how to concatenate two dictionaries to create a new one in Python?
This question already has an answer here:
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
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
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
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)
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