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}?
views:
681answers:
3d4 = 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.
slowest: concatenate the
itemsand calldicton 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
dictconstructor 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
middling: a loop of
updatecalls 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).