编码字符串数组

Question

编辑：[学校作业]

所以，我想只用 0 和 1 对单词进行编码。

0 = 单词不存在
1 = 出现的单词

我的字典对应：

String[] dictionary = {"Hello", "I", "am", "Lukas", "and", "Jonas", "play", "football"};

例如：如果我对这些词进行编码...

String[] listOfWords = {"Hello", "play" "football"};

我必须有以下数组：

int[] wordsEncode = {1,0,0,0,0,0,1,1};

您可以看到“Hello”出现了，“I”“am”“Lukas”“和”“Jonas”没有出现。最后，出现了“play”和“football”。
我们必须保持字典的顺序，这是我代码中的问题。
我真的不知道如何解决这个问题（使用第二个 for 循环？）？
我认为 wordEncode[i] 是我的错误，但如何解决呢？

这是我的代码：

class Dictionary {

    /**
     * Array words Dictionary
     */
    String[] dictionary;
    
    /**
     * Maximum of words MAX_WORDS
     */
    final int MAX_WORDS = 50;
    
    /**
     * Number of words in the dictionary
     */
    int numberWordsDictionary;

    /**
     * Constructor
     */
    Dictionary() {
        dictionary = new String[MAX_WORDS];
        numberWordsDictionary = 0;
    }

int[] encoder(String[] listOfWords) {
    int[] wordsEncode = new int[numberWordsDictionary];
    
    StringBuilder builder = new StringBuilder();
    for(String word : dictionary) {
        builder.append(word);
    }
    String dictionaryString = builder.toString();     
    
    for(int i = 0; i < listOfWords.length; i++) {
        if(dictionaryString.contains(listOfWords[i])) {
            wordsEncode[i] = 1;
        } else {
            wordsEncode[i] = 0;
        }
    }
    return wordsEncode;
}

}

抱歉缩进（与我的 Java IDE 不同）:(
谢谢！

Answer 1

/* This approach is wrong, the combined string could catch words that are 
  part of the ending of one word and part of the beginning of another but 
  not actually a word in the dictionary. For instance, if you had
  "arch" and "attach" in your dictionary, testing for "chat" would return true
*/
/*
    StringBuilder builder = new StringBuilder();
    for(String word : dictionary) {
        builder.append(word);
    }
    String dictionaryString = builder.toString();     
*/    
    for(int i = 0; i < listOfWords.length; i++) {
      boolean found = false;
      for (int j = 0; j < dictionary.length; j++) {
        if (dictionary[j].equalslistOfWords[i]) {
          found = true;
        }
      }
      if (found) {
        wordsEncode[i] = 1;
      } else {
        wordsEncode[i] = 0;
      }
      // you can also do wordsEncode[i] = found ? 1 : 0;
    }
    return wordsEncode;
}

Answer 2

循环输入单词。对于每个输入单词，查看您的目标单词列表是否包含该特定单词。如果是这样，请将 1 添加到您的结果列表中。如果不是，则加零。

我使用更方便的集合，但您可以对数组执行相同的方法。

List< String > input = List.of( "Hello", "I", "am", "Lukas", "and", "Jonas", "play", "football" ) ;
List< String > targets = List.of( "Hello", "play" "football" ) ;
List< Integers > hits = new ArrayList<>() ;
for( String s : input )
{
    int i = targets.contains( s ) ? 1 : 0 ;
    hits.add( i ) ;
}

Answer 3

使用两级嵌套循环，您应该检查dictionary[]的每个元素是否在listOfWords[]中，如果是，则将wordsEncode[]中相应索引处的值更新为1。

import java.util.Arrays;

public class Main {
    public static void main(String[] args) {
        String[] dictionary = { "Hello", "I", "am", "Lukas", "and", "Jonas", "play", "football" };
        String[] listOfWords = { "Hello", "play", "football" };
        int[] wordsEncode = new int[dictionary.length];

        for (int i = 0; i < dictionary.length; i++) {
            boolean found = false;
            for (String s : listOfWords) {
                if (s.equals(dictionary[i])) {
                    found = true;
                    break;
                }
            }
            if (found) {
                wordsEncode[i] = 1;
            }
        }

        // Display the result
        System.out.println(Arrays.toString(wordsEncode));
    }
}

输出：

[1, 0, 0, 0, 0, 0, 1, 1]

Answer 4

您在这里所做的是遍历字典数组并将单词添加到 StringBuilder 以检查您在 listOfWords 数组中获得的某个单词是否在 StringBuilder 中。但是有一个更好的解决方案，您可以创建一个嵌套循环，比较 listOfWords 数组和字典数组的每个元素，如果找到匹配项，它将第二个循环索引处的编码数组值设置为 1：

int[] encoder(String[] listOfWords) {
    int[] wordsEncode = new int[numberWordsDictionary];

    for (int i = 0; i < listOfWords.length; i++) {
        for (int j = 0; j < numberWordsDictionary; j++) {

            if (listOfWords[i].equals(dictionary[j])) {
                wordsEncode[j] = 1;
                break;
            }
        }
    }
    return wordsEncode;
}