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Convert (back and forth) UNIX timestamp to pandas.tslib.Timestamp and datetime for series

I am working with python 3.5.2, pandas 0.18.1 and sqlite3.
In my data base, I have a column unix_time with INT for seconds since 1970. Ideally I want to read my dataframe from sqlite, and then create a time column which would correspond to the datetime or pandas.tslib.Timestamp conversion of the unix_time column that I woul only use for some processing and then drop before saving the dataframe back.
The issue is that when parsing the unix_time column using :
df = pd.read_from_sql_query("SELECT * FROM test", con, parse_dates=['unix_time'])
I obtain pandas.tslib.Timestamp types which is fine for my processing, but then I have to recreate my original unix_time column using :
df['unix_time'][i] = (df['unix_time'][i] - datetime(1970,1,1)).total_seconds()
which is really 'dirty'
First question : Do you have a better way?
I thought about giving up the unix time format and only use datetime format but the to_datetime method from pandas returns in fact pandas.tslib.Timestamp ... And anyway, doing so would force me to iterate over all rows which is a bad solution. (It is impossible to apply to_datetime on something else than a view over a single cell of the dataframe
Second question : Is it possible to apply it on a series?
My last try was with directly using df['time'] = datetime.datetime.fromtimestamp(df['unix_time']) but surprisingly, it also returns pandas.tslib.Timestamp.
In the end, knowing that I can only save unix timestamps or datetimes, my only choices for the moment are :
parsing but then having to convert them back to unix timestamp one by
one.
Or not parse it but have to convert them to pandas.tslib.Timestamp
one by one.
It would be great if I could convert a whole series.
Last question : Is there a way to convert a unix timestamps series to datetime (or at least pandas.tslib.Timestamp), or a pandas.tslib.Timestamp (or datetime) series to unix timestamps?
Thanks
EDIT:
During my processing, I extract a row that I want to append to my dataset. Apparently, the coversion to pandas.tslib.Timestamp appends implicitly when passing from dataframe to serie :
df = pd.DataFrame({'UNX':pd.date_range('2016-01-01', freq='9999S', periods=10).astype(np.int64)//10**9})
df['Date'] = pd.to_datetime(df.UNX, unit='s')
print(df.Date.dtypes)
print(type(df['Date'][0]))
test = df.iloc[0]
print(type(test.Date))
new_df = test.to_frame().transpose() #from here, impossible to do : new_df.to_sql("test", con) because the type for 'Date' is not supported
print(new_df.Date.dtypes)
returns
datetime64[ns]
<class 'pandas.tslib.Timestamp'>
<class 'pandas.tslib.Timestamp'>
object
Is there a way to convert the 'Date' in new_df from pandas.tslib.Timestamp to datetime64[ns] or datetime.datetime (or simply str) ?

IIUC you can do it this way:
In [96]: df = pd.DataFrame({'UNX':pd.date_range('2016-01-01', freq='9999S', periods=10).astype(np.int64)//10**9})
In [97]: df
Out[97]:
UNX
0 1451606400
1 1451616399
2 1451626398
3 1451636397
4 1451646396
5 1451656395
6 1451666394
7 1451676393
8 1451686392
9 1451696391
Convert UNIX epoch to Python datetime:
In [98]: df['Date'] = pd.to_datetime(df.UNX, unit='s')
In [99]: df
Out[99]:
UNX Date
0 1451606400 2016-01-01 00:00:00
1 1451616399 2016-01-01 02:46:39
2 1451626398 2016-01-01 05:33:18
3 1451636397 2016-01-01 08:19:57
4 1451646396 2016-01-01 11:06:36
5 1451656395 2016-01-01 13:53:15
6 1451666394 2016-01-01 16:39:54
7 1451676393 2016-01-01 19:26:33
8 1451686392 2016-01-01 22:13:12
9 1451696391 2016-01-02 00:59:51
Convert datetime to UNIX epoch:
In [100]: df['UNX2'] = df.Date.astype('int64')//10**9
In [101]: df
Out[101]:
UNX Date UNX2
0 1451606400 2016-01-01 00:00:00 1451606400
1 1451616399 2016-01-01 02:46:39 1451616399
2 1451626398 2016-01-01 05:33:18 1451626398
3 1451636397 2016-01-01 08:19:57 1451636397
4 1451646396 2016-01-01 11:06:36 1451646396
5 1451656395 2016-01-01 13:53:15 1451656395
6 1451666394 2016-01-01 16:39:54 1451666394
7 1451676393 2016-01-01 19:26:33 1451676393
8 1451686392 2016-01-01 22:13:12 1451686392
9 1451696391 2016-01-02 00:59:51 1451696391
Check:
In [102]: df.UNX.eq(df.UNX2).all()
Out[102]: True

