基于规则的命名实体识别

/* * ===================================================================================== * * Filename: regex_test.c * * Description: 根据规则模板抽取命名实体 * * Version: 1.0 * Created: 2011年04月06日 10时49分16秒 * Revision: none * Compiler: gcc * * Author: 齐保元 (qby), [e-mail] * Company: ict,gucas * * ===================================================================================== */#include <stdio.h>#include <stdlib.h>#include <sys/types.h>#include <regex.h>#include <string.h>//规则结构体typedef struct _RULE_{ char name[256];//规则的名称,如a+n+b+ double freq;//规则可以推导出是一个命名实体的概率，有王石提供}RULE;//存放所有的正则表达式对象static regex_t regexes[256];//输入，格式为{{"蓝屏","司机"}{"a","n"}},每个词对应一个词性，下标一致static char *input[2][500];//declearFILE *file_open(char* file_name,char* mode);char* read_line (FILE *fp);char* parse_line(char* line,int index,char sign);void process_regex(int count,char*line);static void close_all_regex(int count);voidget_corpse(int count){FILE *fp=file_open("huge.crf_format.txt_0","r");char *line;int i=0,j=0;char *word,*pos;while((line=read_line(fp))!=NULL){//遇到空白行if(strlen(line)==0){ char pos1[500]={'/0'}; j=0; for(j=0;input[1][j]!=NULL && strlen(input[1][j])!=0;j++){ sprintf(pos1,"%s%s",pos1,input[1][j]); //printf("pos1:%s/n",pos1); } process_regex(count,pos1); i=j=0; continue;}word=parse_line(line,0,'/t');pos=parse_line(line,1,'/t');input[0][i++]=word;input[1][j++]=pos;} close_all_regex(count);}/*----------------------------------------------------------------------------- * 对每一行规则进行解析，存入结构体对象 *-----------------------------------------------------------------------------*/RULEparse_rule(char* line){if(line==NULL){perror("空串");exit(1);}RULE rule;memset(rule.name,0,sizeof(rule.name));char *p=line;char *q=line;while(*p==' ')p++;while(*q==' ')q++;while(*q!='/t')q++;//指向regex的结尾int index=0;while(p!=q){if(*p=='+' || *p=='*')sprintf(rule.name,"%s%s",rule.name,"//");sprintf(rule.name,"%s%c",rule.name,p[index]);p++;}return rule;}/*----------------------------------------------------------------------------- * 对所有的正则表达式进行编译，保存到结构提regexes,返回加载的正则表达式的个数 *-----------------------------------------------------------------------------*/static int compile_all_regx(){ int p=0,z=0,cflags=0; char ebuf[300]; regex_t reg; char *pattern;FILE *fp=file_open("template","r");char *line;RULE rule;while((line=read_line(fp))!=NULL){if(strlen(line)==0)continue;printf("第%d行/n",p);/*a0.1b0.2*/rule=parse_rule(line);pattern=rule.name;printf("加载第%d个正则表达式:%s,freq:%f/n",p,pattern,rule.freq);z=regcomp(®,pattern,cflags); if(0!=z){ regerror(z,®,ebuf,sizeof(ebuf)); fprintf(stderr,"%s:pattern '%s'/n",ebuf,pattern); return; } regexes[p]=reg; p++;free(line);}return p;}/*----------------------------------------------------------------------------- * 关闭所有打开的正则表达式 *-----------------------------------------------------------------------------*/static void close_all_regex(int count){ int i=0; for(i=0;i<count;i++){ regfree(®exes[i]); }}/*----------------------------------------------------------------------------- * 打开文件，返回句柄 *-----------------------------------------------------------------------------*/FILE *file_open(char* file_name,char* mode){FILE *fp;if((fp=fopen(file_name,mode))==NULL){fprintf(stderr,"无法打开文件:%s./n",file_name);return NULL;}return fp;}/*----------------------------------------------------------------------------- * 返回一行 *-----------------------------------------------------------------------------*/char*read_line (FILE *fp){if(feof(fp))return NULL;char ch;int max_len=100;char *line=(char*)malloc(max_len);char *new_line;int counter=0;//counterwhile( ((ch=fgetc(fp))!='/n') && !feof(fp)){if(counter>=max_len-1){max_len*=2;new_line=(char*)malloc(max_len);strcpy(new_line,line);free(line);line=new_line;} *(line+(counter++))=ch;}*(line+counter)='/0';return line;}/*----------------------------------------------------------------------------- * 解析，line，要取的下标，分割符 *-----------------------------------------------------------------------------*/char* parse_line(char* line,int index,char sign){char *p=line;char *q=line;while(*p==' ')p++;while(*q==' ')q++;int i=0;char *ret=(char*)calloc(sizeof(char),300);while(1){if(*q!=sign){q++;continue;}if(i<index){i++;p=++q;continue;}else{strncpy(ret,p,q-p);p=++q;break;}if(*q=='/0')break;}return ret;}void process_regex(int count,char *lbuf){ int p=0;char *pattern; int x,z,no=0,cflags=0; char ebuf[1280]={'/0'};char ret[1024];char *curr;regmatch_t pm[10];const size_t nmatch=10;char word[200];int word_i; int ori_i; //对输入在每个正则表达式进行匹配，选取最长的for(p=0;p<count;p++){++no;aif((z=strlen(lbuf))>=0 && lbuf[z-1]=='/n')lbuf[z-1]=0;curr=lbuf; ori_i=0;while(regexec(®exes[p],curr,nmatch,pm,0)==0){if(pm[0].rm_so==-1)break;memset(ret,0x00,sizeof(ret));memset(word,0x00,sizeof(word));memcpy(ret,curr+pm[0].rm_so,pm[0].rm_eo-pm[0].rm_so);for(word_i=pm[0].rm_so+ori_i;word_i<pm[0].rm_eo+ori_i;word_i++){sprintf(word,"%s%s",word,input[0][word_i]);}printf("pos:%s->word:%s/n/n",ret,word);curr+=pm[0].rm_eo; ori_i+=pm[0].rm_eo;}}}int main(int argc,char** argv){ int count=compile_all_regx();//预编译regex get_corpse(count); return EXIT_SUCCESS;}