我有一个功能, binary_range_search, ,这就是这样称为:

my $brs_iterator = binary_range_search(
    target => $range,                   # eg. [1, 200]
    search => $ranges                   # eg. [ {start => 1,   end => 1000},
);                                      #       {start => 500, end => 1500} ]

brs_iterator->() 将在所有$范围重叠的所有 @$范围内迭代。

我想扩展 binary_range_search 能够以多个范围为目标,例如:

target => $target_ranges # eg. [ [1, 200], [50, 300], ... ]
search => $search_ranges # as above

因此,当$ range-> [0]的搜索耗尽时,它应该继续前进到$ ranga-> [1],依此类推。这是所讨论的功能,其原始形式:

sub binary_range_search {
    my %options = @_;
    my $range    = $options{target}  || return;
    my $ranges   = $options{search}  || return;

    my ( $low, $high ) = ( 0, @{$ranges} - 1 );

    while ( $low <= $high ) {

        my $try = int( ( $low + $high ) / 2 );

        $low  = $try + 1, next if $ranges->[$try]{end}   < $range->[0];
        $high = $try - 1, next if $ranges->[$try]{start} > $range->[1];

        my ( $down, $up ) = ($try) x 2;

        my %seen = ();

        my $brs_iterator = sub {

            if (    $ranges->[ $up + 1 ]{end}       >= $range->[0]
                    and $ranges->[ $up + 1 ]{start} <= $range->[1]
                    and !exists $seen{ $up + 1 } )
            {
                $seen{ $up + 1 } = undef;
                return $ranges->[ ++$up ];
            }
            elsif ( $ranges->[ $down - 1 ]{end}       >= $range->[0]
                    and $ranges->[ $down + 1 ]{start} <= $range->[1]
                    and !exists $seen{ $down - 1 }
                    and $down > 0 )
            {
                $seen{ $down - 1 } = undef;
                return $ranges->[ --$down ];
            }
            elsif ( !exists $seen{$try} ) {
                $seen{$try} = undef;
              return $ranges->[$try];
            }
            else {
                return;
            }

        };
        return $brs_iterator;
    }
    return sub { };
}

这是一种标准的二进制搜索策略,直到找到重叠范围为止。然后,它在右侧移动,耗尽它,在左侧移动,耗尽它,最后放弃。理想情况下,应该应该 shift 我想的下一个目标范围,并重做搜索(也许是通过递归?)。我的问题是,我不确定如何通过迭代器结构使该工作。

有帮助吗?

解决方案

我只是将您的迭代器生成包装为for循环,并构建了一系列迭代函数。

根据上下文,我要么返回主迭代器或迭代函数列表。我不确定你想要什么。

use strict;
use warnings;


my $t = [ [1,200], [400,900] ];
my @r = (
    { start =>   1, end =>  100 },
    { start =>   2, end =>  500 },
    { start => 204, end =>  500 },
    { start => 208, end =>  500 },
    { start => 215, end => 1000 },
    { start => 150, end => 1000 },
    { start => 500, end => 1100 },
);

# Get a master iterator that will process each iterator in turn.
my $brs_iterator = binary_range_search(
    targets => $t,  
    search => \@r,
);

# Get an array of iterators
my @brs_iterator = binary_range_search(
    targets => $t,  
    search => \@r,
);



sub binary_range_search {
    my %options = @_;
    my $targets = $options{targets}  || return;
    my $ranges  = $options{search}  || return;


    my @iterators;

    TARGET:
    for my $target ( @$targets ) {

        my ( $low, $high ) = ( 0, $#{$ranges} );

        RANGE_CHECK:
        while ( $low <= $high ) {

            my $try = int( ( $low + $high ) / 2 );

            # Remove non-overlapping ranges
            $low  = $try + 1, next RANGE_CHECK 
                if $ranges->[$try]{end}   < $target->[0];

            $high = $try - 1, next RANGE_CHECK 
                if $ranges->[$try]{start} > $target->[1];

            my ( $down, $up ) = ($try) x 2;

            my %seen = ();

            my $brs_iterator = sub {

                if (    exists $ranges->[$up + 1]
                        and $ranges->[ $up + 1 ]{end}   >= $target->[0]
                        and $ranges->[ $up + 1 ]{start} <= $target->[1]
                        and !exists $seen{ $up + 1 } )
                {
                    $seen{ $up + 1 } = undef;
                    return $ranges->[ ++$up ];
                }
                elsif ( $ranges->[ $down - 1 ]{end}       >= $target->[0]
                        and $ranges->[ $down + 1 ]{start} <= $target->[1]
                        and !exists $seen{ $down - 1 }
                        and $down > 0 )
                {
                    $seen{ $down - 1 } = undef;
                    return $ranges->[ --$down ];
                }
                elsif ( !exists $seen{$try} ) {
                    $seen{$try} = undef;
                  return $ranges->[$try];
                }
                else {
                    return;
                }

