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

Question

Say I have three dicts

d1={1:2,3:4} d2={5:6,7:9} d3={10:8,13:22} 

How do I create a new d4 that combines these three dictionaries? i.e.:

d4={1:2,3:4,5:6,7:9,10:8,13:22} 

Answer 1

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 2

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.