⭐ 欢迎来到虫虫下载站! | 📦 资源下载 📁 资源专辑 ℹ️ 关于我们
⭐ 虫虫下载站

📄 simtexth.cpp

📁 QT 开发环境里面一个很重要的文件
💻 CPP
字号:
/******************************************************************************** Copyright (C) 1992-2006 Trolltech ASA. All rights reserved.**** This file is part of the Qt Linguist of the Qt Toolkit.**** This file may be used under the terms of the GNU General Public** License version 2.0 as published by the Free Software Foundation** and appearing in the file LICENSE.GPL included in the packaging of** this file.  Please review the following information to ensure GNU** General Public Licensing requirements will be met:** http://www.trolltech.com/products/qt/opensource.html**** If you are unsure which license is appropriate for your use, please** review the following information:** http://www.trolltech.com/products/qt/licensing.html or contact the** sales department at sales@trolltech.com.**** This file is provided AS IS with NO WARRANTY OF ANY KIND, INCLUDING THE** WARRANTY OF DESIGN, MERCHANTABILITY AND FITNESS FOR A PARTICULAR PURPOSE.******************************************************************************/#include "simtexth.h"#include "metatranslator.h"#include <QString>#include <QList>#include <string.h>typedef QList<MetaTranslatorMessage> TML;/*  How similar are two texts?  The approach used here relies on co-occurrence  matrices and is very efficient.  Let's see with an example: how similar are "here" and "hither"?  The  co-occurrence matrix M for "here" is M[h,e] = 1, M[e,r] = 1, M[r,e] = 1, and 0  elsewhere; the matrix N for "hither" is N[h,i] = 1, N[i,t] = 1, ...,  N[h,e] = 1, N[e,r] = 1, and 0 elsewhere.  The union U of both matrices is the  matrix U[i,j] = max { M[i,j], N[i,j] }, and the intersection V is  V[i,j] = min { M[i,j], N[i,j] }.  The score for a pair of texts is      score = (sum of V[i,j] over all i, j) / (sum of U[i,j] over all i, j),  a formula suggested by Arnt Gulbrandsen.  Here we have      score = 2 / 6,  or one third.  The implementation differs from this in a few details.  Most importantly,  repetitions are ignored; for input "xxx", M[x,x] equals 1, not 2.*//*  Every character is assigned to one of 20 buckets so that the co-occurrence  matrix requires only 20 * 20 = 400 bits, not 256 * 256 = 65536 bits or even  more if we want the whole Unicode.  Which character falls in which bucket is  arbitrary.  The second half of the table is a replica of the first half, because of  laziness.*/static const int indexOf[256] = {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`   a   b   c   d   e   f   g   h   i   j   k   l   m   n   o    0,  1,  2,  3,  4,  5,  6,  7,  8,  9,  6,  10, 11, 12, 13, 14,//  p   q   r   s   t   u   v   w   x   y   z   {   |   }   ~    15, 12, 16, 17, 18, 19, 2,  10, 15, 7,  19, 2,  6,  7,  10, 0,    0,  0,  0,  0,  0,  0,  0,  0,  0,  0,  0,  0,  0,  0,  0,  0,    0,  0,  0,  0,  0,  0,  0,  0,  0,  0,  0,  0,  0,  0,  0,  0,    0,  2,  6,  7,  10, 12, 15, 19, 2,  6,  7,  10, 12, 15, 19, 0,    1,  3,  4,  5,  8,  9,  11, 13, 14, 16, 2,  6,  7,  10, 12, 15,    0,  1,  2,  3,  4,  5,  6,  7,  8,  9,  6,  10, 11, 12, 13, 14,    15, 12, 16, 17, 18, 19, 2,  10, 15, 7,  19, 2,  6,  7,  10, 0,    0,  1,  2,  3,  4,  5,  6,  7,  8,  9,  6,  10, 11, 12, 13, 14,    15, 12, 16, 17, 18, 19, 2,  10, 15, 7,  19, 2,  6,  7,  10, 0};/*  The entry bitCount[i] (for i between 0 and 255) is the number of bits used to  represent i in binary.*/static const int bitCount[256] = {    0,  1,  1,  2,  1,  2,  2,  3,  1,  2,  2,  3,  2,  3,  3,  4,    1,  2,  2,  3,  2,  3,  3,  4,  2,  3,  3,  4,  3,  4,  4,  5,    1,  2,  2,  3,  2,  3,  3,  4,  2,  3,  3,  4,  3,  4,  4,  5,    2,  3,  3,  4,  3,  4,  4,  5,  3,  4,  4,  5,  4,  5,  5,  6,    1,  2,  2,  3,  2,  3,  3,  4,  2,  3,  3,  4,  3,  4,  4,  5,    2,  3,  3,  4,  3,  4,  4,  5,  3,  4,  4,  5,  4,  5,  5,  6,    2,  3,  3,  4,  3,  4,  4,  5,  3,  4,  4,  5,  4,  5,  5,  6,    3,  4,  4,  5,  4,  5,  5,  6,  4,  5,  5,  6,  5,  6,  6,  7,    1,  2,  2,  3,  2,  3,  3,  4,  2,  3,  3,  4,  3,  4,  4,  5,    2,  3,  3,  4,  3,  4,  4,  5,  3,  4,  4,  5,  4,  5,  5,  6,    2,  3,  3,  4,  3,  4,  4,  5,  3,  4,  4,  5,  4,  5,  5,  6,    3,  4,  4,  5,  4,  5,  5,  6,  4,  5,  5,  6,  5,  6,  6,  7,    2,  3,  3,  4,  3,  4,  4,  5,  3,  4,  4,  5,  4,  5,  5,  6,    3,  4,  4,  5,  4,  5,  5,  6,  4,  5,  5,  6,  5,  6,  6,  7,    3,  4,  4,  5,  4,  5,  5,  6,  4,  5,  5,  6,  5,  6,  6,  7,    4,  5,  5,  6,  5,  6,  6,  7,  5,  6,  6,  7,  6,  7,  7,  8};struct CoMatrix{    /*      The matrix has 20 * 20 = 400 entries.  