Create a new dict with a given dict written to given path-CodePudding

I have a dict, and a list describing a path. I want to create a dict, with the dict written to the path. For example:

dog_path = ["animals", "dog"]

dog_content = {
  "legs": 4,
}

# desired output
animals = {
 "animals": {
   "dog": {
     "legs": 4,
   },
  }, 
}

This is what I have so far. It works for my test cases but thinking perhaps there is a simpler, more pythonic way to do this.

def wrap_dict(path, content):
  if not path:
    return content
  
  res = {}
  d = res
  for k in path[:-1]:
    d[k] = {}
    d = d[k]
  d[path[-1]] = content
  return res

Any suggestions appreciated!

CodePudding user response：

def wrap_dict(path, content):
    return {path[0]: wrap_dict(path[1:], content)} if path else content

wrap_dict(["animals", "dog"], {"legs": 4})
# returns {'animals': {'dog': {'legs': 4}}}

CodePudding user response：

It's perfectly alright, although it would be better if res, d and k had more meaningful names.

Another possible implementation would be using recursion:

def wrap_dict(path, content):
    if not path:
        return content
    return {path[0]: wrap_dict(path[1:], content)}

I think this would be easier to maintain, although I'd guess it would be slightly less performant. If you want to have a path of thousands of items, you might run up against the recursion limit, though.