            };
            push @iterators, $brs_iterator;
            next TARGET;
        }

    }

    # In scalar context return master iterator that iterates over the list of range iterators.
    # In list context returns a list of range iterators.
    return wantarray 
         ? @iterators 
         : sub { 
             while( @iterators ) {
                 if( my $range = $iterators[0]() ) {
                     return $range;
                 }
                 shift @iterators;
             }
             return;
        }; 
}

其他提示

如果您想迭代与搜索范围重叠的所有值,则不需要二进制搜索。

首先是习惯前提:

use warnings;
use strict;

use Carp;

首先,检查我们是否有 targetsearch 参数和每个范围,起点不超过其终点。否则,我们拒绝继续进行。

sub binary_range_search {
  my %arg = @_;

  my @errors;
  my $target = $arg{target} || push @errors => "no target";
  my $search = $arg{search} || push @errors => "no search";

  for (@$target) {
    my($start,$end) = @$_;
    push @errors => "Target start ($start) is greater than end ($end)"
      if $start > $end;
  }

  for (@$search) {
    my($start,$end) = @{$_}{qw/ start end /};
    push @errors => "Search start ($start) is greater than end ($end)"
      if $start > $end;
  }

  croak "Invalid use of binary_range_search:\n",
        map "  - $_\n", @errors
    if @errors;

迭代器本身是维持以下状态的关闭:

  my $i;
  my($ta,$tb);
  my($sa,$sb);
  my $si = 0;

在哪里

  • $i 如果定义是当前重叠范围的下一个值
  • $ta$tb 是当前目标范围的起点和终点
  • $sa$sb 就像以上一样,但对于当前的搜索范围
  • $si 是索引 @$search 并定义当前搜索范围

我们将分配和返回迭代器 $it. 。声明和初始化是分开的,因此迭代器可以在必要时调用自身。

  my $it;
  $it = sub {

如果没有更多的目标范围,或者没有搜索范围,我们就完成了。

    return unless @$target && @$search;

什么时候 $i 已定义,这意味着我们通过增加发现了重叠和迭代 $i 直到它大于当前目标范围或当前搜索范围的终点。

    if (defined $i) {
      # iterating within a target range

      if ($i > $tb || $i > $sb) {
        ++$si;
        undef $i;
        return $it->();
      }
      else {
        return $i++;
      }
    }

否则,我们需要确定下一个目标范围是否重叠任何搜索范围。但是,如果 $i 是未定义的,我们已经考虑了所有搜索范围,我们丢弃了当前的目标范围并重新开始。

    else {
      # does the next target range overlap?

      if ($si >= @$search) {
        shift @$target;
        $si = 0;
        return $it->();
      }

在这里,我们拔出了当前目标范围的起点和终点(始终在 @$target)和当前搜索范围(由 $si).

      ($ta,$tb) = @{ $target->[0] };
      ($sa,$sb) = @{ $search->[$si] }{qw/ start end /};

现在对重叠的测试很简单。对于截然不同的搜索范围,我们忽略并继续前进。否则,我们发现重叠中的最左点,并从那里迭代。

      if ($sb < $ta || $sa > $tb) {
        # disjoint
        ++$si;
        undef $i;
        return $it->();
      }
      elsif ($sa >= $ta) {
        $i = $sa;
        return $i++;
      }
      elsif ($ta >= $sa) {
        $i = $ta;
        return $i++;
      }
    }
  };

最后,我们返回迭代器:

  $it;
}

与您的问题类似的示例

my $it = binary_range_search(
  target => [ [1, 200], [50, 300] ],
  search => [ { start =>   1, end => 1000 },
              { start => 500, end => 1500 },
              { start =>  40, end =>   60 },
              { start => 250, end =>  260 } ],
);

while (defined(my $value = $it->())) {
  print "got $value\n";
}

其内部点的输出是

got 1
[...]
got 200
got 40
[...]
got 60
got 50
[...]
got 300
got 50
[...]
got 60
got 250
[...]
got 260

将其分为两个函数,即在范围上循环并调用实现常规二进制切碎的内部函数的外部函数。

警告:一个非常C ++的偏见答案:

您要做的是定义一种新型的迭代器,即一对通常的迭代器和一个segmemt iTerrator(如果您没有段迭代器,则是一对const指针 / ref to segments,to segments the sevents,to和指向正确段的索引)。您必须定义随机访问迭代器的所有概念(差异,添加整数等)。请记住,至少在C ++ Lingo中,这不是一个真正的随机迭代器,因为添加整数并不是真正恒定的时间。这就是生活。

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