I can’t find it, although I’m sure it’s been answered before: how do I (efficiently) access the 'n’th element of a generator?
a = (i for i in 1:10^6)
a[3]
first(Iterators.drop(a, n-1))
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What do you mean by efficiently? I think it is not possible in O(1), the best would be O(n) as the Kristoffer answer.
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I only meant not like this:
getindex(collect(a), 3)