将一个数组中的所有数字转移到另一个数组中,使用Java。

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英文:

Transfer all numbers from an Array into another in Java

问题

I have a list that is holding multiple numbers in the same index, with all but the last number separated by a comma. I'd like to get all these numbers and transfer them to an array, but I want all individual numbers to be in different indexes in this new array. How can I achieve that? I was unable to find a problem similar to mine and have been struggling with it for a while.

我的列表包含在同一索引中的多个数字,除了最后一个数字以外,其他数字都以逗号分隔。我想将所有这些数字提取并传输到一个数组中,但我希望在这个新数组中,每个单独的数字都在不同的索引中。我该如何实现这一目标?我一直没有找到类似的问题,已经苦恼了一段时间。

I've considered that my way of collecting the data from the data.csv file could be considered bad and I need something completely different.

我考虑过从data.csv文件中收集数据的方式可能不够好,我可能需要完全不同的方法。

英文:

I have a list that is holding multiple numbers in the same index, with all but the last number separated by a comma. I'd like to get all these numbers and transfer them to an array, but I want all individual numbers to be in different indexes in this new array. How can I achieve that? I was unable to find a problem similar to mine and have been struggling with it for a while.

I've considered that my way of collecting the data from the data.csv file could be considered bad and I need something completely different.

My Main class:

public static void main(String[] args) throws FileNotFoundException {
    String file = "data.csv";
    String line;
    List<String> list = new ArrayList<String>();
    int i = 0;
    try (BufferedReader br = new BufferedReader(new FileReader(file))) {
        while((line = br.readLine()) != null){
            //System.out.println(line);
            list.add(i, line);
            i++;
        }
    } catch (Exception e){
        System.out.println(e);
    }

Data.csv file:

20,16,23,20,2,17,9,16,0,6
20,18,16,4,10,9,4,10,16,16
3,6,19,14,14,0,5,22,4,15
19,22,6,18,9,12,15,9,25,14
20,0,1,20,6,19,20,16,15,6
13,1,6,1,4,22,3,11,1,12
10,9,20,7,21,22,3,14,1,8
25,19,21,14,15,16,25,17,6,14
0,20,10,1,3,24,24,5,10,24
2,13,0,9,18,5,21,2,24,21
17,4,16,6,22,14,15,5,10,20
6,15,10,8,10,10,16,22,15,22
2,1,3,25,15,19,0,13,7,17
24,20,6,7,3,3,12,22,23,17
12,10,10,14,2,18,21,5,21,3
13,9,20,11,8,3,23,5,5,3
4,18,4,9,1,2,24,21,23,3
3,3,2,19,6,19,1,10,5,4
13,0,12,19,24,7,10,18,18,1
0,7,10,2,22,20,21,4,3,11

Expected result:

Data[0] = 20
Data[1] = 16
Data[2] = 23
Data[10] = 20
Data[11] = 18

答案1

得分: 2

以下是代码部分的中文翻译:

你正在阅读的这行代码仍然包含很多数字,你想逐个将它们添加到你的列表中。使用 split 将它们拆分成一个值数组。使用 Arrays.asList 将该数组转换为列表,然后使用 addAll 方法将它们全部添加到你的目标列表中。

String line;
List<String> list = new ArrayList<>();
try (BufferedReader br = new BufferedReader(new FileReader(file))) {
    while((line = br.readLine()) != null){
        list.addAll(Arrays.asList(line.split(",")));
    }
} catch (Exception e){
    e.printStackTrace();
}

你始终可以将 List 转换为数组:

String[] values = list.toArray(String[]::new);

如果你想要使用 Java Streams,创建一个 Path,获取行,对于每一行,使用 split 拆分成一个值数组,转换为 List,使用 flatMap 来将 Lists 扁平化为一个值流,然后收集到一个 List 中。

List<String> list = null;
try {
    list = Files.lines(Path.of(file))
        .map(line -> line.split(","))
        .map(Arrays::asList)
        .flatMap(List::stream)
        .collect(Collectors.toList());  // 或者 .toList();  在 Java 16+
} catch (Exception e){
    e.printStackTrace();
}

或者,你可以直接将 Stream 转换为数组,替换对 collecttoList 的调用,使用 toArray

String[] values = Files.lines(Path.of(file))
    .map(line -> line.split(","))
    .map(Arrays::asList)
    .flatMap(List::stream)
    .toArray(String[]::new);
英文:

The line that you're reading still has lots of numbers that you want to add to your list individually. Use split to split them into an array of values. Convert that array to a List with Arrays.asList, them add them all to your target list with the addAll method.

String line;
List&lt;String&gt; list = new ArrayList&lt;&gt;();
try (BufferedReader br = new BufferedReader(new FileReader(file))) {
    while((line = br.readLine()) != null){
        list.addAll(Arrays.asList(line.split(&quot;,&quot;)));
    }
} catch (Exception e){
    e.printStackTrace();
}

You can always convert the List to an array with

String[] values = list.toArray(String[]::new);

If you'd like to use Java Streams, create a Path, get the lines, and for each line, split into an array of values, convert to a List, use flatMap to flatten the Lists to a stream of values, then collect to a List.

List&lt;String&gt; list = null;
try {
    list = Files.lines(Path.of(file))
        .map(line -&gt; line.split(&quot;,&quot;))
        .map(Arrays::asList)
        .flatMap(List::stream)
        .collect(Collectors.toList());  // or .toList();  with Java 16+
} catch (Exception e){
    e.printStackTrace();
}

Or you can convert the Stream directly to an array, replacing the call to collect or toList with toArray:

String[] values = Files.lines(Path.of(file))
    .map(line -&gt; line.split(&quot;,&quot;))
    .map(Arrays::asList)
    .flatMap(List::stream)
    .toArray(String[]::new);

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  • 本文由 发表于 2023年3月21日 02:21:19
  • 转载请务必保留本文链接:https://go.coder-hub.com/75793956.html
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