二维数组中一维数组的搜索索引并逐行返回java

我有一个问题，我有一个一维数组

String[] purposeAll= {"P1","P2"};

二维矩阵与阵列

String[][] purpose = { 
        { "P1", "a", "b", "c", "d", "e", "f", "h" },
        { "P2", "a", "b", "c", "d", "e", "g", "i" },
        { "P3", "l", "m", "n", "o", "p", "q" } };

我需要在数组二维中搜索数组一维的每个索引，结果将得到：

arrayOfPurpose[][]={ 
        { "P1", "a", "b", "c", "d", "e", "f", "h" },
        { "P2", "a", "b", "c", "d", "e", "g", "i" }  };

我的代码是

public static void ArrayOfPurpose(String[][] purpose, String[] purposeAll) {

String[][][] arrayOfPurpose = new String[purposeAll.length][purpose.length][];
    ArrayList<String[]> list = new ArrayList<String[]>();

    for (int i = 0; i < purpose.length; i++) {
        list.add(purpose[i]);
    }
    String[] tmp;
    for (int k = 0; k < purposeAll.length; k++) {
        arrayOfPurpose[k]= new String[purposeAll.length][];
        System.out.print("ok");
        for (int i = 0; i < list.size(); i++) {

            tmp = list.get(i);
            arrayOfPurpose[i]= new String[purposeAll.length][tmp.length];
            for (int j = 0; j < purpose[i].length; j++) {
                 if (tmp[0] == (purposeAll[k])) {
                    arrayOfPurpose[k][i][j] = tmp[j];
                  System.out.println(arrayOfPurpose[k][i][j]);
                }

            }

        }
    }
}

---

非常感谢！！：

Java8流解决方案：

public class blah {

    static String[] purposeAll= {"P1","P2"};

    static String[][] purpose = { 
            { "P1", "a", "b", "c", "d", "e", "f", "h" },
            { "P2", "a", "b", "c", "d", "e", "g", "i" },
            { "P3", "l", "m", "n", "o", "p", "q" } };
    public static String[][] arrayOfPurpose(String[][] purpose, String[] purposeAll) {
        return Arrays.asList(purpose).stream()
            .filter(arr -> Arrays.asList(purposeAll).contains(arr[0]))
            .toArray(String[][]::new);
    }
    public static void main(String[] args) {
        System.out.println(Arrays.deepToString(arrayOfPurpose(purpose, purposeAll)));
        // prints [[P1, a, b, c, d, e, f, h], [P2, a, b, c, d, e, g, i]]
    }
}

---

Java8流解决方案：

public class blah {

    static String[] purposeAll= {"P1","P2"};

    static String[][] purpose = { 
            { "P1", "a", "b", "c", "d", "e", "f", "h" },
            { "P2", "a", "b", "c", "d", "e", "g", "i" },
            { "P3", "l", "m", "n", "o", "p", "q" } };
    public static String[][] arrayOfPurpose(String[][] purpose, String[] purposeAll) {
        return Arrays.asList(purpose).stream()
            .filter(arr -> Arrays.asList(purposeAll).contains(arr[0]))
            .toArray(String[][]::new);
    }
    public static void main(String[] args) {
        System.out.println(Arrays.deepToString(arrayOfPurpose(purpose, purposeAll)));
        // prints [[P1, a, b, c, d, e, f, h], [P2, a, b, c, d, e, g, i]]
    }
}

---

使用问题中的内容，以非Java8的方式，您可以这样做：

for (String valueToCheck : purposeAll) {
    for (String[] values : purpose) {
        if (values[0].equals(valueToCheck)) {
            list.add(values);
        }
    }
}

String[][] arrayOfPurpose = new String[list.size()][];

int index = 0;

for (String[] a : list) {
    arrayOfPurpose[index++] = a;
}


System.out.println("Size of arrayOfPurpose: " + arrayOfPurpose.length);

---

使用问题中的内容，以非Java8的方式，您可以这样做：

for (String valueToCheck : purposeAll) {
    for (String[] values : purpose) {
        if (values[0].equals(valueToCheck)) {
            list.add(values);
        }
    }
}

String[][] arrayOfPurpose = new String[list.size()][];

int index = 0;

for (String[] a : list) {
    arrayOfPurpose[index++] = a;
}


System.out.println("Size of arrayOfPurpose: " + arrayOfPurpose.length);

---

我希望我正确地理解了这个问题

    String[] purposeAll= {"P1","P2"};
    //String[] purposeAll= {"P3","P1"}; Testing Stuff.

