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}
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).