这种方法是我程序的瓶颈。 (或者，至少在大部分时间用完了。)

其目的是根据字符在两者中出现的方式计算两个字符串之间的相似性，或者在一个字符串中出现或不出现在另一个字符串中。

代码是正确的，但我想优化它。您是否看到任何低效率，不良做法或不必要的复杂性？

(不要担心Gene类。这里真正关注的是Genes携带的字符串。)

public static int simpleScore(Gene g0, Gene g1) {

if (g0.equals(g1)) { //identical genes have distance 0

return 0;

}

String frag0 = g0.getDNA();

String frag1 = g1.getDNA();

Map charOccurencesFrag0 = new HashMap(Constants.DNA_CHARS.length);

Map charOccurencesFrag1 = new HashMap(Constants.DNA_CHARS.length);

int multSCOST = 0;

int multDCOST = 0;

int multDDCOST = 0;

for (char c : Constants.DNA_CHARS) { //look through all acceptable DNA chars

charOccurencesFrag0.put(c, StringUtils.countMatches(frag0, "" + c));

charOccurencesFrag1.put(c, StringUtils.countMatches(frag1, "" + c));

//if a char appears in one frag but not the other

if ((frag0.indexOf(c) != -1 && frag1.indexOf(c) == -1) || (frag0.indexOf(c) == -1 && frag1.indexOf(c) != -1)) {

multDDCOST += (charOccurencesFrag0.get(c) + charOccurencesFrag1.get(c));

}

//if a char appears in both frags

if (frag0.indexOf(c) != -1 && frag1.indexOf(c) != -1) {

int diff = Math.abs(charOccurencesFrag0.get(c) - charOccurencesFrag1.get(c));

multDCOST += diff;

multSCOST += Math.min(charOccurencesFrag0.get(c), charOccurencesFrag1.get(c));

}

}

return Constants.SCOST * multSCOST + Constants.DDCOST * multDDCOST + Constants.DCOST * multDCOST;

}根据回复更新了代码

这导致了相当大的加速。感谢所有评论的人。还有其他人有什么想法吗？

public static int simpleScore(Gene g0, Gene g1) {

if (g0.equals(g1)) { //identical genes have distance 0

return 0;

}

String frag0 = g0.getDNA();

String frag1 = g1.getDNA();

int[] charOccurencesFrag0 = countOccurrences(frag0, Constants.DNA_CHARS);

int[] charOccurencesFrag1 = countOccurrences(frag1, Constants.DNA_CHARS);

int multSCOST = 0;

int multDCOST = 0;

int multDDCOST = 0;

for (int ket = 0; ket < Constants.DNA_CHARS.length; ket++) {

//if a char appears in one frag but not the other

if ((charOccurencesFrag0[ket] > 0 && charOccurencesFrag1[ket] == 0) || (charOccurencesFrag0[ket] == 0 && charOccurencesFrag1[ket] >0)){

multDDCOST += charOccurencesFrag0[ket] + charOccurencesFrag1[ket];

}

//if a char appears in both frags

if (charOccurencesFrag0[ket] != 0 && charOccurencesFrag1[ket] != 0){

int diff = Math.abs(charOccurencesFrag0[ket] - charOccurencesFrag1[ket]);

multDCOST += diff;

multSCOST += Math.min(charOccurencesFrag0[ket] , charOccurencesFrag1[ket]);

}

}

return Constants.SCOST * multSCOST + Constants.DDCOST * multDDCOST + Constants.DCOST * multDCOST;

}

// from MAK on SO

private static int[] countOccurrences(String x, char[] validDNAChars){

int[] count=new int[validDNAChars.length];

for(int i=0;i

int index=-1;

for(int j=0;j

if (x.charAt(i)==validDNAChars[j]){

index=j;

}

}

if (index>=0) count[index]++;

}

return count;

}

本文实例讲述了java字符串相似度算法。分享给大家供大家参考。具体实现方法如下：

public class Levenshtein {

private int compare(String str, String target) {

int d[][]; // 矩阵

int n = str.length();

int m = target.length();

int i; // 遍历str的

int j; // 遍历target的

char ch1; // str的

char ch2; // target的

int temp; // 记录相同字符,在某个矩阵位置值的增量,不是0就是1

if (n == 0) {

return m;

}

if (m == 0) {

return n;

}

d = new int[n + 1][m + 1];

for (i = 0; i <= n; i++) { // 初始化第一列

d[i][0] = i;

}

for (j = 0; j <= m; j++) { // 初始化第一行

d[0][j] = j;

}

for (i = 1; i <= n; i++) { // 遍历str

ch1 = str.charAt(i - 1);

// 去匹配target

for (j = 1; j <= m; j++) {

ch2 = target.charAt(j - 1);

if (ch1 == ch2) {

temp = 0;

} else {

temp = 1;

}

// 左边+1,上边+1, 左上角+temp取最小

d[i][j] = min(d[i - 1][j] + 1, d[i][j - 1] + 1, d[i - 1][j - 1] + temp);

}

}

return d[n][m];

}

private int min(int one, int two, int three) {

return (one = one < two ? one : two) < three ? one : three;

}

/**

* 获取两字符串的相似度

*

* @param str

* @param target

*

* @return

*/

public float getSimilarityRatio(String str, String target) {

return 1 - (float) compare(str, target) / Math.max(str.length(), target.length());

}

public static void main(String[] args) {

Levenshtein lt = new Levenshtein();

String str = "ab";

String target = "ac";

System.out.println("similarityRatio=" + lt.getSimilarityRatio(str, target));

}

}

希望本文所述对大家的Java程序设计有所帮助。