Add column with number of days between dates in DataFrame pandas

Assuming these were datetime columns (if they're not apply to_datetime) you can just subtract them:

df['A'] = pd.to_datetime(df['A'])
df['B'] = pd.to_datetime(df['B'])

In [11]: df.dtypes  # if already datetime64 you don't need to use to_datetime
Out[11]:
A    datetime64[ns]
B    datetime64[ns]
dtype: object

In [12]: df['A'] - df['B']
Out[12]:
one   -58 days
two   -26 days
dtype: timedelta64[ns]

In [13]: df['C'] = df['A'] - df['B']

In [14]: df
Out[14]:
             A          B        C
one 2014-01-01 2014-02-28 -58 days
two 2014-02-03 2014-03-01 -26 days

Note: ensure you're using a new of pandas (e.g. 0.13.1), this may not work in older versions.

A list comprehension is your best bet for the most Pythonic (and fastest) way to do this:

[int(i.days) for i in (df.B - df.A)]

1. i will return the timedelta(e.g. '-58 days')
2. i.days will return this value as a long integer value(e.g. -58L)
3. int(i.days) will give you the -58 you seek.

If your columns aren't in datetime format. The shorter syntax would be: df.A = pd.to_datetime(df.A)

To remove the 'days' text element, you can also make use of the dt() accessor for series: https://pandas.pydata.org/pandas-docs/stable/generated/pandas.Series.dt.html

So,

df[['A','B']] = df[['A','B']].apply(pd.to_datetime) #if conversion required
df['C'] = (df['B'] - df['A']).dt.days

which returns:

             A          B   C
one 2014-01-01 2014-02-28  58
two 2014-02-03 2014-03-01  26