Round trip between Pandas Timestamp and Unix Seconds (since 1970-01-01):
date_in = pd.to_datetime("2022-04-07")
# type(date_in) is: pandas._libs.tslibs.timestamps.Timestamp
unix_seconds = date_in.value//10**9
date_out = pd.to_datetime(unix_seconds, unit="s")
Output:
date_in
Out[1]: Timestamp('2021-04-07 00:00:00')
unix_seconds
Out[2]: 1617753600
date_out
Out[3]: Timestamp('2021-04-07 00:00:00')

Pandas - converting datetime64 to Period

I have a Series of dates in datetime64 format.
I want to convert them to a series of Period with a monthly frequency. (Essentially, I want to group dates into months for analytical purposes).
There must be a way of doing this - I just cannot find it quickly.
Note: these dates are not the index of the data frame - they are just a column of data in the data frame.
Example input data (as a Series)
data = pd.to_datetime(pd.Series(['2014-10-01', '2014-10-01', '2014-10-31', '2014-11-15', '2014-11-30', np.NaN, '2014-12-01']))
print (data)
My current kludge/work around looks like
data = pd.to_datetime(pd.Series(['2014-10-01', '2014-10-01', '2014-10-31', '2014-11-15', '2014-11-30', np.NaN, '2014-01-01']))
data = pd.DatetimeIndex(data).to_period('M')
data = pd.Series(data.year).astype('str') + '-' + pd.Series((data.month).astype('int')).map('{:0>2d}'.format)
data = data.where(data != '2262-04', other='No Date')
print (data)

Their are some issues currently (even in master) dealing with NaT in PeriodIndex, so your approach won't work like that. But seems that you simply want to resample; so do this. You can of course specify a function for how if you want.
In [57]: data
Out[57]:
0 2014-10-01
1 2014-10-01
2 2014-10-31
3 2014-11-15
4 2014-11-30
5 NaT
6 2014-12-01
dtype: datetime64[ns]
In [58]: df = DataFrame(dict(A = data, B = np.arange(len(data))))
In [59]: df.dropna(how='any',subset=['A']).set_index('A').resample('M',how='count')
Out[59]:
B
A
2014-10-31 3
2014-11-30 2
2014-12-31 1

import pandas as pd
import numpy as np
datetime import datetime
data = pd.to_datetime(
pd.Series(['2014-10-01', '2014-10-01', '2014-10-31', '2014-11-15', '2014-11-30', np.NaN, '2014-01-01']))
data=pd.Series(['{}-{:02d}'.format(x.year,x.month) if isinstance(x, datetime) else "Nat" for x in pd.DatetimeIndex(data).to_pydatetime()])
0 2014-10
1 2014-10
2 2014-10
3 2014-11
4 2014-11
5 Nat
6 2014-01
dtype: object
Best I could come up with, if the only non datetimes objects possible are floats you can change if isinstance(x, datetime) to if not isinstance(x, float)

Pandas and csv import into dataframe. How to best to combine date anbd date fields into one

