想要计算字符串在Java中的出现次数
想要计算字符串在Java中的出现次数
提问于 2015-03-29 07:17:10
因此我有一个.txt文件，我使用以下命令调用它
String[] data = loadStrings("data/data.txt");该文件已经排序，基本上看起来如下所示：
Animal
Animal
Cat
Cat
Cat
Dog我希望在java中创建一个算法来计算排序列表，而不使用任何像Multisets或Maps/HashMaps这样的库。到目前为止，我已经成功地让它打印出出现频率最高的单词，如下所示：
ArrayList<String> words = new ArrayList();
int[] occurrence = new int[2000];
Arrays.sort(data);
for (int i = 0; i < data.length; i ++ ) {
words.add(data[i]); //Put each word into the words ArrayList
}
for(int i =0; i<data.length; i++) {
occurrence[i] =0;
for(int j=i+1; j<data.length; j++) {
if(data[i].equals(data[j])) {
occurrence[i] = occurrence[i]+1;
}
}
}
int max = 0;
String most_talked ="";
for(int i =0;i<data.length;i++) {
if(occurrence[i]>max) {
max = occurrence[i];
most_talked = data[i];
}
}
println("The most talked keyword is " + most_talked + " occuring " + max + " times.");我想要的不仅仅是出现频率最高的单词，可能是前5名或前10名。希望这已经足够清楚了。感谢您的阅读
回答 3
Stack Overflow用户
回答已采纳
发布于 2015-03-29 07:45:23
既然你说你不想使用某种数据结构，我认为你可以做这样的事情，但它的性能不佳。我通常更喜欢存储索引而不是值。
ArrayList<String> words = new ArrayList();
int[] occurrence = new int[2000];
Arrays.sort(data);
int nwords = 0;
occurrence[nwords]=1;
words.add(data[0]);
for (int i = 1; i < data.length; i ++ ) {
if(!data[i].equals(data[i-1])){ //if a new word is found
words.add(data[i]); //put it into the words ArrayList
nwords++; //increment the index
occurrence[nwords]=0; //initialize its occurrence counter
}
occurrence[nwords]++; //increment the occurrence counter
}
int max;
for(int k=0; k<5; k++){ //loop to find 5 times the most talked word
max = 0; //index of the most talked word
for(int i = 1; i<words.size(); i++) { //for every word
if(occurrence[i]>occurrence[max]) { //if it is more talked than max
max = i; //than it is the new most talked
}
}
println("The most talked keyword is " + words.get(max) + " occuring " + occurence[max] + " times.");
occurence[max]=0;
}每次我找到具有较高出现值的值时，我将其出现计数器设置为0，并再次重复该数组，这重复了5次。
Stack Overflow用户
发布于 2015-03-29 07:29:54
如果你不能使用Guava的Multiset，那么你可以自己实现一个等价的。基本上，您只需要创建一个Map<String, Integer>，它跟踪每个单词(键)的计数(值)。这意味着要改变这一点
ArrayList<String> words = new ArrayList<String>();
// ...
for (int i = 0; i < data.length; i ++ ) {
words.add(data[i]); //Put each word into the words ArrayList
}如下所示：
Map<String, Integer> words = new HashMap<String>();
// ...
for (String word : data) {
Integer count = words.get(word);
words.put(word, (count != null : count.intValue() + 1 ? 1));
}填充地图后，只需使用sort it by the values即可。
如果您也不能使用Map，您可以执行以下操作：
首先，为单词计数创建一个包装类：
public class WordCount implements Comparable<WordCount> {
private String word;
private int count;
public WordCount(String w, int c) {
this.word = w;
this.count = c;
}
public String getWord() {
return word;
}
public int getCount() {
return count;
}
public void incrementCount() {
count++;
}
@Override
public int compareTo(WordCount other) {
return this.count - other.count;
}
}然后，更改代码以在列表中存储WordCount实例(而不是String)：
ArrayList<WordCount> words = new ArrayList<WordCount>();
// ...
for (String word : data) {
WordCount wc = new WordCount(word, 1);
boolean wordFound = false;
for (WordCount existing : words) {
if (existing.getWord().equals(wc.getWord())) {
existing.incrementCount();
wordFound = true;
break;
}
}
if (!wordFound) {
words.add(wc);
}
}最后，填充List之后，只需使用Collections.sort()对其进行排序。这很容易，因为值对象实现了Comparable
Collections.sort(words, Collections.reverseOrder());Stack Overflow用户
发布于 2015-03-29 07:44:52
你可以试试像这样简单的东西..
int count = 0;
for( int i = 0; i < words.size(); i++ ){
System.out.printf("%s: ", words.get( i ));
for( int j = 0; j < words.size(); j++ ) {
if( words.get( i ).equals( words.get( j ) ) )
count++;
}
System.out.printf( "%d\n", count );
}页面原文内容由Stack Overflow提供。腾讯云小微IT领域专用引擎提供翻译支持
原文链接：
https://stackoverflow.com/questions/29323751
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