Glen Rubin <rubing...@gmail.com> writes:

> Can I do the following without using loops??
>
> I have list, e.g.
>
> '(4 6 66 33 26 6 83 5)
>
> I want to partition it so that I get a subset of lists that build up
> to the original:
>
> ( (4) (4 6) (4 6 66) (4 6 66 33) ....)

(reductions conj [] [4 6 66 33 26 6 83 5])
=> ([] [4] [4 6] [4 6 66] [4 6 66 33] [4 6 66 33 26] ...)

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