From pd.date_range(\'2016-01\', \'2016-05\', freq=\'M\', ).strftime(\'%Y-%m\')
, the last month is 2016-04
, but I was expecting it to be 2016
I had a similar problem when using datetime objects in dataframe. I would set the boundaries through .min() and .max() functions and then fill in missing dates using the pd.date_range function. Unfortunately the returned list/df was missing the maximum value.
I found two work arounds for this:
1) Add "closed = None" parameter in the pd.date_range function. This worked in the example below; however, it didn't work for me when working only with dataframes (no idea why).
2) If option #1 doesn't work then you can add one extra unit (in this case a day) using the datetime.timedelta() function. In the case below it over indexed by a day but it can work for you if the date_range function isn't giving you the full range.
import pandas as pd
import datetime as dt
#List of dates as strings
time_series = ['2020-01-01', '2020-01-03', '2020-01-5', '2020-01-6', '2020-01-7']
#Creates dataframe with time data that is converted to datetime object
raw_data_df = pd.DataFrame(pd.to_datetime(time_series), columns = ['Raw_Time_Series'])
#Creates an indexed_time list that includes missing dates and the full time range
#Option No. 1 is to use the closed = None parameter choice.
indexed_time = pd.date_range(start = raw_data_df.Raw_Time_Series.min(),end = raw_data_df.Raw_Time_Series.max(),freq='D',closed= None)
print('indexed_time option #! = ', indexed_time)
#Option No. 2 if the function allows you to extend the time by one unit (in this case day)
#by using the datetime.timedelta function to get what you need.
indexed_time = pd.date_range(start = raw_data_df.Raw_Time_Series.min(),end = raw_data_df.Raw_Time_Series.max()+dt.timedelta(days=1),freq='D')
print('indexed_time option #2 = ', indexed_time)
#In this case you over index by an extra day because the date_range function works properly
#However, if the "closed = none" parameters doesn't extend through the full range then this is a good work around