Comparing dictionary values with a list and returning the key-CodePudding

I am almost there with my code, and I think I need to make a minor change to return the desired output but really I have encountered a 'coder's block'(if there is anything like that). Really appreciate any help: my code:

def no_you_pick(my_dict, my_list):
    output = []
    for i in my_list:
        for k, v in my_dict.items():
            if i in v:
                output.append(k)
                output.sort()
                return output
        else:
            return "Sorry, no restaurants meet your restrictions"

grading_scale={'Moes, Larrys, Curlys': ['vegetarian', 'vegan', 'gluten-free'], 'Hot Chili Pepper': ['vegetarian', 'gluten-free', 'vegan']}
guests_diet = ["vegetarian", "vegan", "gluten-free"]
print(no_you_pick(grading_scale, guests_diet))

Output:

['Moes, Larrys, Curlys']

Desired output:

["Hot Chili Pepper", "Moes, Larrys, Curlys"]

PS: I have also tried to return output in line with the nested for loop but it returns ["Hot Chili Pepper"] *I do not want to use list comprehension as I do not want the else statement printed within [] brackets.

CodePudding user response：

You are returning too early (from within the loop) and you can shorten this, of course:

def no_you_pick(my_dict, my_list):
    output = [k for k, v in my_dict.items() if any(x in v for x in my_list)]
    if output:
        output.sort()  # only sort it once
        return output
    return "Sorry, no restaurants meet your restrictions"

CodePudding user response：

You should add more example of what you expect with different inputs.
First you only get one result in output because you return as soon as you find a match.
Second, you should avoid adding a result multiple times if their dishes match multiple inputs: I think output should be a set.
Finally, why would you return a string when otherwise a list is expected? It's okay to return [] or {}, you can print your warning nonetheless:

def no_you_pick(my_dict, my_list):
  output = set()
  for i in my_list:
      for k, v in my_dict.items():
          if i in v:
              output.add(k)
  if not output:
      print("Sorry, no restaurants meet your restrictions")
  return output
grading_scale={'Moes, Larrys, Curlys': ['vegetarian', 'vegan', 'gluten-free'], 'Hot Chili Pepper': ['vegetarian', 'gluten-free', 'vegan']}
guests_diet = ["vegetarian", "vegan", "gluten-free"]
print(no_you_pick(grading_scale, guests_diet))

CodePudding user response：

The issue in your code is that you are returning the first value you find. Don't forget that calling return exits your function. A solution could be adding the return line outside the forloop.

A simpler way to do this would be:

def no_you_pick(my_dict, my_list):
    output = [x[0] for x in my_dict.items() if set(my_list).issubset(x[1])]
    if len(output) > 0:
        return output
    return "Sorry, no restaurants meet your restrictions"