how to concatenate two dictionaries to create a new one in Python? [duplicate]

Answer 1

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.

Answer 2

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)

Answer 3

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.

Answer 4

1. Slowest and doesn't work in Python3: concatenate the items and call dict 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
   

2. Fastest: exploit the dict constructor to the hilt, then one update:

   $ 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
   

3. 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
   

4. 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).

Answer 5

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), {})