解析两个文件并生成员工数据 [关闭]

Question

假设我们有两个使用逗号分隔值的日志文件。 file1.txt 代表employee id 和employee name，file2.txt 代表他关联的employee id 和projects。 file1 具有唯一条目。 file2 会有多对多的关系。如果新员工没有分配任何项目，则在file2.txt 中没有任何条目。

File1.txt:(EmpId, EmpName)
1,abc
2,ac
3,bc
4,acc
5,abb
6,bbc
7,aac
8,aba
9,aaa

File2.txt: (EmpId, ProjectId)
1,102
2,102
1,103
3,101
5,102
1,103
2,105
2,200
9,102

Find the each employee has been assigned to number of projects. For New employees if they dont have any projects print 0;
Output:
1=3
2=3
3=1
4=0
5=1
6=0
7=0
8=0
9=1

我使用 BufferedReader 从file1 中读取一行并将其与file2 中的每一行进行比较。下面是我的代码，

public static void main(String[] args) throws IOException {
        // TODO Auto-generated method stub
        BufferedReader file1 = new BufferedReader(new FileReader("file1.txt"));
        BufferedReader file2 = new BufferedReader(new FileReader("file2.txt"));
        BufferedReader file3 = new BufferedReader(new FileReader("file2.txt"));
        HashMap<String,Integer> empProjCount = new HashMap<String, Integer>();
        int lines =0;
        while (file2.readLine() != null) 
            lines++;
        String line1 = file1.readLine();
        String[] line_1 = line1.split(",");
        String line2 = file3.readLine();
        String[] line_2 = line2.split(",");
        while(line1 != null && line2 != null)
        {
            int count = 0;
            for(int i=1;i<=lines+1 && line2 != null;i++)
            {
            if(line_1[0].equals(line_2[0]))
            {
                count++;
            }
            line2 = file3.readLine();
            if(line2 != null){
                line_2 = line2.split(",");
            }
            }
            file3 = new BufferedReader(new FileReader("file2.txt"));
            empProjCount.put(line_1[0], count);
            line1 = file1.readLine();
            if(line1 != null) line_1 = line1.split(",");
            line2 = file3.readLine();
            if(line2 != null) line_2 = line2.split(",");
        }
        System.out.println(empProjCount); 

我的问题是，

1. 有什么方法可以优化它小于 O(n^2)，而不使用任何额外的空间？

2. 我使用 3 BufferedReader 读取file2.txt，因为一旦我们读取一行，它就会移动到下一行。有没有其他选项可以标记当前行？

3. 如果我们将其视为一个表，那么查询上述场景的最佳方法是什么？

Answer 1

对于 1：是的。

对于 2：是的：

我会在两次迭代中完成：

1. 迭代 ID (file1) 并初始化映射 (empId, projectCounter)

2. 迭代项目 (file2) 并为每一行更新 (projectCounter++) 地图中的相应条目。

这样，您的执行时间几乎是线性的（对于 file1 和 file2 的大小）。

Answer 2

从file 1 收集所有员工ID 的Map，并将其初始化为包含0 用于项目计数。

    // Build my map of all employees.
    Map<Integer, Integer> employeeProjectCount = Arrays.stream(file1)
            // Get empId - Split on comma, take the first field and convert to integer.
            .map(s -> Integer.valueOf(s.split(",")[0]))
            // Build a Map for the results.
            .collect(Collectors.toMap(
                    // Key is emp ID.
                    empId -> empId,
                    // Value starts at zero.
                    empId -> ZERO
            ));

浏览file 2 数项目。

    // Walk the projects list.
    Arrays.stream(file2)
            // Get empId - Split on comma, take the first field and convert to integer (again).
            .map(s -> Integer.valueOf(s.split(",")[0]))
            // Count the projects.
            .forEach(empId -> employeeProjectCount.put(empId, employeeProjectCount.get(empId)+1));

打印出来：

    // Print it.
    System.out.println(employeeProjectCount);

给予

{1=3, 2=3, 3=1, 4=0, 5=1, 6=0, 7=0, 8=0, 9=1}

顺便说一句：我以String[]s 的身份处理这些文件。

String[] file1 = {
        "1,abc",
        "2,ac",
        "3,bc",
        "4,acc",
        "5,abb",
        "6,bbc",
        "7,aac",
        "8,aba",
        "9,aaa",};
String[] file2 = {
        "1,102",
        "2,102",
        "1,103",
        "3,101",
        "5,102",
        "1,103",
        "2,105",
        "2,200",
        "9,102",
};

Answer 3

使用Files.lines 和正则表达式：

Pattern employeePattern = Pattern.compile("(?<id>\\d+),(?<name>\\s+)");
Set<String> employees = Files.lines(Paths.get("file1.txt"));
    .map(employeePattern::matcher).filter(Matcher::matches)
    .map(m -> m.group("id")).collect(Collectors.toSet());

Pattern projectPattern = Pattern.compile("(?<emp>\\d+),(?<proj>\\d+)");
Map<String,Long> projects = Files.lines(Paths.get("file2.txt"))
    .map(projectPattern::matcher).filter(Matcher::matches)
    .collect(Collectors.groupingBy(m -> m.group("emp"), Collectors.counting());

并打印结果：

employees.stream()
    .map(emp -> emp + "=" + projects.getOrDefault(emp, 0L))
    .forEach(System.out::println);