On Tue, 27 Sep 2016 05:08 pm, Nagy Lc3a1szlc3b3 Zsolt wrote:
> #2. Much more effective version, but requires too many lines:
>
> d= {}
> d.update(d1)
> d.update(d2)
> d.update(d3)
new_dict = {}
for d in (d1, d2, d3):
new_dict.update(d)
Or, if you know for sure there's only three dicts:
new_dict = d1.copy()
new_dict.update(d2, **d3)
Or if you prefer:
new_dict = dict(d1) # same as copying
new_dict.update(d2, **d3)
--
Steve
“Cheer up,” they said, “things could be worse.” So I cheered up, and sure
enough, things got worse.
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