python - Combining two dictionaries by merging its values in a sorted list -

i'm having trouble writing function union_collections consumes 2 dictionaries (d1 , d2) representing 2 book collections. function produces new dictionary containing books present in d1 or d2, while maintaining following rules:

• the produced dictionary should not contain duplicate book titles.
• each list of book titles within produced dictionary should sorted using built- in sort() method.
• cannot use built-in function fromkeys() in solution

these sample collections testing:

collection1 = \          {'f':['flatland', 'five minute mysteries', 'films of 1990s', 'fight club'],          't':['the art of computer programming', 'the catcher in rye'],          'p':['paradise lost', 'professional blackjack', 'paradise regained'],          'c':['calculus in real world', 'calculus revisited', 'cooking one'],          'd':['dancing cats', 'disaster @ midnight']}  collection2 = \         {'f':['flatland', 'films of 1990s'],          'a':['a brief history of time', 'a tale of 2 cities'],          'd':['dealing stress', 'dancing cats'],          't':['the art of computer programming', 'the catcher in rye'],          'p':['power , wealth', 'poker essentials', 'post secret'],          'c':['cat couples', 'calculus', 'calculus revisited',               'cooking one', 'calculus in real world', 'cooking made easy']}` 

an example: unique_collections(collection1, collection2) should produce:

{'f' : ['fight club' , 'films of 1990s', 'five minute mysteries', 'flatland'],  't' : ['the art of computer programming', 'the catcher in rye'],  'p' : ['paradise lost' , 'paradise regained', 'poker essentials', 'post secret' , 'power , wealth', 'professional blackjack'],  'c' : ['calculus' , 'calculus in real world' , 'calculus revisited' , 'cat couples', 'cooking one', 'cooking made easy'],  'd' : ['dancing cats' , 'dealing stress' , 'disaster @ midnight'],  'a' : ['a brief history of time' , 'a tale of 2 cities']}` 

so far i've written:

def union_collections(d1, d2):     union = {}      key in d1 or d2:         if key in d1 , key not in d2: # if key in d1             union[key] = d1[val]          if key in d2 , key not in d1: #             union[key] = d2[val]          if key in d1 , key in d2:             union = dict(list(d1.items()) + list(d2.items()))      return sorted(union.values()) 

this function isn't working , have no idea how fix adhere following requirements.

cannot import modules.

some issues in code -

1. when - union = dict(list(d1.items()) + list(d2.items())) - not think valid, cannot add dictionaries. , not need well, not make sense requirement.

2. sorted() returns sorted list, not in place sorting. , return sorted list of values , not dictionary, need use list.sort() or sorted() function when creating dictionaries directly.

3. for key in d1 or d2 - iterates on keys in d1, need use set(d1.keys()).union(d2.keys()) .

4. d1[val] (d2[val]) - not correct, there not val variable, use d1[key] instead.

for case key found in both dictionary, can add lists of both dictionaries , convert set , list , sort , assign union dictionary. example -

def union_collections(d1, d2):     union = {}      key in set(d1.keys()).union(d2.keys()):         if key in d1 , key not in d2: # if key in d1             union[key] = d1[key]          if key in d2 , key not in d1:              union[key] = d2[key]          if key in d1 , key in d2:             union[key] = sorted(list(set(d1[key] + d2[key])))      return union 

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as asked in comments -

for when key in both dictionaries, there way without use of sets?

the way without sets -

def union_collections(d1, d2):     union = {}      key in set(d1.keys()).union(d2.keys()):         if key in d1 , key not in d2: # if key in d1             union[key] = d1[key]          if key in d2 , key not in d1:              union[key] = d2[key]          if key in d1 , key in d2:             y = []             union[key] = y             x in d1[key]:                 y.append(x)             x in d2[key]:                 if x not in y:                     y.append(x)             y.sort()      return union