CODEWITHSUNDEEP
streamschemeracketinfinite-sequence
Mirzhan Irkegulov
Mirzhan Irkegulov

Reputation: 18055

Infinite ascending sequence in Racket

Is there an analog of Python's itertools.count in Racket? I want to create an infinite stream of evenly spaced numbers. in-naturals is similar to what i want, but does not provide step. I'd want not to reinvent the wheel, but if there's no equivalent function, how to write one? (i presume, generators should be used)

Upvotes: 4

Views: 796

Answers (2)

Michael F
Michael F

Reputation: 31

Making the function yourself can be done in a single line:

(define (stream-from n s) (stream-cons n (stream-from (+ n s) s)))

To test it, you here is an example that prints 100000 numbers:

#lang racket
(require racket/stream)

(define (stream-from n s) (stream-cons n (stream-from (+ n s) s)))

(define (stream-while s p)
  (let ([fst (stream-first s)])
  (if (p fst) (stream-cons fst (stream-while (stream-rest s) p)) empty-stream)))

(define test (stream-while (stream-from 0 1) (λ (x) (< x 100000))))

(stream-for-each println test)

Upvotes: 2

&#211;scar L&#243;pez
&#211;scar L&#243;pez

Reputation: 236004

You can get the same functionality of Python's count using in-range with an infinite end value:

(define (count start step)
  (in-range start +inf.0 step))

For example:

(define s (count 2.5 0.5))

(stream-ref s 0)
=> 2.5
(stream-ref s 1)
=> 3.0
(stream-ref s 2)
=> 3.5
(stream-ref s 3)
=> 4.0

Upvotes: 6

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