有没有办法在检查字符串时包含单词变体(过去时,动名词)?

时间:2013-04-03 17:48:14

标签: javascript

假设我有一系列搜索词,比如 -

var searchTerms = ["blow", "search", "fly", "type"]

和一串字符串,如

var arrayToBeSearched = ["blowing", "searched", "flew", "typed", "blah", "blah","blah"]

检查数组时,是否有一种简单的方法可以包含过去时或其他单词变体?或者我应该只在searchTerms中包含变体?

4 个答案:

答案 0 :(得分:2)

虽然我自己没有使用此框架的经验,但您可以查看lunr.js 该库在浏览器中提供全文搜索,其中一个功能是stemming或匹配“搜索”,“搜索”等对照根词“搜索”。

答案 1 :(得分:2)

存在词形还原算法,例如Porter Stemmer。这会将您的单词映射到它们的词干,然后可以直接比较它们的相等性。该算法被描述为here。完整复制Javascript implementation

// Porter stemmer in Javascript. Few comments, but it's easy to follow against the rules in the original
// paper, in
//
//  Porter, 1980, An algorithm for suffix stripping, Program, Vol. 14,
//  no. 3, pp 130-137,
//
// see also http://www.tartarus.org/~martin/PorterStemmer

// Release 1 be 'andargor', Jul 2004
// Release 2 (substantially revised) by Christopher McKenzie, Aug 2009

var stemmer = (function(){
    var step2list = {
            "ational" : "ate",
            "tional" : "tion",
            "enci" : "ence",
            "anci" : "ance",
            "izer" : "ize",
            "bli" : "ble",
            "alli" : "al",
            "entli" : "ent",
            "eli" : "e",
            "ousli" : "ous",
            "ization" : "ize",
            "ation" : "ate",
            "ator" : "ate",
            "alism" : "al",
            "iveness" : "ive",
            "fulness" : "ful",
            "ousness" : "ous",
            "aliti" : "al",
            "iviti" : "ive",
            "biliti" : "ble",
            "logi" : "log"
        },

        step3list = {
            "icate" : "ic",
            "ative" : "",
            "alize" : "al",
            "iciti" : "ic",
            "ical" : "ic",
            "ful" : "",
            "ness" : ""
        },

        c = "[^aeiou]",          // consonant
        v = "[aeiouy]",          // vowel
        C = c + "[^aeiouy]*",    // consonant sequence
        V = v + "[aeiou]*",      // vowel sequence

        mgr0 = "^(" + C + ")?" + V + C,               // [C]VC... is m>0
        meq1 = "^(" + C + ")?" + V + C + "(" + V + ")?$",  // [C]VC[V] is m=1
        mgr1 = "^(" + C + ")?" + V + C + V + C,       // [C]VCVC... is m>1
        s_v = "^(" + C + ")?" + v;                   // vowel in stem

    return function (w) {
        var     stem,
            suffix,
            firstch,
            re,
            re2,
            re3,
            re4,
            origword = w;

        if (w.length < 3) { return w; }

        firstch = w.substr(0,1);
        if (firstch == "y") {
            w = firstch.toUpperCase() + w.substr(1);
        }

        // Step 1a
        re = /^(.+?)(ss|i)es$/;
        re2 = /^(.+?)([^s])s$/;

        if (re.test(w)) { w = w.replace(re,"$1$2"); }
        else if (re2.test(w)) { w = w.replace(re2,"$1$2"); }

        // Step 1b
        re = /^(.+?)eed$/;
        re2 = /^(.+?)(ed|ing)$/;
        if (re.test(w)) {
            var fp = re.exec(w);
            re = new RegExp(mgr0);
            if (re.test(fp[1])) {
                re = /.$/;
                w = w.replace(re,"");
            }
        } else if (re2.test(w)) {
            var fp = re2.exec(w);
            stem = fp[1];
            re2 = new RegExp(s_v);
            if (re2.test(stem)) {
                w = stem;
                re2 = /(at|bl|iz)$/;
                re3 = new RegExp("([^aeiouylsz])\\1$");
                re4 = new RegExp("^" + C + v + "[^aeiouwxy]$");
                if (re2.test(w)) {  w = w + "e"; }
                else if (re3.test(w)) { re = /.$/; w = w.replace(re,""); }
                else if (re4.test(w)) { w = w + "e"; }
            }
        }

        // Step 1c
        re = /^(.+?)y$/;
        if (re.test(w)) {
            var fp = re.exec(w);
            stem = fp[1];
            re = new RegExp(s_v);
            if (re.test(stem)) { w = stem + "i"; }
        }

        // Step 2
        re = /^(.+?)(ational|tional|enci|anci|izer|bli|alli|entli|eli|ousli|ization|ation|ator|alism|iveness|fulness|ousness|aliti|iviti|biliti|logi)$/;
        if (re.test(w)) {
            var fp = re.exec(w);
            stem = fp[1];
            suffix = fp[2];
            re = new RegExp(mgr0);
            if (re.test(stem)) {
                w = stem + step2list[suffix];
            }
        }

        // Step 3
        re = /^(.+?)(icate|ative|alize|iciti|ical|ful|ness)$/;
        if (re.test(w)) {
            var fp = re.exec(w);
            stem = fp[1];
            suffix = fp[2];
            re = new RegExp(mgr0);
            if (re.test(stem)) {
                w = stem + step3list[suffix];
            }
        }

        // Step 4
        re = /^(.+?)(al|ance|ence|er|ic|able|ible|ant|ement|ment|ent|ou|ism|ate|iti|ous|ive|ize)$/;
        re2 = /^(.+?)(s|t)(ion)$/;
        if (re.test(w)) {
            var fp = re.exec(w);
            stem = fp[1];
            re = new RegExp(mgr1);
            if (re.test(stem)) {
                w = stem;
            }
        } else if (re2.test(w)) {
            var fp = re2.exec(w);
            stem = fp[1] + fp[2];
            re2 = new RegExp(mgr1);
            if (re2.test(stem)) {
                w = stem;
            }
        }

        // Step 5
        re = /^(.+?)e$/;
        if (re.test(w)) {
            var fp = re.exec(w);
            stem = fp[1];
            re = new RegExp(mgr1);
            re2 = new RegExp(meq1);
            re3 = new RegExp("^" + C + v + "[^aeiouwxy]$");
            if (re.test(stem) || (re2.test(stem) && !(re3.test(stem)))) {
                w = stem;
            }
        }

        re = /ll$/;
        re2 = new RegExp(mgr1);
        if (re.test(w) && re2.test(w)) {
            re = /.$/;
            w = w.replace(re,"");
        }

        // and turn initial Y back to y

        if (firstch == "y") {
            w = firstch.toLowerCase() + w.substr(1);
        }

        return w;
    }
})();

答案 2 :(得分:0)

在Javascipt中没有本地方法可以执行此操作。但是,您可以利用REGEX处理数据,以查找您想要的任何模式。例如“ed”,“ing”等。

但是你的帖子中没有提到的是搜索的范围和环境。这是否发生在您的网站/应用程序的特定有限部分?如果它更大,但仍然在客户端,你可能想要(正如其他人提到的那样)使用库或者编写一个执行所需操作的插件。如果这是更整体的东西,你可能想要在服务器端执行,甚至在DB本身执行。

答案 3 :(得分:0)

这是node.js解决方案。请注意,它是根据LGPL许可的。

https://github.com/grachev/node-lemmer