Do you know a much nicer way to take just the n-th item of a lazy range?


import std.stdio, std.array, std.range;
void main() {
    auto fib = recurrence!("a[n-1] + a[n-2]")(1, 1);
    writeln(array(take(fib, 10)).back);
}


In Python I use next(isslice(x, n, n+1)):

>>> from itertools import islice
>>> r = (x*x for x in xrange(10)) # lazy
>>> next(islice(r, 5, 6))
25

Bye,
bearophile

On Tuesday 22 February 2011 18:07:46 bearophile wrote:
> Do you know a much nicer way to take just the n-th item of a lazy range?
> 
> 
> import std.stdio, std.array, std.range;
> void main() {
>     auto fib = recurrence!("a[n-1] + a[n-2]")(1, 1);
>     writeln(array(take(fib, 10)).back);
> }
> 
> In Python I use next(isslice(x, n, n+1)):
> >>> from itertools import islice
> >>> r = (x*x for x in xrange(10)) # lazy
> >>> next(islice(r, 5, 6))
> 
> 25
> 
> Bye,
> bearophile

Assuming that it's a forward range rather than an input range:

auto s = range.save;
s.popFrontN(n - 1);
writeln(s.front);

The problem is that you have to process a lazy range before you can get at any of its elements, and once you've processed an element, it's no longer in the range. So, you pretty much have to save the range and operate on a copy of it. At that point, you can remove the elements prior to the one you care about and then take the one you care about from the front of the range.

- Jonathan M Davis

Jonathan M Davis:

> Assuming that it's a forward range rather than an input range:
> 
> auto s = range.save;
> s.popFrontN(n - 1);
> writeln(s.front);

This program gives:
test.d(5): Error: no property 'popFrontN' for type 'Recurrence!(fun,int,2u)'

import std.stdio, std.array, std.range;
void main() {
    auto fib = recurrence!("a[n-1] + a[n-2]")(1, 1);
    auto s = fib.save;
    s.popFrontN(10 - 1);
    writeln(s.front);
}

Thank you,
bye,
bearophile

On Wednesday 23 February 2011 04:34:28 bearophile wrote:
> Jonathan M Davis:
> > Assuming that it's a forward range rather than an input range:
> > 
> > auto s = range.save;
> > s.popFrontN(n - 1);
> > writeln(s.front);
> 
> This program gives:
> test.d(5): Error: no property 'popFrontN' for type
> 'Recurrence!(fun,int,2u)'
> 
> import std.stdio, std.array, std.range;
> void main() {
>     auto fib = recurrence!("a[n-1] + a[n-2]")(1, 1);
>     auto s = fib.save;
>     s.popFrontN(10 - 1);
>     writeln(s.front);
> }
> 
> Thank you,
> bye,
> bearophile

Okay, so you need to do popFrontN(s, n - 1). I'm too used to arrays which allow you to use that sort of syntax. That and not bothering with [] when slicing seem to be the two biggest problems that I'm seeing when dealing with ranges which aren't arrays. I'm just too used to arrays.

- Jonathan M Davis

Jonathan M Davis:

> Okay, so you need to do popFrontN(s, n - 1).

Right, silly me :-) I need to read error messages.


> I'm too used to arrays which allow you to use that sort of syntax.

Me too.

Thank you,
bye,
bearophile