我是Java Strings的新手，问题是我想要计算String中特定单词的出现次数。假设我的字符串是：
i have a male cat. the color of male cat is Black
现在我不想分开它，所以我想搜索一个单词＆＃34; male cat＆＃34;。它在我的字符串中出现两次！
我正在尝试的是：
int c = 0;
for (int j = 0; j < text.length(); j++) {
if (text.contains("male cat")) {
c += 1;
}
}
System.out.println("counter=" + c);
答案 0 :(得分：35)
您可以使用以下代码：
String in = "i have a male cat. the color of male cat is Black";
int i = 0;
Pattern p = Pattern.compile("male cat");
Matcher m = p.matcher( in );
while (m.find()) {
i++;
}
System.out.println(i); // Prints 2
匹配"male cat"。
while(m.find())
表示，在m找到匹配项时，在循环内执行任何操作。
我将i的值增加i++，所以很明显，这会给出一个字符串male cat的数量。
答案 1 :(得分：10)
如果您只想要"male cat"的计数，那么我会这样做：
String str = "i have a male cat. the color of male cat is Black";
int c = str.split("male cat").length - 1;
System.out.println(c);
如果您想确保"female cat"不匹配，请在拆分正则表达式中使用\\b字边界：
int c = str.split("\\bmale cat\\b").length - 1;
答案 2 :(得分：9)
StringUtils有 CountMatches 方法来计算一个String在另一个String中出现的次数。
String input = "i have a male cat. the color of male cat is Black";
int occurance = StringUtils.countMatches(input, "male cat");
System.out.println(occurance);
答案 3 :(得分：4)
Java 8版本：
public static long countNumberOfOccurrencesOfWordInString(String msg, String target) {
return Arrays.stream(msg.split("[ ,\\.]")).filter(s -> s.equals(target)).count();
}
答案 4 :(得分：3)
Java 8版本。
System.out.println(Pattern.compile("\\bmale cat")
.splitAsStream("i have a male cat. the color of male cat is Black")
.count()-1);
答案 5 :(得分：2)
使用indexOf ...
public static int count(String string, String substr) {
int i;
int last = 0;
int count = 0;
do {
i = string.indexOf(substr, last);
if (i != -1) count++;
last = i+substr.length();
} while(i != -1);
return count;
}
public static void main (String[] args ){
System.out.println(count("i have a male cat. the color of male cat is Black", "male cat"));
}
这将显示：2
count（）的另一个实现，只需一行：
public static int count(String string, String substr) {
return (string.length() - string.replaceAll(substr, "").length()) / substr.length() ;
}
答案 6 :(得分：2)
此static方法确实返回另一个字符串上字符串的出现次数。
/**
* Returns the number of appearances that a string have on another string.
*
* @param source a string to use as source of the match
* @param sentence a string that is a substring of source
* @return the number of occurrences of sentence on source
*/
public static int numberOfOccurrences(String source, String sentence) {
int occurrences = 0;
if (source.contains(sentence)) {
int withSentenceLength = source.length();
int withoutSentenceLength = source.replace(sentence, "").length();
occurrences = (withSentenceLength - withoutSentenceLength) / sentence.length();
}
return occurrences;
}
<强>试验：
String source = "Hello World!";
numberOfOccurrences(source, "Hello World!"); // 1
numberOfOccurrences(source, "ello W"); // 1
numberOfOccurrences(source, "l"); // 3
numberOfOccurrences(source, "fun"); // 0
numberOfOccurrences(source, "Hello"); // 1
public static int numberOfOccurrences(String source, String sentence) {
return (source.contains(sentence)) ? (source.length() - source.replace(sentence, "").length()) / sentence.length() : 0;
}
答案 7 :(得分：2)
为什么不递归？
public class CatchTheMaleCat {
private static final String MALE_CAT = "male cat";
static int count = 0;
public static void main(String[] arg){
wordCount("i have a male cat. the color of male cat is Black");
System.out.println(count);
}
