Sort a dictionary by list of lists

Question

I am trying to sort a dictionary by list of lists. The items in the list of lists are keys in the dictionary. I asked it before but the answers didn't solve the issue.

My input list is:

   mylist= [
    ['why', 'was', 'cinderella', 'late', 'for', 'the', 'ball', 'she', 'forgot', 'to', 'swing', 'the', 'bat'],
    ['why', 'is', 'the', 'little', 'duck', 'always', 'so', 'sad', 'because', 'he', 'always', 'sees', 'a', 'bill', 'in', 'front', 'of', 'his', 'face'],
    ['what', 'has', 'four', 'legs', 'and', 'goes', 'booo', 'a', 'cow', 'with', 'a', 'cold'], 
    ['what', 'is', 'a', 'caterpillar', 'afraid', 'of', 'a', 'dogerpillar'],
    ['what', 'did', 'the', 'crop', 'say', 'to', 'the', 'farmer', 'why', 'are', 'you', 'always', 'picking', 'on', 'me']
    ]

My dictionary somewhat looks like this:

    myDict = {'to': [7, 11, 17, 23, 24, 25, 26, 33, 34, 37, 39, 41, 47, 48, 53, 56], 
    'jam': [20], 'black': [5], 'farmer': [11],
    'woodchuck': [54], 'has': [14, 16, 51], 'who': [16]
    }

My code is:

    def sort_by_increasing_order(mylist, myDict):
    #temp = sorted(myDict, key=myDict.get)
    temp = sorted(myDict, key=lambda tempKey: for tempKey in mylist, reverse=True )
    return temp

Something like:

    sort_by_increasing_order(['d', 'e', 'f'], {'d': [0, 1], 'e': [1, 2, 3], 'f': [4]})
    result: ['f', 'e', 'd']

So for my sample input it would look like:

        sort_by_increasing_order(mylist, myDict)
        >> ['to','woodchuck','has','jam','who','farmer']

The commented line just sorts by the dictionary keys when i try to sort by the list. My approach is not correct. The result should a list with increasing order of the length of indices as mentioned above. Any suggestion.

Answer 1

With reference to @doukremt answer assuming you are aware of decorators.

mydict = {'d': [0, 1], 'e': [1, 2, 3], 'f': [4]}
mylist = [['d', 'e', 'f', 'c'], ['c', 'v', 'd', 'n']]

def convert_to_set(mylist, result_set=None):
    if result_set is None:
        result_set = []
    for item in mylist:
        if isinstance(item, str):
            result_set.append(item)
        if isinstance(item, list):
            convert_to_set(item, result_set)
    return set(result_set)

def list_to_set(f):
    def wrapper(mylist, mydict):
        myset = convert_to_set(mylist)
        result = f(myset, mydict)
        return result
    return wrapper

@list_to_set
def findit(mylist, mydict):
    gen = ((k, mydict[k]) for k in mylist if k in mydict)
    return [k for k, v in sorted(gen, key=lambda p: len(p[1]))]

print findit(mylist, mydict)

Answer 2

>>> D= {'to': [7, 11, 17, 23, 24, 25, 26, 33, 34, 37, 39, 41, 47, 48, 53, 56], 
...     'jam': [20], 'black': [5], 'farmer': [11],
...     'woodchuck': [54], 'has': [14, 16, 51], 'who': [16]
...     }
>>> 
>>> sorted(D, key=lambda k:len(D[k]), reverse=True)
['to', 'has', 'who', 'jam', 'black', 'farmer', 'woodchuck']

For the values

>>> sorted(D.values(), key=len, reverse=True)
[[7, 11, 17, 23, 24, 25, 26, 33, 34, 37, 39, 41, 47, 48, 53, 56], [14, 16, 51], [16], [20], [5], [11], [54]]

For (keys, values)

>>> sorted(D.items(), key=lambda i:len(i[1]), reverse=True)
[('to', [7, 11, 17, 23, 24, 25, 26, 33, 34, 37, 39, 41, 47, 48, 53, 56]), ('has', [14, 16, 51]), ('who', [16]), ('jam', [20]), ('black', [5]), ('farmer', [11]), ('woodchuck', [54])]

Edit: Still not really clear what you are asking for. Your example doesn't seem to care about the length at all, otherwise "has" should come before "woodchuck"? Changing len to max may be what you want

>>> D = {'to': [7, 11, 17, 23, 24, 25, 26, 33, 34, 37, 39, 41, 47, 48, 53, 56], 
...     'jam': [20], 'black': [5], 'farmer': [11],
...     'woodchuck': [54], 'has': [14, 16, 51], 'who': [16]
...     }
>>> 
>>> sorted(D, key=lambda k:max(D[k]), reverse=True)
['to', 'woodchuck', 'has', 'jam', 'who', 'farmer', 'black']
>>> sorted(D.values(), key=max, reverse=True)
[[7, 11, 17, 23, 24, 25, 26, 33, 34, 37, 39, 41, 47, 48, 53, 56], [54], [14, 16, 51], [20], [16], [11], [5]]
>>> sorted(D.items(), key=lambda i:max(i[1]), reverse=True)
[('to', [7, 11, 17, 23, 24, 25, 26, 33, 34, 37, 39, 41, 47, 48, 53, 56]), ('woodchuck', [54]), ('has', [14, 16, 51]), ('jam', [20]), ('who', [16]), ('farmer', [11]), ('black', [5])]

Answer 3

def findit(mylist, mydict):
    gen = ((k, mydict[k]) for k in mylist if k in mydict)
    return [k for k, v in sorted(gen, key=lambda p: len(p[1]))]


>>> findit(['d', 'e', 'f'], {'d': [0, 1], 'e': [1, 2, 3], 'f': [4]})
['f', 'd', 'e']