1

我在一个文本文件中有 1,000 行数据，我希望每一行都是它自己的 float []。

1,1,1,1,1,1
2,2,2,2,2,2
3,3,3,3,3,3

会导致：

 float[0] = {1,1,1,1,1,1}
 float[1] = {2,2,2,2,2,2}
 float[2] = {3,3,3,3,3,3}

这可能吗？我只能找到将整个文件加载到数组中的示例。我尝试对所有数组进行硬编码，但超出了 ~65,000 的字节字符限制

4

4 回答 4

3

尝试以下操作：

// this list will store all the created arrays
List<float[]> arrays = new ArrayList<float[]>();

// use a BufferedReader to get the handy readLine() function
BufferedReader reader = new BufferedReader(new FileReader("myfile.txt"));

// this reads in all the lines. If you only want the first thousand, just
// replace these loop conditions with a regular counter variable
for (String line = reader.readLine(); line != null; line = reader.readLine()) {
    String[] floatStrings = line.split(",");
    float[] floats = new float[floatStrings.length];
    for (int i = 0; i < floats.length; ++i) {
        floats[i] = Float.parseFloat(floatStrings[i]);
    }
    arrays.add(floats);
}

请注意，我没有添加任何异常处理（例如，readLine() 会引发 IOException）。

于 2012-09-21T02:55:44.083 回答
0

  • 使用LineIterator读取每一行而不加载整个文件

  • 对于每一行，使用正则表达式提取(\d\.)+之类的数字，并在使用find()和group()等方法找到的匹配项上进行迭代

于 2012-09-21T02:43:19.847 回答
0 

<body>
<pre>

import java.io.FileReader;

public class Check {

    public static void main(String[] args) {    
        readingfile();
    }
    public static void readingfile() {
        try {
            FileReader read = new FileReader("D:\\JavaWkspace\\numbers.txt");
            int index;
            String nums1 = "";

            while ((index = read.read()) != -1) {
                if (((char) index) != '\n') {
                    nums1 += String.valueOf((char) index);
                }
            }

            System.out.println("Problem statement: Print out the greatest number on each line:\n" + nums1);

            String f = nums1.substring(0, 14);
            String s = nums1.substring(15, 29);
            String t = nums1.substring(30);

            String[] fs = f.split(",");
            int size = fs.length;
            int[] arr = new int[size];
            for (int i = 0; i < size; i++) {
                arr[i] = Integer.parseInt(fs[i]);
            }
            int max = arr[0];
            for (int i = 0; i < arr.length; i++) {
                if (max < arr[i]) {
                    max = arr[i];
                }
            }
            System.out.println("\nGreatest number in the first line is:" + (max));

            String[] sstr = s.split(",");
            int size2 = sstr.length;
            int[] arr2 = new int[size2];
            for (int i = 0; i < size2; i++) {
                arr2[i] = Integer.parseInt(sstr[i]);
            }

            int max2 = arr2[0];
            for (int i = 0; i < arr2.length; i++) {
                if (max2 < arr2[i]) {
                    max2 = arr2[i];
                }
            }
            System.out.println("\nGreatest number in the second line is:" + (max2));

            String[] str3 = t.split(",");
            int size3 = str3.length;
            int[] arr3 = new int[size3];
            for (int i = 0; i < size3; i++) {
                arr3[i] = Integer.parseInt(str3[i]);
            }

            int max3 = arr3[0];
            for (int i = 0; i < arr3.length; i++) {
                if (max3 < arr3[i]) {
                    max3 = arr3[i];
                }
            }
            System.out.println("\nGreatest number in the third line is:" + (max3));
            read.close();
        } catch (Exception e) {
            System.out.println(e);
        }
    }
}
</pre>
</body>

于 2021-11-22T08:23:44.107 回答
-1

使用 .split("\n") 循环文件的行分隔内容，然后将每个结果转换为浮点数组。以下是如何为您将字符串转换为浮点数 => http://www.devdaily.com/java/edu/qanda/pjqa00013.shtml

于 2012-09-21T02:36:18.730 回答