    String[][] purpose = { 
    { "P1", "a", "b", "c", "d", "e", "f", "h" },
    { "P2", "a", "b", "c", "d", "e", "g", "i" },
    { "P3", "l", "m", "n", "o", "p", "q" } };

    List<String[]> arrayOfPurposes = new ArrayList<>();
    //Looping through the purposeAll to check purpose's first 
    //entry and seeing if it equals the first purposeAll value
    for(int i = 0; i < purpose.length; i++){
        for(int j = 0; j < purposeAll.length; j++){
            if(purpose[i][0].equals(purposeAll[j])){
                arrayOfPurposes.add(purpose[i]);
            }
        }
    }

    //One way to get the results into a String[][]
    String[][] result = {arrayOfPurposes.get(0),arrayOfPurposes.get(1)};

    //Displaying to console
    for(int i = 0; i < result.length; i++){
        for(int j = 0; j < result[i].length; j++){
            System.out.print(result[i][j]);
        }
        System.out.println();
    }

---

我觉得您应该使用字符串[]的列表而不是字符串[]，因为结果的初始化非常简单。此外，在大多数情况下，列表更有效率和帮助

我希望我正确理解了这个问题

    String[] purposeAll= {"P1","P2"};
    //String[] purposeAll= {"P3","P1"}; Testing Stuff.

    String[][] purpose = { 
    { "P1", "a", "b", "c", "d", "e", "f", "h" },
    { "P2", "a", "b", "c", "d", "e", "g", "i" },
    { "P3", "l", "m", "n", "o", "p", "q" } };

    List<String[]> arrayOfPurposes = new ArrayList<>();
    //Looping through the purposeAll to check purpose's first 
    //entry and seeing if it equals the first purposeAll value
    for(int i = 0; i < purpose.length; i++){
        for(int j = 0; j < purposeAll.length; j++){
            if(purpose[i][0].equals(purposeAll[j])){
                arrayOfPurposes.add(purpose[i]);
            }
        }
    }

    //One way to get the results into a String[][]
    String[][] result = {arrayOfPurposes.get(0),arrayOfPurposes.get(1)};

    //Displaying to console
    for(int i = 0; i < result.length; i++){
        for(int j = 0; j < result[i].length; j++){
            System.out.print(result[i][j]);
        }
        System.out.println();
    }

---

---

我觉得您应该使用字符串[]的列表而不是字符串[]，因为结果的初始化非常简单。此外，在大多数情况下，列表更有效率和帮助

我只是想说清楚。。。您希望从purposeAll[]中获取每个值，并检查其值是否是二维数组值列表中的第一个值。如果是这样，将其拉出并将整个字符串[]放入arrayOfPurpose？为什么arrayOfPurpose 3d出现在您的代码中？@AdrianLeonhard-这不是3d问题是什么？到目前为止，您的代码有什么问题？@Ascalonian String[][]arrayOfPurpose=新字符串[purposeAll.length][purpose.length][]；我只是想说清楚。。。您希望从purposeAll[]中获取每个值，并检查其值是否是二维数组值列表中的第一个值。如果是这样，将其拉出并将整个字符串[]放入arrayOfPurpose？为什么arrayOfPurpose 3d出现在您的代码中？@AdrianLeonhard-这不是3d问题是什么？到目前为止，您的代码有什么问题？@Ascalonian String[][]arrayOfPurpose=新字符串[purposeAll.length][purpose.length][]；我得到的结果是：P3lmnopq，但不是P1和P2。我得到的结果是：P3lmnopq，但不是P1和P2。没有数组。asList.contains？伙计，这和我的想法差不多，但我面临的问题是无法插入数组[][]：@DaraTith-看看新的评论。我用这种逻辑使它工作起来：-@AdrianLeonhard-有很多方法可以检查。由于OP似乎正在学习这方面的知识，我想我会给出一个更直接的示例no Arrays.asList.contains？buddy，这几乎与mind相似，但我面临的问题是无法插入数组[]：@DaraTith-看看新的评论。我用这种逻辑使它工作起来：-@AdrianLeonhard-有很多方法可以检查。由于OP似乎正在学习这方面的知识，我想我会给出一个更直接的例子