• bearophile (10/13) Feb 22 2011 Do you know a much nicer way to take just the n-th item of a lazy range?
• Jonathan M Davis (11/29) Feb 22 2011 Assuming that it's a forward range rather than an input range:
• bearophile (13/18) Feb 23 2011 This program gives:
• Jonathan M Davis (6/28) Feb 23 2011 Okay, so you need to do popFrontN(s, n - 1). I'm too used to arrays whic...
• bearophile (6/8) Feb 23 2011 Me too.

bearophile <bearophileHUGS lycos.com> writes:

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

 next(islice(r, 5, 6))



25

Bye,
bearophile

Jonathan M Davis <jmdavisProg gmx.com> writes:

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

 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

bearophile <bearophileHUGS lycos.com> writes:

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

Jonathan M Davis <jmdavisProg gmx.com> writes:

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

bearophile <bearophileHUGS lycos.com> writes:

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