Ganesh Pal wrote: > Dear Python friend > > > I have a nested data dictonary in the below format and I need to store > 1000 of entries which are in teh below format > > >>>> X['emp_01']['salary3'] = dict(sex="f", status="single", exp="4", > grade="A",payment="200") >>>> X['emp_01']['salary4'] = dict(sex="f", status="single", exp="4", > grade="A",payment="400") >>>> X['emp_01']['salary5'] = dict(sex="f", status="single", exp="4", > grade="A",payment="400") > > > I only thing thats is changing is payment and I have payment_list as a > list > [100,200,400,500]: > > > The value salary3 ,salary4,salary4 is to be generated in the loop . Iam > trying to optimize the above code , by looping as shown below > > >>>> X = {} >>>> X['emp_01'] ={} >>>> for salary in range(len(payment_list)):
Whenever you feel the urge to write range(len(whatever)) -- resist that temptation, and you'll end up with better Python code ;) > ... X['emp_01'][salary] = dict(sex="f", status="single", exp="4", > grade="A",payment=payment_list[salary]) > ... >>>> X > {'emp_01': {0: {'grade': 'A', 'status': 'single', 'payment': 100, 'exp': > '4', 'sex': 'f'}, 1: {'grade': 'A', 'status': 'single', 'payment': 200, > 'exp': '4', 'sex': 'f'}, 2: {'grade': 'A', 'status': 'single', 'payment': > 400, 'exp': '4', 'sex': 'f'}, 3: {'grade': 'A', 'status': 'single', > 'payment': 500, 'exp': '4', 'sex': 'f'}}} >>>> > > > Any other suggestion , Please let me know I am on python 2.7 and Linux Instead of artificially blowing up your database change its structure. For example: X["emp_01"] = dict( sex="f", status="single", exp="4", grade="A", payments=[100, 200, 400, 500] ) -- https://mail.python.org/mailman/listinfo/python-list