Perhaps this is a more elegant approach to the tactical problem:
simultaneously iterating over displaced versions of the original list. E.g.
_x '(1 1 2 3 5 7 200 201 202) ; shifted right
x '(1 2 3 5 7 200 201 202 203) ; original list
x_ '(2 3 5 7 200 201 202 203 203) ; shifted left
When (= (+ _x 1) x (- x_ 1)) we are mid-run. Otherwise we're at the start
or end of a run. Code below.
A further displacement is needed to deal with extended runs: x__ to skip
elements in long runs.
#lang racket
; Given a strictly ascending list of numbers, summarise runs
; (... b b+1 ... b+n c ...) -> (... a b - b+n c ...)
(define (runify xs)
(define MIN (first xs))
(define MAX (last xs))
(for/list ([_x (cons MIN xs)]
[x xs]
[x_ (append (rest xs) (list MAX))]
[x__ (append (drop xs 2) (list MAX MAX))]
#:unless (= (+ _x 2) x_ (- x__ 1)))
(if (= (+ _x 1) x (- x_ 1))
'-
x)))
; Replace runify-ed lists with lists of lists
; E.g. '(1 - 3 8 12 15 - 100) -> '((1 3) (8) (12) (15 100))
;
(define (chunkify xs [acc null])
(cond [(null? xs) acc]
[(equal? (first xs) '-)
(chunkify (drop xs 2)
(cons (list (caar acc) (second xs))
(rest acc)))]
[else
(chunkify (rest xs) (cons (list (first xs)) acc))]))
(define xs '(1 2 3 5 7 200 201 202 203))
(runify xs) ; '(1 - 3 5 7 200 - 203)
(reverse (chunkify (runify xs))) ; '((1 3) (5) (7) (200 203))
* * *
Racket question: How do I define a sequence generator (in-shifted-list xs
shift) to make runify work in true single pass fashion?
I would like to be able to write the following version of runify, without
generating auxiliary lists:
(define (runify-efficient xs)
(for/list ([_x (in-shifted-list xs -1)]
[x xs]
[x_ (in-shifted-list xs 1)]
[x__ (in-shifted-list xs 2)]
#:unless (= (+ _x 2) x_ (- x__ 1)))
(if (= (+ _x 1) x (- x_ 1))
'-
x)))
Dan
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