Paul Rubin wrote:
> Steven Bethard <[EMAIL PROTECTED]> writes:
>> Cameron Walsh wrote:
>>> Which brings me to the question, would this solution:
>>> B = set(B)
>>> A = B + list(x for x in A if x not in B)
>>> be faster than this solution:
>>> B = set(B)
>>> A.sort(key=B.__contains__, reverse=True)
>> [timings deleted]
>> That said, I'd probably still use the first solution -- it's more
>> immediately obvious why that one works.
>
> Wait a minute, the first example looks wrong, B has gotten replaced by
> a set and then it's added to a list.
Yep. If you look back at the lines I actually timed, they were::
B_set = set(B)
A = B + list(x for x in A if x not in B_set)
and::
B_set = set(B)
A.sort(key=B_set.__contains__, reverse=True)
As you noted, you'll get an error if you try to concatenate B as a set
to the list.
Steve
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