private static boolean wordCount(String str){
if(str.contains(MALE_CAT)){
count++;
return wordCount(str.substring(str.indexOf(MALE_CAT)+MALE_CAT.length()));
}
else{
return false;
}
}
}
答案 8 :(得分：1)
用空字符串替换需要计数的字符串，然后使用不带字符串的长度来计算出现次数。
public int occurrencesOf(String word)
{
int length = text.length();
int lenghtofWord = word.length();
int lengthWithoutWord = text.replace(word, "").length();
return (length - lengthWithoutWord) / lenghtofWord ;
}
答案 9 :(得分：1)
公共类TestWordCount {
public static void main(String[] args) {
int count = numberOfOccurences("Alice", "Alice in wonderland. Alice & chinki are classmates. Chinki is better than Alice.occ");
System.out.println("count : "+count);
}
public static int numberOfOccurences(String findWord, String sentence) {
int length = sentence.length();
int lengthWithoutFindWord = sentence.replace(findWord, "").length();
return (length - lengthWithoutFindWord)/findWord.length();
}
}
答案 10 :(得分：1)
这将有效
int word_count(String text,String key){
int count=0;
while(text.contains(key)){
count++;
text=text.substring(text.indexOf(key)+key.length());
}
return count;
}
答案 11 :(得分：0)
此处填写示例，
package com.test;
import java.util.HashMap;
import java.util.Iterator;
import java.util.Map;
public class WordsOccurances {
public static void main(String[] args) {
String sentence = "Java can run on many different operating "
+ "systems. This makes Java platform independent.";
String[] words = sentence.split(" ");
Map<String, Integer> wordsMap = new HashMap<String, Integer>();
for (int i = 0; i<words.length; i++ ) {
if (wordsMap.containsKey(words[i])) {
Integer value = wordsMap.get(words[i]);
wordsMap.put(words[i], value + 1);
} else {
wordsMap.put(words[i], 1);
}
}
/*Now iterate the HashMap to display the word with number
of time occurance */
Iterator it = wordsMap.entrySet().iterator();
while (it.hasNext()) {
Map.Entry<String, Integer> entryKeyValue = (Map.Entry<String, Integer>) it.next();
System.out.println("Word : "+entryKeyValue.getKey()+", Occurance : "
+entryKeyValue.getValue()+" times");
}
}
}
答案 12 :(得分：0)
对于scala，只有1行
def numTimesOccurrenced（text：String，word：String）= text.split（word）.size-1
答案 13 :(得分：0)
公共类WordCount {
public static void main(String[] args) {
// TODO Auto-generated method stub
String scentence = "This is a treeis isis is is is";
String word = "is";
int wordCount = 0;
for(int i =0;i<scentence.length();i++){
if(word.charAt(0) == scentence.charAt(i)){
if(i>0){
if(scentence.charAt(i-1) == ' '){
if(i+word.length()<scentence.length()){
if(scentence.charAt(i+word.length()) != ' '){
continue;}
}
}
else{
continue;
}
}
int count = 1;
for(int j=1 ; j<word.length();j++){
i++;
if(word.charAt(j) != scentence.charAt(i)){
break;
}
else{
count++;
}
}
if(count == word.length()){
wordCount++;
}
}
}
System.out.println("The word "+ word + " was repeated :" + wordCount);
}
}
答案 14 :(得分：0)
简单的解决方案就在这里 -
下面的代码使用HashMap，因为它将维护键和值。所以这里的键将是单词，值将被计数（在给定的字符串中出现一个单词）。
public class WordOccurance
{
public static void main(String[] args)
{
HashMap<String, Integer> hm = new HashMap<>();
String str = "avinash pande avinash pande avinash";
//split the word with white space
String words[] = str.split(" ");
for (String word : words)
{
//If already added/present in hashmap then increment the count by 1
if(hm.containsKey(word))
{
hm.put(word, hm.get(word)+1);
}
else //if not added earlier then add with count 1
{
