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copied!<p>Requires numpy >= 1.7. This is for pandas 0.13 (releasing soon). See docs <a href="http://pandas.pydata.org/pandas-docs/dev/timeseries.html#time-deltas" rel="nofollow">here</a></p> <pre><code>In [3]: df = DataFrame(dict(A = Timestamp('20130101'), B = Timestamp('20130101')+ pd.to_timedelta(list(range(5)),unit='D'))) In [4]: df Out[4]: A B 0 2013-01-01 00:00:00 2013-01-01 00:00:00 1 2013-01-01 00:00:00 2013-01-02 00:00:00 2 2013-01-01 00:00:00 2013-01-03 00:00:00 3 2013-01-01 00:00:00 2013-01-04 00:00:00 4 2013-01-01 00:00:00 2013-01-05 00:00:00 [5 rows x 2 columns] In [5]: df.dtypes Out[5]: A datetime64[ns] B datetime64[ns] dtype: object In [6]: df['C'] = df['B']-df['A'] In [7]: df Out[7]: A B C 0 2013-01-01 00:00:00 2013-01-01 00:00:00 00:00:00 1 2013-01-01 00:00:00 2013-01-02 00:00:00 1 days, 00:00:00 2 2013-01-01 00:00:00 2013-01-03 00:00:00 2 days, 00:00:00 3 2013-01-01 00:00:00 2013-01-04 00:00:00 3 days, 00:00:00 4 2013-01-01 00:00:00 2013-01-05 00:00:00 4 days, 00:00:00 [5 rows x 3 columns] In [8]: df.dtypes Out[8]: A datetime64[ns] B datetime64[ns] C timedelta64[ns] dtype: object In [9]: df['C'].astype('timedelta64[s]') Out[9]: 0 0 1 86400 2 172800 3 259200 4 345600 Name: C, dtype: float64 </code></pre> <p>In 0.12 you can do this</p> <pre><code>In [1]: df = DataFrame(dict(A = Timestamp('20130101'), B = [Timestamp('20130101')+timedelta(days=i) for i in range(5) ])) In [2]: df['C'] = df['B']-df['A'] In [3]: Series(df['C'].values / np.timedelta64(1,'s')) Out[3]: 0 0 1 86400 2 172800 3 259200 4 345600 dtype: float64 </code></pre>

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