I have a csv file that I am trying to import into pandas.
There are two columns of intrest. date and hour and are the first two cols.
E.g.
date,hour,...
10-1-2013,0,
10-1-2013,0,
10-1-2013,0,
10-1-2013,1,
10-1-2013,1,
How do I import using pandas so that that hour and date is combined or is that best done after the initial import?
df = DataFrame.from_csv('bingads.csv', sep=',')
If I do the initial import how do I combine the two as a date and then delete the hour?
Thanks

Define your own date_parser:
In [291]: from dateutil.parser import parse
In [292]: import datetime as dt
In [293]: def date_parser(x):
.....: date, hour = x.split(' ')
.....: return parse(date) + dt.timedelta(0, 3600*int(hour))
In [298]: pd.read_csv('test.csv', parse_dates=[[0,1]], date_parser=date_parser)
Out[298]:
date_hour a b c
0 2013-10-01 00:00:00 1 1 1
1 2013-10-01 00:00:00 2 2 2
2 2013-10-01 00:00:00 3 3 3
3 2013-10-01 01:00:00 4 4 4
4 2013-10-01 01:00:00 5 5 5

Apply read_csv instead of read_clipboard to handle your actual data:
>>> df = pd.read_clipboard(sep=',')
>>> df['date'] = pd.to_datetime(df.date) + pd.to_timedelta(df.hour, unit='D')/24
>>> del df['hour']
>>> df
date ...
0 2013-10-01 00:00:00 NaN
1 2013-10-01 00:00:00 NaN
2 2013-10-01 00:00:00 NaN
3 2013-10-01 01:00:00 NaN
4 2013-10-01 01:00:00 NaN
[5 rows x 2 columns]

Take a look at the parse_dates argument which pandas.read_csv accepts.
You can do something like:
df = pandas.read_csv('some.csv', parse_dates=True)
# in which case pandas will parse all columns where it finds dates
df = pandas.read_csv('some.csv', parse_dates=[i,j,k])
# in which case pandas will parse the i, j and kth columns for dates

Since you are only using the two columns from the cdv file and combining those into one, I would squeeze into a series of datetime objects like so:
import pandas as pd
from StringIO import StringIO
import datetime as dt
txt='''\
date,hour,A,B
10-1-2013,0,1,6
10-1-2013,0,2,7
10-1-2013,0,3,8
10-1-2013,1,4,9
10-1-2013,1,5,10'''
def date_parser(date, hour):
dates=[]
for ed, eh in zip(date, hour):
month, day, year=list(map(int, ed.split('-')))
hour=int(eh)
dates.append(dt.datetime(year, month, day, hour))
return dates
p=pd.read_csv(StringIO(txt), usecols=[0,1],
parse_dates=[[0,1]], date_parser=date_parser, squeeze=True)
print p
Prints:
0 2013-10-01 00:00:00
1 2013-10-01 00:00:00
2 2013-10-01 00:00:00
3 2013-10-01 01:00:00
4 2013-10-01 01:00:00
Name: date_hour, dtype: datetime64[ns]

How do I convert strings in a Pandas data frame to a 'date' data type?

I have a Pandas data frame, one of the column contains date strings in the format YYYY-MM-DD
For e.g. '2013-10-28'
At the moment the dtype of the column is object.
How do I convert the column values to Pandas date format?

Essentially equivalent to #waitingkuo, but I would use pd.to_datetime here (it seems a little cleaner, and offers some additional functionality e.g. dayfirst):
In [11]: df
Out[11]:
a time
0 1 2013-01-01
1 2 2013-01-02
2 3 2013-01-03
In [12]: pd.to_datetime(df['time'])
Out[12]:
0 2013-01-01 00:00:00
1 2013-01-02 00:00:00
2 2013-01-03 00:00:00
Name: time, dtype: datetime64[ns]
In [13]: df['time'] = pd.to_datetime(df['time'])
In [14]: df
Out[14]:
a time
0 1 2013-01-01 00:00:00
1 2 2013-01-02 00:00:00
2 3 2013-01-03 00:00:00
Handling ValueErrors
If you run into a situation where doing
df['time'] = pd.to_datetime(df['time'])
Throws a
ValueError: Unknown string format
That means you have invalid (non-coercible) values. If you are okay with having them converted to pd.NaT, you can add an errors='coerce' argument to to_datetime:
df['time'] = pd.to_datetime(df['time'], errors='coerce')

