how to concatenate two dictionaries to create a new one in Python? [duplicate]
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Slowest and doesn't work in Python3: concatenate the
items
and calldict
on the resulting list:$ python -mtimeit -s'd1={1:2,3:4}; d2={5:6,7:9}; d3={10:8,13:22}' \ 'd4 = dict(d1.items() + d2.items() + d3.items())' 100000 loops, best of 3: 4.93 usec per loop
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Fastest: exploit the
dict
constructor to the hilt, then oneupdate
:$ python -mtimeit -s'd1={1:2,3:4}; d2={5:6,7:9}; d3={10:8,13:22}' \ 'd4 = dict(d1, **d2); d4.update(d3)' 1000000 loops, best of 3: 1.88 usec per loop
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Middling: a loop of
update
calls on an initially-empty dict:$ python -mtimeit -s'd1={1:2,3:4}; d2={5:6,7:9}; d3={10:8,13:22}' \ 'd4 = {}' 'for d in (d1, d2, d3): d4.update(d)' 100000 loops, best of 3: 2.67 usec per loop
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Or, equivalently, one copy-ctor and two updates:
$ python -mtimeit -s'd1={1:2,3:4}; d2={5:6,7:9}; d3={10:8,13:22}' \ 'd4 = dict(d1)' 'for d in (d2, d3): d4.update(d)' 100000 loops, best of 3: 2.65 usec per loop
I recommend approach (2), and I particularly recommend avoiding (1) (which also takes up O(N) extra auxiliary memory for the concatenated list of items temporary data structure).
d4 = dict(d1.items() + d2.items() + d3.items())
alternatively (and supposedly faster):
d4 = dict(d1)
d4.update(d2)
d4.update(d3)
Previous SO question that both of these answers came from is here.
You can use the update()
method to build a new dictionary containing all the items:
dall = {}
dall.update(d1)
dall.update(d2)
dall.update(d3)
Or, in a loop:
dall = {}
for d in [d1, d2, d3]:
dall.update(d)
Here's a one-liner (imports
don't count :) that can easily be generalized to concatenate N dictionaries:
Python 3
from itertools import chain
dict(chain.from_iterable(d.items() for d in (d1, d2, d3)))
and:
from itertools import chain
def dict_union(*args):
return dict(chain.from_iterable(d.items() for d in args))
Python 2.6 & 2.7
from itertools import chain
dict(chain.from_iterable(d.iteritems() for d in (d1, d2, d3))
Output:
>>> from itertools import chain
>>> d1={1:2,3:4}
>>> d2={5:6,7:9}
>>> d3={10:8,13:22}
>>> dict(chain.from_iterable(d.iteritems() for d in (d1, d2, d3)))
{1: 2, 3: 4, 5: 6, 7: 9, 10: 8, 13: 22}
Generalized to concatenate N dicts:
from itertools import chain
def dict_union(*args):
return dict(chain.from_iterable(d.iteritems() for d in args))
I'm a little late to this party, I know, but I hope this helps someone.
Use the dict constructor
d1={1:2,3:4}
d2={5:6,7:9}
d3={10:8,13:22}
d4 = reduce(lambda x,y: dict(x, **y), (d1, d2, d3))
As a function
from functools import partial
dict_merge = partial(reduce, lambda a,b: dict(a, **b))
The overhead of creating intermediate dictionaries can be eliminated by using thedict.update()
method:
from functools import reduce
def update(d, other): d.update(other); return d
d4 = reduce(update, (d1, d2, d3), {})