On 30May2017 12:06, Peter Otten <__pete...@web.de> wrote:
Cameron Simpson wrote:
As written it should be a bit slower: to construct a set each member get
tested for presence. The cost is in making the set, not in searching it.
No, CPython is a bit smarter than that:
dis.dis('if m in {"1", "January"}: pass')
1 0 LOAD_NAME 0 (m)
2 LOAD_CONST 3 (frozenset({'1', 'January'}))
Ah, I was wonderig about things like that during today...
[ More cool info snipped... ]
Thanks,
Cameron Simpson <c...@zip.com.au>
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