
Freq: 17T, dtype: int64
NOTE:
If you'd like to .groupby a certain column then use the following syntax:
df.groupby(pd.Grouper(key="my_col", freq="3M"))
I have a pandas dataframe like as shown below
df = pd.DataFrame({'login_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','','10/11/1990'],
'DURATION':[21,30,200,34,45,np.NaN})
I would like to add DURATION values to the login_date column
The DURATION is represented in Days type
If there is NA in DURATION column, just replace it with 0.
So, I tried the below
df['DURATION'] = df['DURATION'].fillna(0)
df['login_date'] = pd.to_datetime(df['login_date'])
df['DURATION'] = df['DURATION'].astype('Int64')
df['logout_Date'] = df['login_date'] + pd.offsets.DateOffset(days=df['DURATION'])
However, this results in an error as shown below
TypeError: Invalid type <class 'pandas.core.series.Series'>. Must be int or float.
But I have already converted my DURATION column to int64 type.
How to add a column of values to my logout_date column
Try:
df["logout_date"] = pd.to_datetime(df["login_date"]) + df["DURATION"].fillna(0).apply(lambda x: pd.Timedelta(days=x))
print(df)
Prints:
login_date DURATION logout_date
0 5/7/2013 09:27:00 AM 21.0 2013-05-28 09:27:00
1 09/08/2013 11:21:00 AM 30.0 2013-10-08 11:21:00
2 06/06/2014 08:00:00 AM 200.0 2014-12-23 08:00:00
3 06/06/2014 05:00:00 AM 34.0 2014-07-10 05:00:00
4 45.0 NaT
5 10/11/1990 NaN 1990-10-11 00:00:00
I have a dataframe 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'],
'end_date':['5/12/2013 09:27:00 AM',np.nan,'06/11/2014 08:00:00 AM',np.nan,'12/16/2011 10:00:00','10/18/2012 00:00:00',np.nan],
'type':['O','I','O','O','I','O','I']})
df.start_date = pd.to_datetime(df.start_date)
df['end_date'] = pd.to_datetime(df.end_date)
I would like to fillna() under the end_date column based on two approaches below
a) If NA is found in any row except last row of that person, fillna by copying the value from next row
b) If NA is found in the last row of that person fillna by adding 10 days to his start_date (because there is no next row for that person to copy from. So, we give random value of 10 days)
The rules a and b only for persons with type=I.
For persons with type=O, just fillna by copying the value from start_date.
This is what I tried. You can see am writing the same code line twice.
df['end_date'] = np.where(df['type'].str.contains('I'),pd.DatetimeIndex(df['end_date'].bfill()),pd.DatetimeIndex(df.start_date.dt.date))
df['end_date'] = np.where(df['type'].str.contains('I'),pd.DatetimeIndex(df['start_date'] + pd.DateOffset(10)),pd.DatetimeIndex(df.start_date.dt.date))
Any elegant and efficient way to write this as I have to apply this on a big data with 15 million rows?
I expect my output to be like as shown below
Solution
s1 = df.groupby('person_id')['start_date'].shift(-1)
s1 = s1.fillna(df['start_date'] + pd.DateOffset(days=10))
s1 = df['end_date'].fillna(s1)
s2 = df['end_date'].fillna(df['start_date'])
df['end_date'] = np.where(df['type'].eq('I'), s1, s2)
Explanations
Group the dataframe on person_id and shift the column start_date one units upwards.
>>> df.groupby('person_id')['start_date'].shift(-1)
0 2013-09-08 11:21:00
1 2014-06-06 08:00:00
2 2014-06-06 05:00:00
3 NaT
4 2012-10-13 00:00:00
5 2012-12-13 11:45:00
6 NaT
Name: start_date, dtype: datetime64[ns]
Fill the NaN values in the shifted column with the values from start_date column after adding an offset of 10 days
>>> s1.fillna(df['start_date'] + pd.DateOffset(days=10))
0 2013-09-08 11:21:00
1 2014-06-06 08:00:00
2 2014-06-06 05:00:00
3 2014-06-16 05:00:00
4 2012-10-13 00:00:00
5 2012-12-13 11:45:00
6 2012-12-23 11:45:00
Name: start_date, dtype: datetime64[ns]
Now fill the NaN values in end_date column with the above series s1
>>> df['end_date'].fillna(s1)
0 2013-05-12 09:27:00
1 2014-06-06 08:00:00
2 2014-06-11 08:00:00
3 2014-06-16 05:00:00
4 2011-12-16 10:00:00
5 2012-10-18 00:00:00
6 2012-12-23 11:45:00
Name: end_date, dtype: datetime64[ns]
Similarly fill the NaN values in end_date column with the values from start_date column to create a series s2
>>> df['end_date'].fillna(df['start_date'])
0 2013-05-12 09:27:00
1 2013-09-08 11:21:00
2 2014-06-11 08:00:00
3 2014-06-06 05:00:00
4 2011-12-16 10:00:00
5 2012-10-18 00:00:00
6 2012-12-13 11:45:00
Name: end_date, dtype: datetime64[ns]
Then use np.where to select the values from s1 / s2 based on the condition where the type is I or O
>>> df
person_id start_date end_date type
