changing local namespace of a function

[Kent Johnson]

Bo Peng wrote:
> Dear list,
> 
> I have many dictionaries with the same set of keys and I would like to 
> write a function to calculate something based on these values. For 
> example, I have
> 
> a = {'x':1, 'y':2}
> b = {'x':3, 'y':3}
> 
> def fun(dict):
>   dict['z'] = dict['x'] + dict['y']
> 
> fun(a) and fun(b) will set z in each dictionary as the sum of x and y.
> 
> My function and dictionaries are a lot more complicated than these so I 
> would like to set dict as the default namespace of fun. Is this 
> possible? The ideal code would be:
> 
> def fun(dict):
>   # set dict as local namespace
>   # locals() = dict?
>   z = x + y

You can part way there using keyword arguments. You just have to use dictionary syntax for changing 
values in the dictionary:

  >>> def f(d, x=None, y=None):
  ...   d['z'] = x + y
  ...
  >>> a = {'x':1, 'y':2}
  >>> b = {'x':3, 'y':3}
  >>>
  >>> f(a, **a)
  >>> a
{'y': 2, 'x': 1, 'z': 3}
  >>> f(b, **b)
  >>> b
{'y': 3, 'x': 3, 'z': 6}

Kent