我必须比较两个4000-5000个字符的字符串。
我需要结果百分比（即70％ - 80％匹配），在java。
请建议我解决它。
此致
答案 0 :(得分：6)
以下是比较两个字符串并以0到100的整数形式获取结果的代码。
/**
*
* @author WARLOCK
*/
public class LockMatch {
public static void main(String arg[]) {
//---Provide source and target strings to lock_match function to compare--//
System.out.println("Your Strings are Matched="+lock_match("The warlock","The warlock powered by WTPL")+"%");
}
public static int lock_match(String s, String t) {
int totalw = word_count(s);
int total = 100;
int perw = total / totalw;
int gotperw = 0;
if (!s.equals(t)) {
for (int i = 1; i <= totalw; i++) {
if (simple_match(split_string(s, i), t) == 1) {
gotperw = ((perw * (total - 10)) / total) + gotperw;
} else if (front_full_match(split_string(s, i), t) == 1) {
gotperw = ((perw * (total - 20)) / total) + gotperw;
} else if (anywhere_match(split_string(s, i), t) == 1) {
gotperw = ((perw * (total - 30)) / total) + gotperw;
} else {
gotperw = ((perw * smart_match(split_string(s, i), t)) / total) + gotperw;
}
}
} else {
gotperw = 100;
}
return gotperw;
}
public static int anywhere_match(String s, String t) {
int x = 0;
if (t.contains(s)) {
x = 1;
}
return x;
}
public static int front_full_match(String s, String t) {
int x = 0;
String tempt;
int len = s.length();
//----------Work Body----------//
for (int i = 1; i <= word_count(t); i++) {
tempt = split_string(t, i);
if (tempt.length() >= s.length()) {
tempt = tempt.substring(0, len);
if (s.contains(tempt)) {
x = 1;
break;
}
}
}
//---------END---------------//
if (len == 0) {
x = 0;
}
return x;
}
public static int simple_match(String s, String t) {
int x = 0;
String tempt;
int len = s.length();
//----------Work Body----------//
for (int i = 1; i <= word_count(t); i++) {
tempt = split_string(t, i);
if (tempt.length() == s.length()) {
if (s.contains(tempt)) {
x = 1;
break;
}
}
}
//---------END---------------//
if (len == 0) {
x = 0;
}
return x;
}
public static int smart_match(String ts, String tt) {
char[] s = new char[ts.length()];
s = ts.toCharArray();
char[] t = new char[tt.length()];
t = tt.toCharArray();
int slen = s.length;
//number of 3 combinations per word//
int combs = (slen - 3) + 1;
//percentage per combination of 3 characters//
int ppc = 0;
if (slen >= 3) {
ppc = 100 / combs;
}
//initialising an integer to store the total % this class genrate//
int x = 0;
//declaring a temporary new source char array
char[] ns = new char[3];
//check if source char array has more then 3 characters//
if (slen < 3) {
} else {
for (int i = 0; i < combs; i++) {
for (int j = 0; j < 3; j++) {
ns[j] = s[j + i];
}
if (cross_full_match(ns, t) == 1) {
x = x + 1;
}
}
}
x = ppc * x;
return x;
}
/**
*
* @param s
* @param t
* @return
*/
public static int cross_full_match(char[] s, char[] t) {
int z = t.length - s.length;
int x = 0;
if (s.length > t.length) {
return x;
} else {
for (int i = 0; i <= z; i++) {
for (int j = 0; j <= (s.length - 1); j++) {
if (s[j] == t[j + i]) {
// x=1 if any charecer matches
x = 1;
} else {
// if x=0 mean an character do not matches and loop break out
x = 0;
break;
}
}
if (x == 1) {
break;
}
}
}
return x;
}
public static String split_string(String s, int n) {
int index;
String temp;
temp = s;
String temp2 = null;
int temp3 = 0;
for (int i = 0; i < n; i++) {
int strlen = temp.length();
index = temp.indexOf(" ");
if (index < 0) {
index = strlen;
}
temp2 = temp.substring(temp3, index);
temp = temp.substring(index, strlen);
temp = temp.trim();
}
return temp2;
}
public static int word_count(String s) {
int x = 1;
int c;
s = s.trim();
if (s.isEmpty()) {
x = 0;
} else {
if (s.contains(" ")) {
for (;;) {
x++;
c = s.indexOf(" ");
s = s.substring(c);
s = s.trim();
if (s.contains(" ")) {
} else {
break;
}
}
}
}
return x;
}
}
只需将两个字符串作为lock_match（string1，string2）的参数提供，它将返回匹配的整数值。如果字符串的大小较大，则增加总名称变量大小 在代码中。 像int total = 1000 然后结果将从0到1000给出。 此代码区分大小写。 大写或小写两个字符串以摆脱这个问题。 源代码位于：lock match
答案 1 :(得分：1)
您可以使用Apache Commons Lang 3。
Maven依赖：
<dependency>
<groupId>org.apache.commons</groupId>
<artifactId>commons-lang3</artifactId>
<version>${commons.lang.version}</version>
</dependency>
@Test
public void test_stringDistance() throws Exception {
String teamName = "Partizn Belgrade";
String propositionName = "Partizan Belgrade";
// This one seems better
double distance = StringUtils.getJaroWinklerDistance(teamName, propositionName);
System.out.println(distance);
}
这是打印输出百分比，越大越好（100％是确切的）
答案 2 :(得分：0)
org.apache.commons.lang3.StringUtils.getJaroWinklerDistance(first, second) 从 commons-lang3:3.6 开始弃用
使用 new org.apache.commons.text.similarity.JaroWinklerDistance().apply(left, right) 而左和右分别代表第一和第二。见下面的maven依赖
<dependency>
<groupId>org.apache.commons</groupId>
<artifactId>commons-text</artifactId>
<version>{text-version}</version>
</dependency>