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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).