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Simple solution using itertools that preserves order (latter dicts have precedence)

import itertools as it
merge = lambda *args: dict(it.chain.from_iterable(it.imap(dict.iteritems, args)))

And it's usage:

>>> x = {'a':1, 'b': 2}
>>> y = {'b':10, 'c': 11}
>>> merge(x, y)
{'a': 1, 'b': 10, 'c': 11}

>>> z = {'c': 3, 'd': 4}
>>> merge(x, y, z)
{'a': 1, 'b': 10, 'c': 3, 'd': 4}

Simple solution using itertools that preserves order (latter dicts have precedence)

# py2
from itertools import chain, imap
merge = lambda *args: dict(chain.from_iterable(imap(dict.iteritems, args)))

# py3
from itertools import chain
merge = lambda *args: dict(chain.from_iterable(map(dict.items, args)))

And it's usage:

>>> x = {'a':1, 'b': 2}
>>> y = {'b':10, 'c': 11}
>>> merge(x, y)
{'a': 1, 'b': 10, 'c': 11}

>>> z = {'c': 3, 'd': 4}
>>> merge(x, y, z)
{'a': 1, 'b': 10, 'c': 3, 'd': 4}

Simple solution using itertools that preserves order (latter dicts have precedence)

import itertools as it
merge = lambda *args: dict(it.chain.from_iterable(it.imap(dict.iteritems, args)))

And it's usage:

x = {'a':1, 'b': 2}
y = {'b':10, 'c': 11}
merge(x, y)
 
>>> {'a': 1, 'b': 10, 'c': 11}

z = {'c': 3, 'd': 4}
merge(x, y, z)
 
>>> {'a': 1, 'b': 10, 'c': 3, 'd': 4}

Simple solution using itertools that preserves order (latter dicts have precedence)

import itertools as it
merge = lambda *args: dict(it.chain.from_iterable(it.imap(dict.iteritems, args)))

And it's usage:

>>> x = {'a':1, 'b': 2}
>>> y = {'b':10, 'c': 11}
>>> merge(x, y)
{'a': 1, 'b': 10, 'c': 11}

>>> z = {'c': 3, 'd': 4}
>>> merge(x, y, z)
{'a': 1, 'b': 10, 'c': 3, 'd': 4}

Simple solution using itertools that preserves order (latter dicts have precedence)

import itertools as it
 
def merge(*args):
    return dict(it.chain.from_iterable(it.imap(dict.iteritems, args)))

And it's usage:

x = {'a':1, 'b': 2}
y = {'b':10, 'c': 11}
merge(x, y)

>>> {'a': 1, 'b': 10, 'c': 11}

z = {'c': 3, 'd': 4}
merge(x, y, z)

>>> {'a': 1, 'b': 10, 'c': 3, 'd': 4}

Simple solution using itertools that preserves order (latter dicts have precedence)

import itertools as it
merge = lambda *args: dict(it.chain.from_iterable(it.imap(dict.iteritems, args)))

And it's usage:

x = {'a':1, 'b': 2}
y = {'b':10, 'c': 11}
merge(x, y)

>>> {'a': 1, 'b': 10, 'c': 11}

z = {'c': 3, 'd': 4}
merge(x, y, z)

>>> {'a': 1, 'b': 10, 'c': 3, 'd': 4}