How about something like
def index(self, x):
return next(i for i, a in enumerate(self) if a == x)
Including start and end:
def index(self, x, start=0, end=-1):
return next(i for i, a in tuple(enumerate(self))[start:end] if a == x)
------- Original Message -------
On Friday, March 11th, 2022 at 4:02 PM, Rob Cliffe <rob.cli...@btinternet.com>
wrote:
> On 11/03/2022 19:30, wfdc wrote:
>
>>> I could (I believe) write "count" as an (inefficient) 1-liner, but not
>>> "index". I suggest it's harder than you think. (Try it!)
>>
>> How much harder? Can you post your candidate?
>
> It was you that said it could be a 1-liner. The burden of proof is on you, if
> you still want to argue the point.
> Rob Cliffe
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