Add column with number of days between dates in Da

Question

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 2014-02-03  2014-03-01

I've tried the following, but get an error when I try to include this in a for loop...

import datetime
date1=df['A'][0]
date2=df['B'][0]
mdate1 = datetime.datetime.strptime(date1, "%Y-%m-%d").date()
rdate1 = datetime.datetime.strptime(date2, "%Y-%m-%d").date()
delta =  (mdate1 - rdate1).days
print delta

What should I do?

Answer 1

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.

Answer 2

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)

Answer 3

How about this:

times['days_since'] = max(list(df.index.values))  
times['days_since'] = times['days_since'] - times['months']  
times

Answer 4

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

df['C'] = (df['B'] - df['A']).dt.days

Which returns

             A          B   C
one 2014-01-01 2014-02-03  33
two 2014-02-03 2014-03-01  26