hm.put(word, 1);
}
}
//Iterate over the hashmap
Set<Entry<String, Integer>> entry = hm.entrySet();
for (Entry<String, Integer> entry2 : entry)
{
System.out.println(entry2.getKey() + " "+entry2.getValue());
}
}
}
答案 15 :(得分：0)
public int occurrencesOf(String word) {
int length = text.length();
int lenghtofWord = word.length();
int lengthWithoutWord = text.replaceAll(word, "").length();
return (length - lengthWithoutWord) / lenghtofWord ;
}
答案 16 :(得分：0)
一旦找到该术语，您需要将其从正在处理的字符串中删除，以便再次解决不同的问题，使用indexOf()和substring()，您不需要要做的包含检查长度时间
答案 17 :(得分：0)
我在这里有另一种方法：
String description = "hello india hello india hello hello india hello";
String textToBeCounted = "hello";
// Split description using "hello", which will return
//string array of words other than hello
String[] words = description.split("hello");
// Get number of characters words other than "hello"
int lengthOfNonMatchingWords = 0;
for (String word : words) {
lengthOfNonMatchingWords += word.length();
}
// Following code gets length of `description` - length of all non-matching
// words and divide it by length of word to be counted
System.out.println("Number of matching words are " +
(description.length() - lengthOfNonMatchingWords) / textToBeCounted.length());
答案 18 :(得分：0)
我们可以通过多种方式计算子串的出现： -
public class Test1 {
public static void main(String args[]) {
String st = "abcdsfgh yfhf hghj gjgjhbn hgkhmn abc hadslfahsd abcioh abc a ";
count(st, 0, "a".length());
}
public static void count(String trim, int i, int length) {
if (trim.contains("a")) {
trim = trim.substring(trim.indexOf("a") + length);
count(trim, i + 1, length);
} else {
System.out.println(i);
}
}
public static void countMethod2() {
int index = 0, count = 0;
String inputString = "mynameiskhanMYlaptopnameishclMYsirnameisjasaiwalmyfrontnameisvishal".toLowerCase();
String subString = "my".toLowerCase();
while (index != -1) {
index = inputString.indexOf(subString, index);
if (index != -1) {
count++;
index += subString.length();
}
}
System.out.print(count);
}}
答案 19 :(得分：0)
子串的出现有很多种方式，主题有两种： -
public class Test1 {
public static void main(String args[]) {
String st = "abcdsfgh yfhf hghj gjgjhbn hgkhmn abc hadslfahsd abcioh abc a ";
count(st, 0, "a".length());
}
public static void count(String trim, int i, int length) {
if (trim.contains("a")) {
trim = trim.substring(trim.indexOf("a") + length);
count(trim, i + 1, length);
} else {
System.out.println(i);
}
}
public static void countMethod2() {
int index = 0, count = 0;
String inputString = "mynameiskhanMYlaptopnameishclMYsirnameisjasaiwalmyfrontnameisvishal".toLowerCase();
String subString = "my".toLowerCase();
while (index != -1) {
index = inputString.indexOf(subString, index);
if (index != -1) {
count++;
index += subString.length();
}
}
System.out.print(count);
}}
答案 20 :(得分：0)
这应该是一个更快的非正则表达式解决方案 （注意 - 不是Java程序员）
String str = "i have a male cat. the color of male cat is Black";
int found = 0;
int oldndx = 0;
int newndx = 0;
while ( (newndx=str.indexOf("male cat", oldndx)) > -1 )
{
found++;
oldndx = newndx+8;
}
答案 21 :(得分：0)
如果您找到要搜索的字符串，则可以继续查找该字符串的长度（如果您在aaaa中搜索aa，则将其视为2次）。
int c=0;
String found="male cat";
for(int j=0; j<text.length();j++){
if(text.contains(found)){
c+=1;
j+=found.length()-1;
}
}
System.out.println("counter="+c);
答案 22 :(得分：0)
字符串在循环遍历时始终包含该字符串。你不想要++，因为现在正在做的只是获取字符串的长度，如果它包含＆＃34; ＆＃34;雄猫＆＃34;
您需要indexOf（）/ substring（）
有点得到我说的话吗？