Python에서 두 사전을 병합하는 방법에 대한 튜토리얼

 x = {'a': 1, 'b': 2}
 y = {'b': 3, 'c': 4}

>>> z
{'a': 1, 'b': 3, 'c': 4}

z = {**x, **y}

z = x.copy()
z.update(y)

def merge_two_dicts(x, y):
 """Given two dicts, merge them into a new dict as a shallow copy."""
 z = x.copy()
 z.update(y)
 return z

 z = merge_two_dicts(x, y)

def merge_dicts(*dict_args):
 """
 Given any number of dicts, shallow copy and merge into a new dict,
 precedence goes to key value pairs in latter dicts.
 """
 result = {}
 for dictionary in dict_args:
 result.update(dictionary)
 return result

z = merge_dicts(a, b, c, d, e, f, g)

 z = dict(x.items() + y.items())

>>> c = dict(a.items() + b.items())
Traceback (most recent call last):
 File "<stdin>", line 1, in <module>
TypeError: unsupported operand type(s) for +: &#39;dict_items&#39; and 
&#39;dict_items&#39;

>>> x = {'a': []}
>>> y = {'b': []}
>>> dict(x.items() | y.items())
Traceback (most recent call last):
 File "<stdin>", line 1, in <module>
TypeError: unhashable type: &#39;list&#39;

>>> x = {'a': 2}
>>> y = {'a': 1}
>>> dict(x.items() | y.items())
{'a': 2}

z = dict(x, **y)

>>> c = dict(a, **b)
Traceback (most recent call last):
 File "<stdin>", line 1, in <module>
TypeError: keyword arguments must be strings

{k: v for d in dicts for k, v in d.items()}

 dict((k, v) for d in dicts for k, v in d.items())

import itertools
z = dict(itertools.chain(x.iteritems(), y.iteritems()))

>>> min(timeit.repeat(lambda: merge_two_dicts(x, y)))
0.5726828575134277
>>> min(timeit.repeat(lambda: {k: v for d in (x, y) for k, v in d.items()} ))
1.163769006729126
>>> min(timeit.repeat(lambda: dict(itertools.chain(x.iteritems(),y.iteritems()))))
1.1614501476287842
>>> min(timeit.repeat(lambda: dict((k, v) for d in (x, y) for k, v in d.items())))
2.2345519065856934

>>> min(timeit.repeat(lambda: {**x, **y}))
0.4094954460160807
>>> min(timeit.repeat(lambda: merge_two_dicts(x, y)))
0.7881555100320838
>>> min(timeit.repeat(lambda: {k: v for d in (x, y) for k, v in d.items()} ))
1.4525277839857154
>>> min(timeit.repeat(lambda: dict(itertools.chain(x.items(), y.items()))))
2.3143140770262107
>>> min(timeit.repeat(lambda: dict((k, v) for d in (x, y) for k, v in d.items())))
3.2069112799945287