I want to subtract dates in \'A\' from dates in \'B\' and add a new column with the difference.
df
A B
one 2014-01-01 2014-02-28
two 2014-
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.
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
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)]
If your columns aren't in datetime format. The shorter syntax would be: df.A = pd.to_datetime(df.A)
How about this:
times['days_since'] = max(list(df.index.values))
times['days_since'] = times['days_since'] - times['months']
times