无法计算字符串的每个部分中有多少个唯一日期可用

Question

我使用换行符 ('\n') 将字符串分成三部分。我想要实现的输出：计算字符串的每个部分中有多少个唯一日期可用。
根据下面的代码，第一部分包含两个唯一日期，第二部分包含两个，第三部分包含三个唯一日期。所以输出应该是这样的：2,2,3,
但是在运行下面的代码之后，我得到了这个输出：5,5,5,5,1,3,1,
我如何获得输出：2,2,3,
提前致谢。

        String strH;
        String strT = null;
        StringBuilder sbE = new StringBuilder();
        String strA = "2021-03-02,2021-03-02,2021-03-02,2021-03-02,2021-03-02,2021-03-11,2021-03-11,2021-03-11,2021-03-11,2021-03-11," + '\n' +
                "2021-03-07,2021-03-07,2021-03-07,2021-03-07,2021-03-07,2021-03-15,2021-03-15,2021-03-15,2021-03-15,2021-03-15," + '\n' +
                "2021-03-02,2021-03-09,2021-03-07,2021-03-09,2021-03-09,";

        String[] strG = strA.split("\n");
        for(int h=0; h<strG.length; h++){
            strH = strG[h];
            String[] words=strH.split(",");
            int wrc=1;
            for(int i=0;i<words.length;i++) {
                for(int j=i+1;j<words.length;j++) {
                    if(words[i].equals(words[j])) {
                        wrc=wrc+1;
                        words[j]="0";
                    }
                }
                if(words[i]!="0"){
                    sbE.append(wrc).append(",");
                    strT = String.valueOf(sbE);
                }
                wrc=1;
            }
        }
        Log.d("TAG", "Output: "+strT);

Answer 1

我会在这里使用一个集合来计算重复项：

String strA = "2021-03-02,2021-03-02,2021-03-02,2021-03-02,2021-03-02,2021-03-11,2021-03-11,2021-03-11,2021-03-11,2021-03-11" + "\n" +
              "2021-03-07,2021-03-07,2021-03-07,2021-03-07,2021-03-07,2021-03-15,2021-03-15,2021-03-15,2021-03-15,2021-03-15" + "\n" +
              "2021-03-02,2021-03-09,2021-03-07,2021-03-09,2021-03-09";
String[] lines = strA.split("\n");
List<Integer> counts = new ArrayList<>();
for (String line : lines) {
    counts.add(new HashSet<String>(Arrays.asList(line.split(","))).size());
}
System.out.println(counts);  // [2, 2, 3]

请注意，我通过删除每行的尾随逗号对 strA 输入进行了较小的清理。

Answer 2

使用 Java 8 个流，这可以在一个语句中完成：

String strA = "2021-03-02,2021-03-02,2021-03-02,2021-03-02,2021-03-02,2021-03-11,2021-03-11,2021-03-11,2021-03-11,2021-03-11," + '\n' +
              "2021-03-07,2021-03-07,2021-03-07,2021-03-07,2021-03-07,2021-03-15,2021-03-15,2021-03-15,2021-03-15,2021-03-15," + '\n' +
              "2021-03-02,2021-03-09,2021-03-07,2021-03-09,2021-03-09,";

String strT = Pattern.compile("\n").splitAsStream(strA)
        .map(strG -> String.valueOf(Pattern.compile(",").splitAsStream(strG).distinct().count()))
        .collect(Collectors.joining(","));

System.out.println(strT); // 2,2,3

注意Pattern.compile("\n").splitAsStream(strA)也可以写成Arrays.stream(strA.split("\n"))，这样写起来更短，但是会创建一个不必要的中间数组。看个人喜好哪个好。

String strT = Arrays.stream(strA.split("\n"))
        .map(strG -> String.valueOf(Arrays.stream(strG.split(",")).distinct().count()))
        .collect(Collectors.joining(","));

第一个版本可以通过只编译一次正则表达式来进一步微优化：

Pattern patternComma = Pattern.compile(",");
String strT = Pattern.compile("\n").splitAsStream(strA)
        .map(strG -> String.valueOf(patternComma.splitAsStream(strG).distinct().count()))
        .collect(Collectors.joining(","));

无法计算字符串的每个部分中有多少个唯一日期可用

Can not count how many number of unique date are available in every part of string

java

android

arrays

string

stringbuilder