If you're using all of racket, here's a way to do it with heavy usage of
higher order functions
#lang racket
(define (cumulative-sum xs)
(map (curry foldl + 0)
(build-list (length xs)
(compose (curry take xs)
add1))))
On Thu, Jan 22, 2015 at 6:34 AM, Matthias Felleisen <matth...@ccs.neu.edu>
wrote:
>
> #lang racket
>
> ;; [Listof Number] -> [Listof Number]
> ;; given (a b c ...) produce (a (+ a b) (+ a b c) ...)
> (define (cumulative-sum l #;"final private:" [so-far 0])
> (cond
> [(empty? l) '()]
> [else (define next (+ so-far (first l)))
> (cons next (cumulative-sum (rest l) next))]))
>
> (module+ main
> (cumulative-sum '(1 2 3)))
>
> ;; The comments are HtDP-style, but the optional entry parameter is not
> available in the teaching languages.
>
>
> On Jan 22, 2015, at 8:53 AM, Alexandr M <rus...@gmail.com> wrote:
>
> > Hello,
> >
> > I am trying to replicate functionality of this code in Python:
> >
> > https://gist.github.com/m-blog/22b7c5d31b3839ffba50#file-cumsum-py
> >
> > in Racket without usage of any libraries.
> >
> > What would the most optimal / concise way to do it in Racket?
> >
> > Thank you.
> >
> > --
> > Best regards,
> > Alex
> > ____________________
> > Racket Users list:
> > http://lists.racket-lang.org/users
>
>
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