使用Perl Regex解析语法树
使用Perl Regex解析语法树
提问于 2010-12-23 13:03:38
也许正则表达式不是解析它的最好方法，如果不是，请告诉我。无论如何，下面是语法树的一些示例：
(S (CC and))
(SBARTMP (IN once) (NP otherstuff))
(S (S (NP blah (VP blah)) (CC then) (NP blah (VP blah (PP blah))) )) 无论如何，我要做的就是拉出连接符( and，then，once等)和它对应的头部(CC，IN，CC)，我已经知道每个语法树的头部(CC，in，CC)，这样它就可以作为一个锚，我还需要检索它的父级(第一个是S，第二个是SBARTMP，第三个是S)，以及它的兄弟(如果有的话)和它的兄弟(在第一个none中，在第二个左侧兄弟中，第三个在左侧和右侧兄弟中)。任何高于父级的值都不包括在内
my $pos = "(\\\w|-)*";
my $sibling = qr{\s*(\\((?:(?>[^()]+)|(?1))*\\))\s*};
my $connective = "once";
my $re = qr{(\(\w*\s*$sibling*\s*\\(IN\s$connective\\)\s*$sibling*\s*\))}; 此代码适用于以下内容：
my $test1 = "(X (SBAR-TMP (IN once) (S sdf) (S sdf)))";
my $test2 = "(X (SBAR-TMP (IN once))";
my $test3 = "(X (SBAR-TMP (IN once) (X as))";
my $test4 = "(X (SBAR-TMP (X adsf) (IN once))"; 它将丢弃顶部的X，并保留其他所有内容，但是，一旦兄弟项中嵌入了内容，它就不匹配，因为正则表达式不会深入。
my $test = "(X (SBAR-TMP (IN once) (MORE stuff (MORE stuff))))"; 我不确定如何解释这一点。我对Perl的扩展模式比较陌生，刚刚开始学习它。为了阐明正则表达式的作用:它查找两个括号和大写字母/组合中的连接词，查找以两个括号结束的相同格式的完整父对象，然后查找所有括号都配对的任意数量的兄弟对象。
回答 3
Stack Overflow用户
回答已采纳
发布于 2010-12-27 00:31:59
要只获取离锚定连接符最近的“父级”，可以将其作为具有失败的递归父级来执行，也可以直接执行。(由于某些原因，我不能编辑我的其他帖子，一定是cookies被删除了)。
use strict;
use warnings;
my $connective = qr/ \((?:IN|CC)\s(?:once|and|then)\)/x;
my $sibling = qr/
\s*
(
(?! $connective )
\(
(?:
(?> (?: [^()]+ ) )
| (?-1)
)*
\)
)
\s*
/x;
my $regex1 = qr/
\( ( [\w-]+ \s* $sibling* \s* $connective \s* $sibling* ) \) #1
/x;
my $regex2 = qr/
( #1
\( \s*
( #2
[\w-]+ \s*
(?> $sibling* \s* $connective (?(R)(*FAIL)) \s* $sibling*
| (?1)
)
)
\s*
\)
)
/x;
my $sample = qq/
(X (SBAR-TMP (IN once) (S sdf) (S sdf)))
(X (SBAR-TMP (IN once))
(X (SBAR-TMP (IN once) (X as))
(X (SBAR-TMP (X adsf) (IN once))
(X (SBAR-TMP (IN once) (MORE stuff (MORE stuff))))
(S (CC and))
(SBARTMP (IN once) (NP otherstuff))
(S (S (NP blah (VP blah)) (CC then) (NP blah (VP blah (PP blah))) ))
/;
while ($sample =~ /$regex1/xg) {
print "Found: $1\n";
}
print '-' x 20, "\n";
while ($sample =~ /$regex2/xg) {
print "Found: $2\n";
}
__END__Stack Overflow用户
发布于 2010-12-23 21:26:36
你为什么放弃这个，你差一点就成功了。试试这个：
use strict;
use warnings;
my $connective = qr/(?: \((?:IN|CC)\s(?:once|and|then)\) )/x;
my $sibling = qr/
\s*
(
(?!$connect)
\(
(?:
(?> (?: [^()]+ ) )
| (?-1)
)*
\)
)
\s*
/x;
my $regex = qr/
( #1
\(
\s* [\w-]+ \s*
(?> $sibling* \s* $connective \s* $sibling*
| (?1)
)
\s*
\)
)
/x;
my @tests = (
'(X (SBAR-TMP (IN once) (S sdf) (S sdf)))',
'(X (SBAR-TMP (IN once))',
'(X (SBAR-TMP (IN once) (X as))',
'(X (SBAR-TMP (X adsf) (IN once))',
);
for my $sample (@tests)
{
while ($sample =~ /$regex/xg) {
print "Found: $1\n";
}
}
my $another =<<EOS;
(S (CC and))
(SBARTMP (IN once) (NP otherstuff))
(S
(S
(NP blah
(VP blah)
)
(CC then)
(NP blah
(VP blah
(PP blah)
)
)
)
)
EOS
print "\n---------\n";
while ($another =~ /$regex/xg) {
print "\nFound:\n$1\n";
}结束
Stack Overflow用户
发布于 2010-12-24 16:38:52
这应该也行得通
use strict;
use warnings;
my $connective = qr/(?: \((?:IN|CC)\s(?:once|and|then)\) )/x;
my $sibling = qr/
(?: \s*
(
(?!$connective)
\(
(?:
(?> (?: [^()]+ ) )
| (?-1)
)*
\)
)
\s* )
/x;
my $regex = qr/
( #1
\( \s*
( #2
[\w-]+ \s*
(?> $sibling* \s* $connective (?(R)(*FAIL)) \s* $sibling*
| (?1)
)
)
\s*
\)
)
/x;
my @tests = (
'(X (SBAR-TMP (IN once) (S sdf) (S sdf)))',
'(X (SBAR-TMP (IN once))',
'(X (SBAR-TMP (IN once) (X as))',
'(X (SBAR-TMP (X adsf) (IN once))',
'(X (SBAR-TMP (IN once) (MORE stuff (MORE stuff))))',
);
for my $sample (@tests)
{
while ($sample =~ /$regex/xg) {
print "Found: $2\n";
}
}
my $another = "
(S (CC and))
(SBARTMP (IN once) (NP otherstuff))
(S (S (NP blah (VP blah)) (CC then) (NP blah (VP blah (PP blah))) ))
";
print "\n---------\n";
while ($another =~ /$regex/xg) {
print "\nFound:\n$2\n";
}
__END__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/4518981
复制相关文章
相似问题
腾讯云开发者
