Pandas cannot convert input to timestamp error

Question

I have two variables in the data set beginning date (format datetime64[ns]) and end date(format datetime64[ns]). I'm using following code to get the dates between beginning date and end date.

pd.date_range(start = data['beginning_date'], end = data['end_date'], freq = 'D')

but it's throwing following error.

cannot convert input to timestamp

why I'm getting above error. I tried changing as below, but it doesn't work.

pd.date_range(start = data['beginning_date'], end = data['end_date'], freq = 'D').astype('datetime')

and also i want each day as separate record, for example: beginning_date = 01APR2015 and end_date = 30APR2015, i want each day as separate record as below.

 01APR2015
 02APR2015 etc

How can I get it as a separate record?

Thanks in Advance.

Answer 1

Assuming you have the following DF:

In [30]: df
Out[30]:
  beginning_date   end_date
0     2013-12-22 2014-01-01
1     2009-12-14 2009-12-28
2     2010-12-31 2011-01-11

I guess you tried to use series instead of scalar values when calling pd.date_range() method:

In [31]: pd.date_range(df.beginning_date, df.end_date)
...
skipped
...
TypeError: Cannot convert input to Timestamp

So try this instead:

In [32]: pd.date_range(df.beginning_date.min(), df.end_date.max())
Out[32]:
DatetimeIndex(['2009-12-14', '2009-12-15', '2009-12-16', '2009-12-17', '2009-12-18', '2009-12-19', '2009-12-20', '2009-12-21', '2009-12-22',
 '2009-12-23',
               ...
               '2013-12-23', '2013-12-24', '2013-12-25', '2013-12-26', '2013-12-27', '2013-12-28', '2013-12-29', '2013-12-30', '2013-12-31',
 '2014-01-01'],
              dtype='datetime64[ns]', length=1480, freq='D')

Answer 2

If you format the input first in this way, then the conversion would work.

pd.date_range(start = '30-APR-2015', end = '05-MAY-2015', freq = 'D')

And the output would be

DatetimeIndex(['2015-04-30', '2015-05-01', '2015-05-02', '2015-05-03',
           '2015-05-04', '2015-05-05'],
          dtype='datetime64[ns]', freq='D')

Answer 3

Если кто придет сюда с этой проблемой из 2к20, то решение - применение lambda

If anyone comes here with this problem from 2k20, the solution is to use lambda...

f = lambda x: len(pd.date_range(start=x[0],end=x[1],freq='M'))

df_clear['difference'] = df_clear[['start_month','finish_month']].apply(f, axis=1)