### Question :

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.

##
Answer #1:

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)

##
Answer #2:

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
```

##
Answer #3:

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

##
Answer #4:

To convert any type of data into days just use Timedelta().days:

```
pd.Timedelta(1985, unit='Y').days
84494
```

##
Answer #5:

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.