Change multiple keys from a dictionary by performing a timedelta operation in Python

Question

I have a dictionary that is an integer. I arbitrarily changed one of the keys to a date, and I need to change the other keys.

Sample data:

{'C-STD-B&M-SUM': {datetime.date(2015, 7, 12): 0, -1: 0.21484699999999998, -2: 0.245074, -3: 0.27874} 

Expected Result:

 {'C-STD-B&M-SUM': {datetime.date(2015, 7, 12): 0, datetime.date(2015, 7, 11): 0.21484699999999998, datetime.date(2015, 7, 10): 0.245074, datetime.date(2015, 7, 9): 0.27874} 

Current code:

 def change_start_date(dictionary_with_temporal_distribution): unsw_mid_year_end_date = datetime.date(2015, 7, 12) dictionary_with_temporal_distribution['C-STD-B&M-SUM'][unsw_mid_year_end_date] = dictionary_with_temporal_distribution['C-STD-B&M-SUM'][0] del dictionary_with_temporal_distribution['C-STD-B&M-SUM'][0] for k, v in dictionary_with_temporal_distribution['C-STD-B&M-SUM'].items(): 

Answer 1

You can try something like -

 def change_start_date(dictionary_with_temporal_distribution): unsw_mid_year_end_date = datetime.date(2015, 7, 12) for k in list(dictionary_with_temporal_distribution['C-STD-B&M-SUM'].keys()): dictionary_with_temporal_distribution['C-STD-B&M-SUM'][unsw_mid_year_end_date + timedelta(days=k)] = dictionary_with_temporal_distribution['C-STD-B&M-SUM'][k] del dictionary_with_temporal_distribution['C-STD-B&M-SUM'][k] 

Answer 2

You can use speech understanding syntax and convert and replace keys:

 dct = {datetime.date(2015, 7, 12): 0, -1: 0.21484699999999998, -2: 0.245074, -3: 0.27874} def offset(offset, base): """Applies an offset in days to a base date. If the offset is already a date it is returned as is.""" if type(offset) == datetime.date: return offset return base + datetime.timedelta(offset) def offset_keys(dct, base): """Takes a dict and runs offset(key, base) on all keys""" return { offset(k, base): v for k, v in dct.items() } pprint(offset_keys(dct, datetime.date(2015, 7, 12))) 

 {datetime.date(2015, 7, 9): 0.27874, datetime.date(2015, 7, 10): 0.245074, datetime.date(2015, 7, 11): 0.21484699999999998, datetime.date(2015, 7, 12): 0}