扫描有序节点标识符的数据
扫描有序节点标识符的数据
提问于 2012-12-15 21:43:00
考虑以下数据：
12 45 64
12 45 76
12 37 39 87
12 67 90
12 39 60 在这个例子中，只有十个不同的数字。如果我有大量的数据，如何在Perl中计算它？
我有一个连接从12到45，从45到64，但不是从12到64。
我们没有从45到12的路线，所以12的邻居是4 (45，37，67，39)，39是2 (87，60)
如何计算数据中所有值的邻域？
编辑
另一个要求是，我们要忽略任何指向自身的值。例如，假设我们有以下文件：
1 4 3
1 2 2
2 6 7在本例中，1的邻域必须是(4，2)
4的邻里必须是3。
2的邻里必须是6(而不是2)
我的意思是，我们必须删除匹配和重复。
回答 4
Stack Overflow用户
回答已采纳
发布于 2012-12-15 22:21:06
使用Graph::Directed模块：
use Graph::Directed qw( );
my $graph = Graph::Directed->new();
while (<>) {
my @points = split;
$graph->add_edge(@points[$_-1, $_])
for 1..$#points;
}
for my $vertex ($graph->vertices()) {
my @successors = grep $_ != $vertex, $graph->successors($vertex);
print("$vertex has ".@successors." successors: @successors\n");
}输入：
1 4 3
1 2 2
2 6 7输出：
6 has 1 successors: 7
4 has 1 successors: 3
2 has 1 successors: 6
1 has 2 successors: 4 2
3 has 0 successors:
7 has 0 successors:Stack Overflow用户
发布于 2012-12-15 22:01:02
my $data = <<'END_DATA';
12 45 64
12 45 76
12 37 39 87
12 67 90
12 39 60
END_DATA
my @lines = split/\n+/, $data;
# map number the list of numbers following it in the sequence
my %neighborhoods = ();
for my $line ( @lines ) {
my @nums = split m/\s+/,$line;
for my $i ( 0 .. $#nums - 1 ) {
$neighborhoods{$nums[$i]}{$nums[$i+1]} = 1;
}
}
foreach my $num ( sort keys %neighborhoods ) {
print "num [$num] neighboorhood (" .
( join "-", keys %{$neighborhoods{$num}} ) .
") count [" . ( scalar keys %{$neighborhoods{$num}} ) .
"]\n";
}输出：
num [12] neighboorhood (67-39-37-45) count [4]
num [37] neighboorhood (39) count [1]
num [39] neighboorhood (60-87) count [2]
num [45] neighboorhood (64-76) count [2]
num [67] neighboorhood (90) count [1]Stack Overflow用户
发布于 2012-12-15 22:08:54
如果我正确地理解了您，您想要计算图上每个节点附近的节点数。我觉得这能做你想做的事。
我已经修改了我的代码，因为您已经解释了从节点到其本身的向量应该被忽略。
use v5.10;
use warnings;
my %routes;
while (<DATA>) {
my @nodes = /\d+/g;
$routes{$_} //= {} for @nodes;
while (@nodes >= 2) {
my ($from, $to) = @nodes;
$routes{$from}{$to}++ unless $from == $to;
shift @nodes;
}
}
for my $key (sort { $a <=> $b } keys %routes) {
my $val = $routes{$key};
printf "%d - neighbourhood size %d",
$key,
scalar keys %$val;
printf " (%s)", join ', ', keys %$val if %$val;
print "\n";
}
__DATA__
12 45 64
12 45 76
12 37 39 87
12 67 90
12 39 60
1 4 3
1 2 2
2 6 7输出
1 - neighbourhood size 2 (4, 2)
2 - neighbourhood size 1 (6)
3 - neighbourhood size 0
4 - neighbourhood size 1 (3)
6 - neighbourhood size 1 (7)
7 - neighbourhood size 0
12 - neighbourhood size 4 (67, 39, 37, 45)
37 - neighbourhood size 1 (39)
39 - neighbourhood size 2 (60, 87)
45 - neighbourhood size 2 (64, 76)
60 - neighbourhood size 0
64 - neighbourhood size 0
67 - neighbourhood size 1 (90)
76 - neighbourhood size 0
87 - neighbourhood size 0
90 - neighbourhood size 0页面原文内容由Stack Overflow提供。腾讯云小微IT领域专用引擎提供翻译支持
原文链接：
https://stackoverflow.com/questions/13896486
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