Use astype
In [31]: df
Out[31]:
a time
0 1 2013-01-01
1 2 2013-01-02
2 3 2013-01-03
In [32]: df['time'] = df['time'].astype('datetime64[ns]')
In [33]: df
Out[33]:
a time
0 1 2013-01-01 00:00:00
1 2 2013-01-02 00:00:00
2 3 2013-01-03 00:00:00

I imagine a lot of data comes into Pandas from CSV files, in which case you can simply convert the date during the initial CSV read:
dfcsv = pd.read_csv('xyz.csv', parse_dates=[0]) where the 0 refers to the column the date is in.
You could also add , index_col=0 in there if you want the date to be your index.
See https://pandas.pydata.org/pandas-docs/stable/reference/api/pandas.read_csv.html

Now you can do df['column'].dt.date
Note that for datetime objects, if you don't see the hour when they're all 00:00:00, that's not pandas. That's iPython notebook trying to make things look pretty.

If you want to get the DATE and not DATETIME format:
df["id_date"] = pd.to_datetime(df["id_date"]).dt.date

Another way to do this and this works well if you have multiple columns to convert to datetime.
cols = ['date1','date2']
df[cols] = df[cols].apply(pd.to_datetime)

It may be the case that dates need to be converted to a different frequency. In this case, I would suggest setting an index by dates.
#set an index by dates
df.set_index(['time'], drop=True, inplace=True)
After this, you can more easily convert to the type of date format you will need most. Below, I sequentially convert to a number of date formats, ultimately ending up with a set of daily dates at the beginning of the month.
#Convert to daily dates
df.index = pd.DatetimeIndex(data=df.index)
#Convert to monthly dates
df.index = df.index.to_period(freq='M')
#Convert to strings
df.index = df.index.strftime('%Y-%m')
#Convert to daily dates
df.index = pd.DatetimeIndex(data=df.index)
For brevity, I don't show that I run the following code after each line above:
print(df.index)
print(df.index.dtype)
print(type(df.index))
This gives me the following output:
Index(['2013-01-01', '2013-01-02', '2013-01-03'], dtype='object', name='time')
object
<class 'pandas.core.indexes.base.Index'>
DatetimeIndex(['2013-01-01', '2013-01-02', '2013-01-03'], dtype='datetime64[ns]', name='time', freq=None)
datetime64[ns]
<class 'pandas.core.indexes.datetimes.DatetimeIndex'>
PeriodIndex(['2013-01', '2013-01', '2013-01'], dtype='period[M]', name='time', freq='M')
period[M]
<class 'pandas.core.indexes.period.PeriodIndex'>
Index(['2013-01', '2013-01', '2013-01'], dtype='object')
object
<class 'pandas.core.indexes.base.Index'>
DatetimeIndex(['2013-01-01', '2013-01-01', '2013-01-01'], dtype='datetime64[ns]', freq=None)
datetime64[ns]
<class 'pandas.core.indexes.datetimes.DatetimeIndex'>

For the sake of completeness, another option, which might not be the most straightforward one, a bit similar to the one proposed by #SSS, but using rather the datetime library is:
import datetime
df["Date"] = df["Date"].apply(lambda x: datetime.datetime.strptime(x, '%Y-%d-%m').date())

