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Editorial Team
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Asked: 2026-05-26T10:02:36+00:00

Let’s say I have a generator that I want to pull the 10th element

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Let’s say I have a generator that I want to pull the 10th element from but ignore the first 9. The generator is a function I’ve written that looks something like this:

def myGenerator(arg1, arg2):
    for i in arg1:
        myState = doSomeWork(i, arg2)
        yield expensiveOperation(myState)

Now I can use it and grab the 10th index out of it like this:

myGen = myGenerator(list1, list2)
tenthElement = next(itertools.islice(myGen,10,11))

I’m wondering if this runs expensiveOperation ten times, or just once?
(EDIT: it calls it 10 times, but this next part is what I’m interested in.)
Is there any way to optimize away the other 9 calls to expensiveOperation since they are discarded? (edited for clarity)

I can think of several other solutions that don’t involve using a generator function and would give exactly what I want, but the syntax isn’t as clean as just turning an iterative function into a generator by replacing return with yield.

EDIT:
I’m not necessarily trying to solve this specific problem so much as looking for a way to inexpensively “scroll” a generator. In the real case I’m currently working with, I don’t actually know which index I want when I call myGenerator for the first time. I may grab the 15th index, then the 27th, then the 82nd. I could probably figure a way to slice to the 15th on the first call, but then I need to scroll 12 more the next time around.

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1 Answer

  1. Editorial Team

    The generator is isolated from its consumer — it doesn’t know what is being throw away. So, yes, it does the expensive operation at every step.

    I would just move the expensive operation outside the generator:

    def myGenerator(arg1, arg2):
    for i in arg1:
        myState = doSomeWork(i, arg2)
        yield myState

myGen = myGenerator(list1, list2)
tenthElement = expensiveOperation(next(itertools.islice(myGen,10,11)))
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