dictionary help

[MRAB]

Krishna Pacifici wrote:
> Thanks for the help.
> 
> Actually this is part of a much larger project, but I have unfortunately 
> pigeon-holed myself into needing to do these things without a whole lot 
> of flexibility.
> 
> To give a specific example I have the following dictionary where I need 
> to remove values that are duplicated with other values and remove values 
> that are duplicates of the keys, but still retain it as a dictionary.  
> Each value is itself a class with many attributes that I need to call 
> later on in the program, but I cannot have duplicates because it would 
> mess up some estimation part of my model.
> 
> d =
> {36: [35, 37, 26, 46], 75: [74, 76, 65, 85], 21: [20, 22, 11, 31], 22: 
> [21, 23, 12, 32], 26: [25, 27, 16, 36], 30: [20, 31, 40]}
> 
> So I want a new dictionary that would get rid of the duplicate values of 
> 21, 22, 36 and 20 and give me back a dictionary that looked like this:
> 
> new_d=
> {36: [35, 37, 26, 46], 75: [74, 76, 65, 85], 21: [20, 11, 31], 22: [23, 
> 12, 32], 26: [25, 27, 16], 30: [40]}
> 
> I understand that a dictionary may not be the best approach, but like I 
> said I have sort of pigeon-holed myself by the way that I am simulating 
> my data and the estimation model that I am using.  Any suggestions or 
> comments about the above problem would be greatly appreciated.
> 
 >>> d = {36: [35, 37, 26, 46], 75: [74, 76, 65, 85], 21: [20, 22, 11,
31], 22: [21, 23, 12, 32], 26: [25, 27, 16, 36], 30: [20, 31, 40]}
 >>> new_d = {}
 >>> seen = set(d.keys())
 >>> for k, v in d.items():
...     new_d[k] = [x for x in v if x not in seen]
...     seen |= set(new_d[k])
...
 >>> new_d
{36: [35, 37, 46], 75: [74, 76, 65, 85], 21: [20, 11, 31], 22: [23, 12,
32], 26: [25, 27, 16], 30: [40]}