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word = words[i]; i++)\n\t\t{\n\t\t\tif (word.p)\n\t\t\t{\n\t\t\t\t// 仅对未识别的词进行匹配\n\t\t\t\tret.push(word);\n\t\t\t\tcontinue;\n\t\t\t}\n\t\t\tlet nameinfo = this.matchName(word.w);\n\t\t\tif (nameinfo.length < 1)\n\t\t\t{\n\t\t\t\tret.push(word);\n\t\t\t\tcontinue;\n\t\t\t}\n\t\t\t// 分离出人名\n\t\t\tlet lastc = 0;\n\t\t\tfor (let ui = 0, url; url = nameinfo[ui]; ui++)\n\t\t\t{\n\t\t\t\tif (url.c > lastc)\n\t\t\t\t{\n\t\t\t\t\tret.push(this.debugToken({\n\t\t\t\t\t\tw: word.w.substr(lastc, url.c - lastc),\n\t\t\t\t\t}, {\n\t\t\t\t\t\t[this.name]: false,\n\t\t\t\t\t}, true));\n\t\t\t\t}\n\n\t\t\t\tret.push(this.debugToken({\n\t\t\t\t\tw: url.w,\n\t\t\t\t\tp: POSTAG.A_NR\n\t\t\t\t}, {\n\t\t\t\t\t[this.name]: true,\n\t\t\t\t}, true));\n\t\t\t\tlastc = url.c + url.w.length;\n\t\t\t}\n\t\t\tlet lastn = nameinfo[nameinfo.length - 1];\n\t\t\tif (lastn.c + lastn.w.length < word.w.length)\n\t\t\t{\n\t\t\t\tret.push(this.debugToken({\n\t\t\t\t\tw: word.w.substr(lastn.c + lastn.w.length),\n\t\t\t\t}, {\n\t\t\t\t\t[this.name]: false,\n\t\t\t\t}, true));\n\t\t\t}\n\t\t}\n\t\treturn ret;\n\t}\n\n\t/**\n\t * 匹配包含的人名，并返回相关信息\n\t *\n\t * @param {string} text 文本\n\t * @param {int} cur 开始位置\n\t * @return {array} 返回格式 {w: '人名', c: 开始位置}\n\t */\n\tmatchName(text: string, cur: number = 0): IWord[]\n\t{\n\t\tif (isNaN(cur)) cur = 0;\n\t\tlet ret: IWord[] = [];\n\t\twhile (cur < text.length)\n\t\t{\n\t\t\t//debug('cur=' + cur + ', ' + text.charAt(cur));\n\t\t\tlet name: string = null;\n\t\t\t// 复姓\n\t\t\tlet f2 = text.substr(cur, 2);\n\t\t\tif (f2 in FAMILY_NAME_2)\n\t\t\t{\n\t\t\t\tlet n1 = text.charAt(cur + 2);\n\t\t\t\tlet n2 = text.charAt(cur + 3);\n\t\t\t\tif (n1 in DOUBLE_NAME_1 && n2 in DOUBLE_NAME_2)\n\t\t\t\t{\n\t\t\t\t\tname = f2 + n1 + n2;\n\t\t\t\t}\n\t\t\t\telse if (n1 in SINGLE_NAME)\n\t\t\t\t{\n\t\t\t\t\tname = f2 + n1 + (n1 == n2 ? n2 : '');\n\t\t\t\t}\n\t\t\t}\n\t\t\t// 单姓\n\t\t\tlet f1 = text.charAt(cur);\n\t\t\tif (name === null && f1 in FAMILY_NAME_1)\n\t\t\t{\n\t\t\t\tlet n1 = text.charAt(cur + 1);\n\t\t\t\tlet n2 = text.charAt(cur + 2);\n\t\t\t\tif (n1 in DOUBLE_NAME_1 && n2 in DOUBLE_NAME_2)\n\t\t\t\t{\n\t\t\t\t\tname = f1 + n1 + n2;\n\t\t\t\t}\n\t\t\t\telse if (n1 in SINGLE_NAME)\n\t\t\t\t{\n\t\t\t\t\tname = f1 + n1 + (n1 == n2 ? n2 : '');\n\t\t\t\t}\n\t\t\t}\n\t\t\t// 检查是否匹配成功\n\t\t\tif (name === null)\n\t\t\t{\n\t\t\t\tcur++;\n\t\t\t}\n\t\t\telse\n\t\t\t{\n\t\t\t\tret.push({ w: name, c: cur });\n\t\t\t\tcur += name.length;\n\t\t\t}\n\t\t}\n\t\treturn ret;\n\t}\n}\n\n// ======================================================================\n\n// debug(matchName('刘德华和李娜娜、司马光、上官飞飞'));\n// debug(matchName('李克'));\n\nexport const init = ChsNameTokenizer.init.bind(ChsNameTokenizer) as typeof ChsNameTokenizer.init;\n\nexport default ChsNameTokenizer;\n"]} |