假设我有一系列搜索词，例如-
var searchTerms = ["blow", "search", "fly", "type"]
和一个字符串数组，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？
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1829 次
4 回答
4
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是否相等。该算法在此处描述。完整复制Javascript 实现：
// 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;
}
})();
于 2013-04-03T18:06:00.800 回答
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，其功能之一是stemming将“搜索”、“搜索”等与根词匹配搜索'。
于 2013-04-03T17:54:53.137 回答
0
这是一个 node.js 解决方案。请注意，它是在 LGPL 下获得许可的。
于 2013-04-03T17:54:44.433 回答
-1
在 Javascipt 中没有本地方法可以执行此操作。但是，您可以做的是利用 REGEX 来处理您的数据以查找几乎任何您想要的模式。例如“ed”、“ing”等。
但是您的帖子中没有提到的是您搜索的范围和情况。这是否发生在您的站点/应用程序的特定有限部分？如果它更大，但仍然在客户端，您可能希望（正如其他人提到的）使用库或编写执行所需操作的插件。如果这是更单一的东西，您可能希望在服务器端甚至在数据库本身执行此操作。
于 2013-04-03T17:53:59.907 回答