在 Java 中对数组中的字符串进行分类和分组

Question

我需要找出对字符串数组进行分类和分组的最佳方法。假设我有数组：

Resources[] = {tester1,tester2,solverC1,solverC2,solverS2,solverS1,tester3,tester4,system}

实际上它的字符串要多得多，但它就是这样的想法。我需要得到答案，在这种情况下我有资源：

Resource: tester || Quantity: 4
Resource: system || Quantity: 1
Resource: solver || Quantity: 4

使用 Levenshtein 距离是最好的主意吗？如果是，也许有些人有想法将数组分组并获取名称（没有数字或字母）并计算它们的数量？

Answer 1

您可以使用流进行映射：

String Resources[] = 
   new String[]{"tester1",
                "tester2","solverC1","solverC2",
                "solverS2","solverS1",
                "tester3","tester4","system"};

Map<String, List<String>> groups = Arrays.asList(Resources)
    .stream().collect(
            Collectors.groupingBy(
                    t -> t.contains("tester") ? "tester" : 
                        (t.contains("solver") ? "solver" : "system"),
                    Collectors.counting()));

这将 return 一张地图，结果如下：

{system=1, tester=4, solver=4}

可以使用以下方法使其更清洁：

private static String group(String t) {
    return t.contains("tester") ? "tester" : (t.contains("solver") ? "solver" : "system");
}

因此可以使用以下方式构建收集器：

Collectors.groupingBy(Main::group, , Collectors.counting())

Answer 2

您可以使用 java 流来获得带有值的漂亮地图。例如：

List<String>  res= Arrays.asList("tester1","tester2","solverC1","solverC2","solverS2","solverS1","tester1","tester4","system");

Map<String,Long> result=res.stream().collect( Collectors.groupingBy(s-> s.replaceAll("\d","").toString(),  Collectors.counting()));

将保留以下值：system="1", tester="4", solverS="2", solverC="2"

我使用了仅从字符串中删除数字的 groupBy 函数，但您可以在此处定义您想要的任何规则：

Collectors.groupingBy(s-> s.replaceAll("\d","").toString()

取决于您是否希望 solverS 不同于 solverC

工作示例：

public class MainClass {
    public static void main(String[] args) {
        System.out.println(new Date()+": Let's start our Whosebug helper project!");

                List<String>  res= Arrays.asList("tester1","tester2","solverC1","solverC2","solverS2","solverS1","tester1","tester4","system");


                Map<String,Long> reuslts=res.stream().collect( Collectors.groupingBy(s-> s.replaceAll("\d","").toString(),  Collectors.counting()));

                 StringBuilder sb = new StringBuilder();
                Iterator<Entry<String, Long>> iter = reuslts.entrySet().iterator();
                while (iter.hasNext()) {
                    Entry<String, Long> entry = iter.next();
                    sb.append(entry.getKey());
                    sb.append('=').append('"');
                    sb.append(entry.getValue());
                    sb.append('"');
                    if (iter.hasNext()) {
                        sb.append(',').append(' ');
                    }
                }
                System.out.println(sb.toString());

    }   

}

// It should print system="1", tester="4", solverS="2", solverC="2"

在 Java 中对数组中的字符串进行分类和分组

Classifying and grouping strings in array in Java

java

grouping

similarity

levenshtein-distance