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我必须比较两个字符串,即多少百分比str1包含str2。计算字符串距离/ JavaScript差异?
str1="Hello wolrd"
str2="hello"
意味着distance(str1, str2)
应该返回50%..就是这样。
我必须比较两个字符串,即多少百分比str1包含str2。计算字符串距离/ JavaScript差异?
str1="Hello wolrd"
str2="hello"
意味着distance(str1, str2)
应该返回50%..就是这样。
请用下面的代码尝试。
<script>
var str1 = 'helloworld';
var str2 = 'hello';
var percent = 0;
var newstr = str1.search(str2);
if(newstr != -1){
percent = (str2.length * 100)/str1.length;
alert(percent+'%');
}else{
alert(percent+'%');
}
</script>
如果我理解你的问题,我敢肯定,这将帮助你。
谢谢。
我不是数学家,不能告诉你一个神奇的公式这样的事情,但是这个例子让你开始使用递归 - 请代码中的注释仔细看:
/* same_words array should hold the following:
["john", "bbc", "hot", "red,", "xyz,", "apples,", "and", "likes"]
NOTES:
- The more words you add, the less the percentage will be... that's how percent works.
- milk and MILK are the same word, but: MILK and MILK, aren't the same because of the comma (same goes for periods).
*/
var jSmith = 'JOHN smith LIKES bananas, HOT chocolate, AND APPLES, nonsense, RED, 321, XYZ, BBC or npr',
jSmithArray = jSmith.trim().toLowerCase().split(' '); //Makes an array of words for jSmith
var jDoe = 'JOHN doe LIKES oranges, milk, AND APPLES, XYZ, 123, RED, green, HOT and... cnn sucks, BBC rocks',
jDoeArray = jDoe.trim().toLowerCase().split(' '); //Makes an array of words for jDoe
var bothJohns = jSmithArray.concat(jDoeArray); //console.log(bothJohns);
var c = 0; //counter
var same_words = []; //collection container
//collects all the words that occur in both sentences
function collectSimilarWords(word) {
same_words.push(word);
}
//The w parameter holds the word "john" for the 1st time.
//The fn parameter holds always the collectSimilarWords() function.
function recur(w, fn) {
for (var p in jSmithArray) { //Every property of the jSmithArray Array holds a word string.
if (w) {
if (w == jSmithArray[p]) { //If the word we passed in as a parameter is the same word as one of the jSmithArray elements...
fn(jSmithArray[p]); //...then pass this word to the collectSimilarWords() function.
}
c += 1; //Increase c so we can move to...
recur(jDoeArray[c], collectSimilarWords); //...the next word recursively.
}
}
}
//Call recur() and pass in the first word of the jDoeArray as the 1st param, and a function as the 2nd param.
recur(jDoeArray[c], collectSimilarWords);
function calcWhatever(samewords) { //Feel free to calculate however you want.
var percent = ((samewords.length * 100)/bothJohns.length).toFixed(2);
var msg = "Out of Smith\'s plus Doe's words (total: " + bothJohns.length + "), about " + percent + "% (or " + same_words.length + ") of them are found in both of their sentences!";
return msg;
}
//Hopefuly setTimeout() will log the same_words array when the recursive function has finished.
window.setTimeout(function() {
console.log(same_words);
console.log( calcWhatever(same_words) );
}, 1000);
代替一个for in
循环,你可以很容易地使用for循环,如下所示:
for (var p = 0; p<jSmithArray.length; p++) { /* ... */ }
你想实现什么规则? IE浏览器。是否区分大小写?你是否正在计算总字数中有多少单词匹配并以百分比形式显示? – psynnott 2012-04-16 06:31:50
“字符串距离”是进行中数学调查的一个复杂且活跃的领域。测量它的方法有多种,例如Levenshtein距离等,具有不同的优点和缺点,具体取决于应用。你能否提供有关申请的详细信息?这是用有用的东西来回答这种性质问题的唯一方法。 – 2012-04-16 06:42:13
http://en.wikibooks.org/wiki/Algorithm_Implementation/Strings/Levenshtein_distance#JavaScript – 2012-04-16 06:45:15