88 lines
3.0 KiB
Java
88 lines
3.0 KiB
Java
package Others;
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import java.io.*;
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import java.util.*;
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/* display the most frequent K words in the file and the times it appear
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in the file – shown in order (ignore case and periods) */
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public class TopKWords {
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static class CountWords {
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private String fileName;
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public CountWords(String fileName) {
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this.fileName = fileName;
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}
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public Map<String, Integer> getDictionary() {
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Map<String, Integer> dictionary = new HashMap<>();
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FileInputStream fis = null;
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try {
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fis = new FileInputStream(fileName); // open the file
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int in = 0;
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String s = ""; // init a empty word
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in = fis.read(); // read one character
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while (-1 != in) {
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if (Character.isLetter((char) in)) {
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s += (char) in; //if get a letter, append to s
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} else {
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// this branch means an entire word has just been read
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if (s.length() > 0) {
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// see whether word exists or not
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if (dictionary.containsKey(s)) {
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// if exist, count++
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dictionary.put(s, dictionary.get(s) + 1);
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} else {
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// if not exist, initiate count of this word with 1
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dictionary.put(s, 1);
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}
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}
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s = ""; // reInit a empty word
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}
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in = fis.read();
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}
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return dictionary;
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} catch (IOException e) {
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e.printStackTrace();
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} finally {
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try {
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// you always have to close the I/O streams
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fis.close();
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} catch (IOException e) {
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e.printStackTrace();
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}
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}
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return null;
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}
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}
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public static void main(String[] args) {
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// you can replace the filePath with yours
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CountWords cw = new CountWords("/Users/lisanaaa/Desktop/words.txt");
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Map<String, Integer> dictionary = cw.getDictionary(); // get the words dictionary: {word: frequency}
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// we change the map to list for convenient sort
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List<Map.Entry<String, Integer>> list = new ArrayList<>(dictionary.entrySet());
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// sort by lambda valueComparator
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list.sort(Comparator.comparing(
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m -> m.getValue())
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);
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Scanner input = new Scanner(System.in);
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int k = input.nextInt();
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while (k > list.size()) {
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System.out.println("Retype a number, your number is too large");
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input = new Scanner(System.in);
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k = input.nextInt();
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}
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for (int i = 0; i < k; i++) {
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System.out.println(list.get(list.size() - i - 1));
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}
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}
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}
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