0 101 2013-05-07 09:27:00 2013-05-12 09:27:00 O
1 101 2013-09-08 11:21:00 2014-06-06 08:00:00 I
2 101 2014-06-06 08:00:00 2014-06-11 08:00:00 O
3 101 2014-06-06 05:00:00 2014-06-06 05:00:00 O
4 202 2011-12-11 10:00:00 2011-12-16 10:00:00 I
5 202 2012-10-13 00:00:00 2012-10-18 00:00:00 O
6 202 2012-12-13 11:45:00 2012-12-23 11:45:00 I
I have a series that looks like this
2014 7 2014-07-01 -0.045417
8 2014-08-01 -0.035876
9 2014-09-02 -0.030971
10 2014-10-01 -0.027471
11 2014-11-03 -0.032968
12 2014-12-01 -0.031110
2015 1 2015-01-02 -0.028906
2 2015-02-02 -0.035563
3 2015-03-02 -0.040338
4 2015-04-01 -0.032770
5 2015-05-01 -0.025762
6 2015-06-01 -0.019746
7 2015-07-01 -0.018541
8 2015-08-03 -0.028101
9 2015-09-01 -0.043237
10 2015-10-01 -0.053565
11 2015-11-02 -0.062630
12 2015-12-01 -0.064618
2016 1 2016-01-04 -0.064852
I want to be able to get the value from a date. Something like:
myseries.loc('2015-10-01') and it returns -0.053565
The index are tuples in the form (2016, 1, 2016-01-04)
You can do it like this:
In [32]:
df.loc(axis=0)[:,:,'2015-10-01']
Out[32]:
value
year month date
2015 10 2015-10-01 -0.053565
You can also pass slice for each level:
In [39]:
df.loc[(slice(None),slice(None),'2015-10-01'),]
Out[39]:
value
year month date
2015 10 2015-10-01 -0.053565|
Or just pass the first 2 index levels:
In [40]:
df.loc[2015,10]
Out[40]:
value
date
2015-10-01 -0.053565
Try xs:
print s.xs('2015-10-01',level=2,axis=0)
#year datetime
#2015 10 -0.053565
#Name: series, dtype: float64
print s.xs(7,level=1,axis=0)
#year datetime
#2014 2014-07-01 -0.045417
#2015 2015-07-01 -0.018541
#Name: series, dtype: float64
I have a column in a pandas data frame looking like:
test1.Received
Out[9]:
0 01/01/2015 17:25
1 02/01/2015 11:43
2 04/01/2015 18:21
3 07/01/2015 16:17
4 12/01/2015 20:12
5 14/01/2015 11:09
6 15/01/2015 16:05
7 16/01/2015 21:02
8 26/01/2015 03:00
9 27/01/2015 08:32
10 30/01/2015 11:52
This represents a time stamp as Day Month Year Hour Minute. I would like to rearrange the date as Year Month Day Hour Minute. So that it would look like:
test1.Received
Out[9]:
0 2015/01/01 17:25
1 2015/01/02 11:43
...
Just use pd.to_datetime:
In [33]:
import pandas as pd
pd.to_datetime(df['date'])
Out[33]:
index
0 2015-01-01 17:25:00
1 2015-02-01 11:43:00
2 2015-04-01 18:21:00
3 2015-07-01 16:17:00
4 2015-12-01 20:12:00
5 2015-01-14 11:09:00
6 2015-01-15 16:05:00
7 2015-01-16 21:02:00
8 2015-01-26 03:00:00
9 2015-01-27 08:32:00
10 2015-01-30 11:52:00
Name: date, dtype: datetime64[ns]
In your case:
pd.to_datetime(test1['Received'])
should just work
If you want to change the display format then you need to parse as a datetime and then apply `datetime.strftime:
In [35]:
import datetime as dt
pd.to_datetime(df['date']).apply(lambda x: dt.datetime.strftime(x, '%m/%d/%y %H:%M:%S'))
Out[35]:
index
0 01/01/15 17:25:00
1 02/01/15 11:43:00
2 04/01/15 18:21:00
3 07/01/15 16:17:00
4 12/01/15 20:12:00
5 01/14/15 11:09:00
6 01/15/15 16:05:00
7 01/16/15 21:02:00
8 01/26/15 03:00:00
9 01/27/15 08:32:00
10 01/30/15 11:52:00
Name: date, dtype: object
So the above is now showing month/day/year, in your case the following should work:
pd.to_datetime(test1['Received']).apply(lambda x: dt.datetime.strftime(x, '%y/%m/%d %H:%M:%S'))
EDIT
it looks like you need to pass param dayfirst=True to to_datetime:
In [45]:
pd.to_datetime(df['date'], format('%d/%m/%y %H:%M:%S'), dayfirst=True).apply(lambda x: dt.datetime.strftime(x, '%m/%d/%y %H:%M:%S'))
Out[45]:
index
0 01/01/15 17:25:00
1 01/02/15 11:43:00
2 01/04/15 18:21:00
3 01/07/15 16:17:00
4 01/12/15 20:12:00
5 01/14/15 11:09:00
6 01/15/15 16:05:00
7 01/16/15 21:02:00
8 01/26/15 03:00:00
9 01/27/15 08:32:00
10 01/30/15 11:52:00
Name: date, dtype: object
Pandas has this in-built, you can specify your datetime format
http://pandas.pydata.org/pandas-docs/stable/generated/pandas.to_datetime.html.
use infer_datetime_format
>>> import pandas as pd
>>> i = pd.date_range('20000101',periods=100)
>>> df = pd.DataFrame(dict(year = i.year, month = i.month, day = i.day))
>>> pd.to_datetime(df.year*10000 + df.month*100 + df.day, format='%Y%m%d')
0 2000-01-01
1 2000-01-02
...
98 2000-04-08
99 2000-04-09
Length: 100, dtype: datetime64[ns]
you can use the datetime functions to convert from and to strings.
# converts to date
datetime.strptime(date_string, 'DD/MM/YYYY HH:MM')
and
# converts to your requested string format
datetime.strftime(date_string, "YYYY/MM/DD HH:MM:SS")