Alternative way to use setdefault() using dictionary comprehension?-CodePudding

I have a nested dictionary that was created from a nested list where the first item in the nested list would be the outer key and outer value would be a dictionary which is the next two items. The following code is working great using the two setdefault() functions because it just adds to the nested dictionary when it sees a duplicate key of the outer. I was just wondering how you could do this same logic using dictionary comprehension?

dict1 = {}
list1 = [[1, 2, 6],
         [1, 3, 7],
         [2, 5, 8],
         [2, 8, 9]]
    
for i in list1:
    dict1.setdefault(i[0], {}).setdefault(i[1], i[2])

"""
OUTPUT: 1 {2: 6, 3: 7} 
        2 {5: 8, 8: 9}
"""

CodePudding user response：

I actually tried to achieve that result and failed.
The comprehension overwrites the new entries.

After, giving this idea a look, I found a similar post in which it is stated it is not possible:
https://stackoverflow.com/questions/11276473/append-to-a-dict-of-lists-with-a-dict-comprehension

I believe Amber's answer best sumarizes what the conclusion with my failed attempt with dict comprehensions:

No - dict comprehensions are designed to generate non-overlapping keys with each iteration; they don't support aggregation. For this particular use case, a loop is the proper way to accomplish the task efficiently (in linear time)

CodePudding user response：

Use the loop because it's very readable and efficient. Not all code has to be a one-liner.

Having said that, it's possible. It abuses syntax, extremely unreadable, inefficient, and generally just plain bad code (don't do it!)

out = {k: dict(v) for k,v in [g for g in [{}] if not 
                              any(g.setdefault(key, []).append(v) 
                                  for key, *v in list1)][0].items()}

Output:

{1: {2: 6, 3: 7}, 2: {5: 8, 8: 9}}