This requires 50 bytes, or 13      words.  Some operations are performed on words for more efficiency.    */    union {    quint8 b[52];    quint32 w[13];    };    CoMatrix() { memset( b, 0, 52 ); }    CoMatrix( const char *text ) {        char c = '\0', d;        memset( b, 0, 52 );        /*          The Knuth books are not in the office only for show; they help make          loops 30% faster and 20% as readable.        */        while ( (d = *text) != '\0' ) {            setCoocc( c, d );            if ( (c = *++text) != '\0' ) {                setCoocc( d, c );                text++;            }        }    }    void setCoocc( char c, char d ) {        int k = indexOf[(uchar) c] + 20 * indexOf[(uchar) d];        b[k >> 3] |= k & 0x7;    }    int worth() const {        int w = 0;        for ( int i = 0; i < 50; i++ )            w += bitCount[b[i]];        return w;    }};static inline CoMatrix reunion( const CoMatrix& m, const CoMatrix& n ){    CoMatrix p;    for ( int i = 0; i < 13; i++ )    p.w[i] = m.w[i] | n.w[i];    return p;}static inline CoMatrix intersection( const CoMatrix& m, const CoMatrix& n ){    CoMatrix p;    for ( int i = 0; i < 13; i++ )    p.w[i] = m.w[i] & n.w[i];    return p;}StringSimilarityMatcher::StringSimilarityMatcher(const QString &stringToMatch){    m_cm = new CoMatrix( stringToMatch.toLatin1().constData() );    m_length = stringToMatch.length();}int StringSimilarityMatcher::getSimilarityScore(const QString &strCandidate){    CoMatrix cmTarget( strCandidate.toLatin1().constData() );    int targetLen = strCandidate.length();    int delta = qAbs( m_length - targetLen );        int score = ( (intersection(*m_cm, cmTarget).worth() + 1) << 10 ) /        ( reunion(*m_cm, cmTarget).worth() + (delta << 1) + 1 );    return score;}StringSimilarityMatcher::~StringSimilarityMatcher(){    delete m_cm;}/** * Checks how similar two strings are. * The return value is the score, and a higher score is more similar than one with a low score. * Linguist considers a score over 190 to be a good match. * \sa StringSimilarityMatcher */int getSimilarityScore(const QString &str1, const char* str2){    CoMatrix cmTarget( str2 );    int targetLen = qstrlen( str2 );    CoMatrix cm( str1.toLatin1().constData() );    int delta = qAbs( (int) str1.length() - targetLen );    int score = ( (intersection(cm, cmTarget).worth() + 1) << 10 ) /        ( reunion(cm, cmTarget).worth() + (delta << 1) + 1 );    return score;}CandidateList similarTextHeuristicCandidates( const MetaTranslator *tor,                        const char *text,                        int maxCandidates ){    QList<int> scores;    CandidateList candidates;    TML all = tor->translatedMessages();    foreach ( MetaTranslatorMessage mtm, all ) {        if ( mtm.type() == MetaTranslatorMessage::Unfinished ||             mtm.translation().isEmpty() )            continue;        QString s = tor->toUnicode( mtm.sourceText(), mtm.utf8() );        int score = getSimilarityScore(s, text);        if ( (int) candidates.count() == maxCandidates &&             score > scores[maxCandidates - 1] )            candidates.removeAt( candidates.size()-1 );        if ( (int) candidates.count() < maxCandidates && score >= textSimilarityThreshold ) {            Candidate cand( s, mtm.translation() );            int i;            for ( i = 0; i < (int) candidates.size(); i++ ) {                if ( score >= scores.at(i) ) {                    if ( score == scores.at(i) ) {                        if ( candidates.at(i) == cand )                            goto continue_outer_loop;                    } else {                        break;                    }                }            }            scores.insert( i, score );            candidates.insert( i, cand );        }        continue_outer_loop:        ;    }    return candidates;}

⌨️ 快捷键说明

复制代码 Ctrl + C
搜索代码 Ctrl + F
全屏模式 F11
切换主题 Ctrl + Shift + D
显示快捷键 ?
增大字号 Ctrl + =
减小字号 Ctrl + -