On Tuesday, November 25, 2003, at 11:48 AM, MOYSE Gilles (Cetelem) wrote:
Hi.
You should define expressions.
To define expressions, you first have to define an expression file.
An expression file contains one expressions per line.
For instance :
time_out
expert_system
...
You can use any character to specify the "expression link". Here, I use the
underscore (_).
Then, you have to build an expression loader. You can store expressions in
recursives HashMap.
Such HashMap must be built so that HashMap.get("word1") = HashMap, and
(HashMap.get("word1")).get("word2") = null, if you want to code the
expression "word1_word2".
In other words 'HashMap.get("a_word")' returns a hashMap containing all the
successors of the word 'a_word'.
So, if your expression file looks like that : time_out expert_system expert_in_information
you'll have to build a loader which returns a HashMap H so that : H.keySet() = {"time", "expert"} ((HashMap)H.get("time")).keySet = {"out"} ((HashMap)H.get("time")).get("out") = null // null indicates the end of the expression ((HashMap)H.get("expert")).keySet = {"system", "in"} ((HashMap)H.get("expert")).get("system") = null ((HashMap)((HashMap)H.get("expert")).get("in")).keySet() = {"information"} ((HashMap)((HashMap)H.get("expert")).get("in")).get("information") = null
These recursives HashMaps code the following tree : time - out - null system --- expert - null |- in - information- null
Such an expression loader may be designed this way :
public static HashMap getExpressionMap( File wordfile ) { HashMap result = new HashMap(); try { String line = null; LineNumberReader in = new LineNumberReader(new FileReader(wordfile)); HashMap hashToAdd = null; while ((line = in.readLine()) != null) { if (line.startsWith(FILE_COMMENT_CHARACTER)) continue;
if (line.trim().length() == 0) continue;
StringTokenizer stok = new
StringTokenizer(line, " \t_");
String curTok = "";
HashMap currentHash = result;
// Test wether the expression contains 2 at
least words or not
if (stok.countTokens() < 2)
{
System.err.println("Warning : '" +
line + "' in file '" + wordfile.getAbsolutePath() + "' line " +
in.getLineNumber() +
" is not an expression.\n\tA
valid expression contains at least 2 words.");
continue;
}
while (stok.hasMoreTokens())
{
curTok = stok.nextToken();
if
(curTok.startsWith(FILE_COMMENT_CHARACTER)) // if comment at the end of the
line, break
break;
if (stok.hasMoreTokens())
hashToAdd = new HashMap(6);
else
hashToAdd = (HashMap)null;
if
(!(currentHash.containsKey(curTok)))
currentHash.put(curTok,
hashToAdd);
currentHash =
(HashMap)currentHash.get(curTok);
}
}
return result;
}
// On error, use an empty table
catch ( Exception e )
{
System.err.println("While processing '" +
wordfile.getAbsolutePath() + "' : " + e.getMessage());
e.printStackTrace();
return new HashMap();
}
}
Then, you must build a filter with 2 FIFO stacks : one is the expression
stack, the other is the default stack.
Then, you define a 'curMap' variable, initially pointing onto the HashMap
returned by the ExpressionFileLoader.
When you receive a token, you check wether it is null or not;
If it is, you check if the standard stack is null or not.
If it is not, you pop a token from the default stack and you
return it.
If it is, you return null
If it is not (the token is not null), you check whether it is
contained in the HashMap or not (curMap.containsKey(token)).
If it is not contained and you were building an expression,
you pop all the terms in the expression stack to push them in the default
stack (so as not to loose information)
If it is not contained and the default stack is empty, you
return the token.
If it is not conatined and the default stack is not empty,
you return the poped token from the default stack and you push the current
token.
If the token is contained in the curMap, then the token MAY be the
first element of an expression.
You push the token in the expression stack, and you dive
into the next level in your expression tree (curMap = curMap.get("token"))
If the next level (now, curMap), is null, then you have
completed your expression. You can pop all the tokens from the expresion
stack to concatenate them, separated by underscores, and push the resulting
String as a token on the default heap (so as to keep the correct tokens
oreder in the token stream)
You also set a flag "I'm in an expression" (expr_falg)
With this short algorithm, you will build expressions, i.e., if an
expression is detected, it will be returned as such, and if it is not, it
will return the tokens, unmodified, in their orginal order.
Hope this will help.
Gilles Moyse
-----Message d'origine----- De : Dragan Jotanovic [mailto:[EMAIL PROTECTED] Envoyé : mardi 25 novembre 2003 12:42 À : Lucene Users List Objet : Tokenizing text custom way
Hi. I need to tokenize text while indexing but I don't want space to be
delimiter. Delimiter should be my custom character (for example comma). I
understand that I would probably need to implement my own analyzer, but
could someone help me where to start. Is there any other way to do this
without writing custom analyzer?
This is what I want to achieve. If I have some text that will be indexed like following:
man, people, time out, sun
and if I enter 'time' as a search word, I don't want to get "time out" in
results. I need exact keyword matching. I would achieve this if I tokenize
"time out" as one token while idexing.
Maybe someone had similar problem? If someone knows how to handle this, please help me.
Dragan Jotanovic
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