如何比较两个集合然后用组合字符串过滤到新集合？

Question

我正在构建一些用于比较 2 哈希集的短代码。

SET 1 = noRek : [1234567892, 1234567891, 1234567890]

SET 2 = Source : [1234567890U0113, 1234567894B0111, 1234567890U0112, 1234567891B0111, 1234567890U0115, 1234567890U0114, 1234567892B0113, 1234567893B0111, 1234567890U0111, 1234567890B0111, 1234567892B0112, 1234567892B0111]

public class diff {
    public static void main(String args[]) {

        String filename = "C:\abc.txt";
        String filename2 = "C:\xyz.txt";
        HashSet<String> al = new HashSet<String>();
        HashSet<String> al1 = new HashSet<String>();
        HashSet<String> source = new HashSet<String>();
        HashSet<String> noRek = new HashSet<String>();
        HashSet<String> diff1 = new HashSet<String>();
        HashSet<String> diff2 = new HashSet<String>();
        String str = null;
        String str2 = null;
        Integer digitRek = 10;
        Integer digitTransaksi = 15;
        //GET REKDATA FROM TARGET
        try {
            String message = new Scanner(new File(filename2)).useDelimiter("\Z").next();
            for (int i = 0; i < message.length(); i += digitRek) {
               noRek.add(message.substring(i, Math.min(i + digitRek, message.length())));
            }
            System.out.println("noRek : " + noRek);
        } catch (Exception e) {
            e.printStackTrace();
        }
        try {
            String message2 = new Scanner(new File(filename)).useDelimiter("\Z").next();

            for (int i = 0; i < message2.length(); i += digitTransaksi) {
               source.add(message2.substring(i, Math.min(i + digitTransaksi, message2.length())));
            }
            System.out.println("Source : " + source);
        } catch (Exception e) {
            e.printStackTrace();
        }

        for (String str3 : source) {
            if (source.contains(noRek.substring(digitRek)) {
                diff1.add(str3);
            }

        }

       System.out.println("Final : " + diff1);

    }

我excpet diff1的输出是这样的

    SET 3 = [1234567890U0111, 1234567890U0112, 1234567890U0113,1234567890U0114, 1234567890U0115, 1234567890B0111, 1234567891B0111, 1234567892B0113, 1234567892B0112, 1234567892B0111]

但实际输出与 SET 2 相同。

简单来说，我需要将 SET 2 与组合进行比较，前 10 位是帐号，然后下一个字符 1 位是代码，然后是自动生成的其余数字。也就是说长度组合SET 2是15位，组合SET 1是10位，那么集合1就是账号的数据，我需要从集合2中的账号获取所有交易。

SET 1是所有账户的数据和 SET 2是交易组合的数据

Answer 1

您可以使用流和 filter

解决此问题

Set<String> diff1 = source.stream().filter(str -> {
    if (str.length() > 10) {
        String account = str.substring(0, 10);
        return noRek.contains(account);
    }
    return false;
}).collect(Collectors.toSet());

如何比较两个集合然后用组合字符串过滤到新集合？

How to compare two set then filter to new set with combination string?

java

hashset