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August 05, 2014 Taking from infinite forward ranges | ||||
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Is there a way to take a bounded rage from a infinite forward range? Given the Fibonacci sequence: auto fib = recurrence!("a[n-1] + a[n-2]")(1, 1); I can take the first n elements: take(fib, 10); But say I want all positive elements below 50000 in value (there are eight such values [2, 8, 34, 144, 610, 2584, 10946, 46368]), how would I "take" them? Of course I could filter the range, leaving only positive values, and then take(fib, 8). But what if I didn't know there were 8, how could I take them from there filtered range? Currently I do this: foreach(e; fib) { if (e >= val) break; // so something with e } or while((e = fib.front()) < n) { // do something with e fib.popFront(); } Is there a better way? |
August 05, 2014 Re: Taking from infinite forward ranges | ||||
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Posted in reply to Andrew Edwards | On Tuesday, 5 August 2014 at 01:23:19 UTC, Andrew Edwards wrote:
> Is there a way to take a bounded rage from a infinite forward range?
>
> Given the Fibonacci sequence:
>
> auto fib = recurrence!("a[n-1] + a[n-2]")(1, 1);
>
> I can take the first n elements:
>
> take(fib, 10);
>
> But say I want all positive elements below 50000 in value (there are eight such values [2, 8, 34, 144, 610, 2584, 10946, 46368]), how would I "take" them? Of course I could filter the range, leaving only positive values, and then take(fib, 8). But what if I didn't know there were 8, how could I take them from there filtered range?
>
> Currently I do this:
>
> foreach(e; fib)
> {
> if (e >= val) break;
> // so something with e
> }
>
> or
>
> while((e = fib.front()) < n)
> {
> // do something with e
> fib.popFront();
> }
>
> Is there a better way?
I'd use std.algorithm.until:
void main()
{
import std.algorithm, std.range, std.stdio;
auto fib_until_50k = recurrence!("a[n-1] + a[n-2]")(1, 1)
.until!(a => a > 50_000);
writeln(fib_until_50k);
}
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August 05, 2014 Re: Taking from infinite forward ranges | ||||
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Posted in reply to Brad Anderson | On 8/5/14, 10:28 AM, Brad Anderson wrote:
> On Tuesday, 5 August 2014 at 01:23:19 UTC, Andrew Edwards wrote:
>> Is there a way to take a bounded rage from a infinite forward range?
>
> I'd use std.algorithm.until:
>
Precisely what I was looking for. Thanks.
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