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

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).

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)