Python Time differencing in pandas

Question

Lets say I have an array of date:

df=pd.DataFrame({'Date': [ '5/4/1985', '6/13/1983', '6/13/1982', '12/28/1987'], 'Name': ['J','K','L','M']})

and I want to find the difference between them and today in years, weeks, or quarters (four in one year).

So far I know I would use pd.datetime.now() to represent today.

CT Zhu · Accepted Answer · 2014-02-09 21:23:27Z

2

To get the number of nano second difference, Just do:

In [60]:
pd.datetime.now()-pd.to_datetime(df['Date'])
Out[60]:
0   10508 days, 15:56:08.609000
1   11199 days, 15:56:08.609000
2   11564 days, 15:56:08.609000
3    9540 days, 15:56:08.609000
Name: Date, dtype: timedelta64[ns]

To get the difference in days or years (etc):

q=pd.datetime.now()-pd.to_datetime(df['Date'])
array(q).astype('timedelta64[D]').astype(int) #in days
array(q).astype('timedelta64[Y]').astype(int) #in years

edited Feb 9, 2014 at 21:23

answered Feb 9, 2014 at 20:56

CT Zhu

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4 Comments

user3084006 Over a year ago

is there a way to get years, quarters or weeks instead of days

CT Zhu Over a year ago

Sorry, the default is not in days but in nanoseconds. Corrected that. I don't know if you can get quarters, there are 4 types available for timedelta64: D W M Y, for day, week, month and year respectively.

user3084006 Over a year ago

So I guess the only way to change it to quarters is to change to months then divide by 3?

CT Zhu Over a year ago

I guess that will be better (than calculated it from the number of weeks or days). There may be more elegant way of doing it, I don't know.

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