# Column Non-Null Count Dtype
--- ------ -------------- -----
0 startDay 110526 non-null object
1 endDay 110526 non-null object
import pandas as pd
df['startDay'] = pd.to_datetime(df.startDay)
df['endDay'] = pd.to_datetime(df.endDay)
# Column Non-Null Count Dtype
--- ------ -------------- -----
0 startDay 110526 non-null datetime64[ns]
1 endDay 110526 non-null datetime64[ns]

Try to convert one of the rows into timestamp using the pd.to_datetime function and then use .map to map the formular to the entire column

Pandas Timedelta in Days

I have a dataframe in pandas called 'munged_data' with two columns 'entry_date' and 'dob' which i have converted to Timestamps using pd.to_timestamp.I am trying to figure out how to calculate ages of people based on the time difference between 'entry_date' and 'dob' and to do this i need to get the difference in days between the two columns ( so that i can then do somehting like round(days/365.25). I do not seem to be able to find a way to do this using a vectorized operation. When I do munged_data.entry_date-munged_data.dob i get the following :
internal_quote_id
2 15685977 days, 23:54:30.457856
3 11651985 days, 23:49:15.359744
4 9491988 days, 23:39:55.621376
7 11907004 days, 0:10:30.196224
9 15282164 days, 23:30:30.196224
15 15282227 days, 23:50:40.261632
However i do not seem to be able to extract the days as an integer so that i can continue with my calculation.
Any help appreciated.

Using the Pandas type Timedelta available since v0.15.0 you also can do:
In[1]: import pandas as pd
In[2]: df = pd.DataFrame([ pd.Timestamp('20150111'),
pd.Timestamp('20150301') ], columns=['date'])
In[3]: df['today'] = pd.Timestamp('20150315')
In[4]: df
Out[4]:
date today
0 2015-01-11 2015-03-15
1 2015-03-01 2015-03-15
In[5]: (df['today'] - df['date']).dt.days
Out[5]:
0 63
1 14
dtype: int64

You need 0.11 for this (0.11rc1 is out, final prob next week)
In [9]: df = DataFrame([ Timestamp('20010101'), Timestamp('20040601') ])
In [10]: df
Out[10]:
0
0 2001-01-01 00:00:00
1 2004-06-01 00:00:00
In [11]: df = DataFrame([ Timestamp('20010101'),
Timestamp('20040601') ],columns=['age'])
In [12]: df
Out[12]:
age
0 2001-01-01 00:00:00
1 2004-06-01 00:00:00
In [13]: df['today'] = Timestamp('20130419')
In [14]: df['diff'] = df['today']-df['age']
In [16]: df['years'] = df['diff'].apply(lambda x: float(x.item().days)/365)
In [17]: df
Out[17]:
age today diff years
0 2001-01-01 00:00:00 2013-04-19 00:00:00 4491 days, 00:00:00 12.304110
1 2004-06-01 00:00:00 2013-04-19 00:00:00 3244 days, 00:00:00 8.887671
You need this odd apply at the end because not yet full support for timedelta64[ns] scalars (e.g. like how we use Timestamps now for datetime64[ns], coming in 0.12)

Not sure if you still need it, but in Pandas 0.14 i usually use .astype('timedelta64[X]') method
http://pandas.pydata.org/pandas-docs/stable/timeseries.html (frequency conversion)
df = pd.DataFrame([ pd.Timestamp('20010101'), pd.Timestamp('20040605') ])
df.ix[0]-df.ix[1]
Returns:
0 -1251 days
dtype: timedelta64[ns]
(df.ix[0]-df.ix[1]).astype('timedelta64[Y]')
Returns:
0 -4
dtype: float64
Hope that will help

Let's specify that you have a pandas series named time_difference which has type
numpy.timedelta64[ns]
One way of extracting just the day (or whatever desired attribute) is the following:
just_day = time_difference.apply(lambda x: pd.tslib.Timedelta(x).days)
This function is used because the numpy.timedelta64 object does not have a 'days' attribute.

To convert any type of data into days just use pd.Timedelta().days:
pd.Timedelta(1985, unit='Y').days
84494