HashBasedTable表<string，string，double> HashMap <string，double>

Question

我有HashMap<String,Double>hm1和guava table Table<String, String, Double> employeeYearsOfService

HashMap hm1

fatima |0.97
AT&T   |0.96

表employeeYearsOfService

 Google={Bill Smith=1.75, Stacy Lerner=11.5}, 
 Microsoft={Bill Smith=13.2,Stacy Lerner=3.5}, 
 AT&T={Bill Smith=2.0, Stacy Lerner=1.4},
 fatima={Bill Smith=1.0, Stacy Lerner=2.0}

表格研究

fatima={Bill Smith=1.0, Stacy Lerner=2.0}
AT&T={Bill Smith=2.0, Stacy Lerner=1.4}

我希望通过创建新表Table<String, String, Double> results = HashBasedTable.create()包含一行employeeYearsOfService来获得此结果 谁与HashMap hm1具有相同的密钥（这是我的问题）

这张照片是为了解莫  

我的代码

表番石榴

    Table<String, String, Double> employeeYearsOfService = 
    HashBasedTable.create();

    employeeYearsOfService.put("AT&T", "Stacy Lerner", 1.4);
    employeeYearsOfService.put("Microsoft", "Stacy Lerner", 3.5);
    employeeYearsOfService.put("Microsoft", "Bill Smith", 13.2);
    employeeYearsOfService.put("Google", "Stacy Lerner", 11.5);

    employeeYearsOfService.put("AT&T", "Bill Smith", 2.0);
    employeeYearsOfService.put("Google", "Bill Smith", 1.75);
    employeeYearsOfService.put("fatima", "Bill Smith", 1.0);
    employeeYearsOfService.put("fatima", "Stacy Lerner", 2.0);

hashmap hm1

HashMap<String, Double> hm = new HashMap<String, Double>();
    HashMap<String, Double> hm1 = new HashMap<String, Double>();
    System.out.println(employeeYearsOfService);
    for (String key : employeeYearsOfService.rowKeySet()) {

        for (Entry<String, Double> employee : 
        employeeYearsOfService.row(key).entrySet()) {
            sum += employee.getValue() * operatCible.get(k);
            sum2 += employee.getValue() * employee.getValue();
            vect1 += operatCible.get(k) * operatCible.get(k);
            Result = (sum / (sqrt(sum2) * sqrt(vect1)));
            k++;

        }

        hm.put(key, Result);

        k = 0;
        sum = 0.0;
        sum2 = 0.0;
        vect1 = 0.0;
        Result = 0.0;

    }
    System.out.println(hm);
    Set<Entry<String, Double>> set = hm.entrySet();
    List<Entry<String, Double>> list = new ArrayList<Entry<String, Double>>
     (set);
    Collections.sort(list, new Comparator<Map.Entry<String, Double>>() {
        public int compare(Map.Entry<String, Double> o1,
                Map.Entry<String, Double> o2) {
            return o2.getValue().compareTo(o1.getValue());
        }
    });

    System.out.println(list);
    System.out.println("le K nn");
    for (Entry<String, Double> entry : list.subList(0, 2)) {
        hm1.put(entry.getKey(), entry.getValue());


    }

新表

循环

Table<String, String, Double> results = HashBasedTable.create();
    System.out.println(hm1);
    for (Entry<String, Double> entry : list) {
       if(entry.getKey().equals(employeeYearsOfService.rowKeySet())){
           results.put(employeeYearsOfService.row(entry.getKey())));
           // how i do it
       }


    }

非常感谢

Answer 1

一种方法是从表中删除您不想要的元素：

employeeYearsOfService.rowKeySet()
    .removeIf(key -> !hm1.containsKey(key));

我在这里使用Table.rowKeySet method从表中获取行键集。这个集合有界到原始表，这意味着当从该集合中删除一个元素时，将从表中删除整行（具有相同的键）。这就是我对Collection.removeIf method所做的事情，如果true地图中没有该密钥，则其谓词会返回hm1。