Use df. dates1-df. dates2 to find the difference between the two dates and then convert the result in the form of months.
You can use simple date arithmetic to find the number of days between two dates in Python. Define the 2 dates between which you want to find the difference in days. Then subtract these dates to get a timedelta object and examine the day's property of this object to get the required result.
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
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)]
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
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