turn dictionary values into paths-CodePudding

I have this dictionary/json file:

{ 
  'animal': {
        'cat': {
            'Persian': {}, 
            'British_Shorthair': {} 
        }
        'dog': {
            'Pug': {},
            'Pitbull': {}
        }
   }
  'vehicle': {
        'car': {
            'Mercedes': {},
            'BMW': {}
        }
}

And I want turn this into a list of paths, meaning sth like this:

[ [animal, cat, persian], [animal, cat, british_shorthait], 
[animal, dog, pug], [animal, dog, pitbull], [vehicle, car, mercedes], 
[vehicle, car, bmw] ]

My file is much bigger than the example, but the structure is the same. They're all 3 levels, meaning all the paths are the same length.

Any ideas how to do this in just a few lines of code? I used multiple for-loops to get a decent solution, but it's quite inconvenient

CodePudding user response：

What is the problem with using multiple for-loops? In this way, you could achieve the result in 'a few lines of code'.

my_dict = { 
  'animal': {
        'cat': {
            'Persian': {}, 
            'British_Shorthair': {} 
        },
        'dog': {
            'Pug': {},
            'Pitbull': {}
        }
   },
  'vehicle': {
        'car': {
            'Mercedes': {},
            'BMW': {}
        }
    }
}


my_list = []
for level1 in my_dict.keys():
  for level2 in my_dict[level1].keys():
    for level3 in my_dict[level1][level2].keys():
      my_list.append([level1, level2, level3])

print(my_list)

>>> 
[['animal', 'cat', 'Persian'], ['animal', 'cat', 'British_Shorthair'], ['animal', 'dog', 'Pug'], ['animal', 'dog', 'Pitbull'], ['vehicle', 'car', 'Mercedes'], ['vehicle', 'car', 'BMW']]

Or, using a list comprehension you could even do it in one line:

[[level1, level2, level3] for level1 in my_dict.keys() for level2 in my_dict[level1].keys() for level3 in my_dict[level1][level2].keys()]

CodePudding user response：

This solution works with structures with any number of levels (and it is short anyway).

def foo(_dict):
    if not _dict:
        return [[]]
    return [[k, *p] for k,v in _dict.items() for p in foo(v)]