LevensteinDistance - 共享郎3.0 API [英] LevensteinDistance - Commons Lang 3.0 API

问题描述

通过共享郎API我可以计算通过<一两个字符串之间的相似性href=\"http://commons.apache.org/lang/api-2.3/org/apache/commons/lang/StringUtils.html#getLevenshteinDistance%28java.lang.String,%20java.lang.String%29\">LevensteinDistance.其结果是需要改变一个串到另一个变化的数量。我希望的结果是从0到1，在那里它会更容易识别的字符串之间的相似度的范围内。其结果将是接近0很大的相似性。这可能吗？

With Commons Lang api I can calculate the similarity between two strings through the LevensteinDistance. The result is the number of changes needed to change one string into another. I wish the result was within the range from 0 to 1, where it would be easier to identify the similarity between the strings. The result would be closer to 0 great similarity. Is it possible?

下面我使用的例子：

public class TesteLevenstein {

    public static void main(String[] args) {      

        int distance1 = StringUtils.getLevenshteinDistance("Boat", "Coat");
        int distance2 = StringUtils.getLevenshteinDistance("Remember", "Alamo");
        int distance3 = StringUtils.getLevenshteinDistance("Steve", "Stereo");

        System.out.println("distance(Boat, Coat): " + distance1);
        System.out.println("distance(Remember, Alamo): " + distance2);
        System.out.println("distance(Steve, Stereo): " + distance3);        

    }
}

谢谢！

推荐答案

只是一些数除以。现在的问题是什么号码？可能对于给定的一对串的最大可能距离。我认为这是较长的字符串的长度（即所有的人物都不同，加上一些更多的添加，用较短的字符串相比较）。

Just divide by some number. The question is what number? Probably the maximum possible distance for the given pair of strings. I think that's the length of the longer string (ie all the characters are different, plus a few more were added, compared with the shorter string).

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