合并 Python 字典

Question

I am trying to merge the following python dictionaries as follow:

dict1= {'paul':100, 'john':80, 'ted':34, 'herve':10}
dict2 = {'paul':'a', 'john':'b', 'ted':'c', 'peter':'d'}

output = {'paul':[100,'a'],
          'john':[80, 'b'],
          'ted':[34,'c'],
          'peter':[None, 'd'],
          'herve':[10, None]}

Is there an efficient way to do this?

Answer 1

output = {k: [dict1[k], dict2.get(k)] for k in dict1}
output.update({k: [None, dict2[k]] for k in dict2 if k not in dict1})

Answer 2

This will work:

{k: [dict1.get(k), dict2.get(k)] for k in set(dict1.keys() + dict2.keys())}

Output:

{'john': [80, 'b'], 'paul': [100, 'a'], 'peter': [None, 'd'], 'ted': [34, 'c'], 'herve': [10, None]}

Answer 3

In Python2.7 or Python3.1 you can easily generalise to work with any number of dictionaries using a combination of list, set and dict comprehensions!

>>> dict1 = {'paul':100, 'john':80, 'ted':34, 'herve':10}
>>> dict2 = {'paul':'a', 'john':'b', 'ted':'c', 'peter':'d'}
>>> dicts = dict1,dict2
>>> {k:[d.get(k) for d in dicts] for k in {k for d in dicts for k in d}}
{'john': [80, 'b'], 'paul': [100, 'a'], 'peter': [None, 'd'], 'ted': [34, 'c'], 'herve': [10, None]}

Python2.6 doesn't have set comprehensions or dict comprehensions

>>> dict1 = {'paul':100, 'john':80, 'ted':34, 'herve':10}
>>> dict2 = {'paul':'a', 'john':'b', 'ted':'c', 'peter':'d'}
>>> dicts = dict1,dict2
>>> dict((k,[d.get(k) for d in dicts]) for k in set(k for d in dicts for k in d))
{'john': [80, 'b'], 'paul': [100, 'a'], 'peter': [None, 'd'], 'ted': [34, 'c'], 'herve': [10, None]}

Answer 4

output = {k:[dict1.get(k),dict2.get(k)] for k in dict1.keys() | dict2.keys()}

 output = dict((k,[dict1.get(k),dict2.get(k)]) for k in set(dict1.keys() + dict2.keys()))

Answer 5

Using chain.from_iterable (from itertools) you can avoid the list/dict/set comprehension with:

dict(chain.from_iterable(map(lambda d: d.items(), list_of_dicts])))

It can be more or less convenient and readable than double comprehension, depending on your personal preference.