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从文本文件中读取整数并将其放入已排序的数组中

[英]Reading integers from a text file and putting them into a sorted array

I've been scratching my head on this all day trying to figure out how to do this. 我整天都在挠头,试图弄清楚该怎么做。 I have tried multiple different ways, but I feel I've been going about this all wrong. 我尝试了多种不同的方法,但是我觉得我一直在做这件事。 My assignment is to read 50 integers from a .txt file and put the contents into a sorted array, then list the highest/lowest/average number, but I can hardly get past step one. 我的任务是从.txt文件中读取50个整数,然后将内容放入已排序的数组中,然后列出最高/最低/平均数,但我几乎无法超越第一步。

These are the 50 numbers in the text file 这是文本文件中的50个数字

64 61 169 113 81 61 206 176 39 100 22 200 128 152 59 165 67 116 165 72 26 149 58 204 188 69 203 94 96 134 83 122 192 85 62 159 35 162 95 92 126 66 66 203 187 18 132 182 181 175 64 61 169 113 81 61 206 176 39 100 22 200 128 152 59 165 67 116 165 72 26 149 58 204 188 69 203 94 96 134 83 122 192 85 62 159 35 162 95 92 126 66 66 203 187 18 132 182 181 175

In this file I've managed to get the "proj8" file to at least print. 在此文件中,我设法至少要打印“ proj8”文件。

import java.util.Scanner;
import java.io.IOException;
import java.util.ArrayList;
import java.util.*;
import java.io.BufferedReader;
import java.io.FileReader;
import java.util.List;
import java.io.*;

public class ect7{

  public static void main(String[] args) throws IOException {

    BufferedReader br = new BufferedReader(new FileReader("proj8.txt"));

    String line = br.readLine();

    while ((line = br.readLine()) != null) {
      System.out.println(line);
    }
    br.close();
  } 
}

In this file I've managed to get the "proj8" file to print and somewhat order them, but insanely so. 在这个文件中,我设法打印了“ proj8”文件并对其进行了某种排序,但如此疯狂。

import java.util.Scanner;
import java.io.IOException;
import java.util.ArrayList;
import java.util.*;
import java.io.BufferedReader;
import java.io.FileReader;
import java.util.List;
import java.io.*;

public class ect73{

public static void main(String[] args) throws IOException {

  //int [] myArr = new int[50];  
  //FileReader fr = new FileReader("proj8.txt");    
  BufferedReader br = new BufferedReader(new FileReader("proj8.txt"));
  List<String> lines = new ArrayList<String>();
  String line = null;

  //String line = br.readLine();

   while ((line = br.readLine()) != null) {
     lines.add(line);
     Collections.sort(lines);
     //System.out.println(line);
     System.out.println(lines);
   }   
   br.close();
   //return lines.toArray(new String[lines.size()]);
  }
}

I know I'm doing this all wrong but I have no idea how to do this right. 我知道我做错了一切,但我不知道如何正确做。 I need to be able to input the .txt file into a sorted array in integer and then list the highest/lowest/average number. 我需要能够将.txt文件输入整数形式的排序数组,然后列出最高/最低/平均数。 Any help is good help, but the more simple the code the better. 任何帮助都是好的帮助,但是代码越简单越好。

you are trying to sort lines of the file which contains several numbers separated by space. 您正在尝试对文件的行进行排序,该行包含由空格分隔的多个数字。 you have to split each line into numbers and for each number you have to parse it as an Integer and store those in the ArrayList . 您必须将每一行拆分为数字,并且对于每个数字,您都必须将其解析为Integer并将其存储在ArrayList finally you can sort the ArrayList . 最后,您可以对ArrayList进行排序。

    BufferedReader br = new BufferedReader(new FileReader("proj8.txt"));
    List<Integer> numbers = new ArrayList<Integer>();
    String line = null;

     //String line = br.readLine();

     while ((line = br.readLine()) != null) {
         String []strNumbers = line.split(" ");
         for(String strNumber : strNumbers){
             numbers.add(Integer.parseInt(strNumber));
         }

         //System.out.println(line);            
     }   
     br.close();

     Collections.sort(numbers);
     System.out.println(numbers);

You're on the right track but looking at your file it looks like it contains more than one number per line so you need to handle this. 您的路线正确,但查看文件似乎每行包含一个以上的数字,因此您需要进行处理。 You also need to convert them to numbers or they will be sorted as strings (meaning that "10" comes before "2"). 您还需要将它们转换为数字,否则它们将按字符串排序(这意味着“ 10”在“ 2”之前)。

I'm not going to give you the whole answer but here are a few methods/classes you can use to solve this: 我不会给您完整的答案,但是您可以使用一些方法/类来解决这个问题:

You won't have to use all of these, but they should be able to help you find a solution. 您不必使用所有这些,但是它们应该能够帮助您找到解决方案。

You dont have to go with complex methods all you have to do is to open the file create a scanner object. 您不必使用复杂的方法,只需打开文件即可创建扫描程序对象。 1 for loop to construct the array from the text file 1 for loop to find the lowest/highest value and the sum of all values. 1 for循环从文本文件构造数组1 for循环查找最低/最高值以及所有值的总和。 then you compute the average and you output all results. 然后计算平均值,然后输出所有结果。

import java.io.File;
import java.io.FileNotFoundException;
import java.util.Scanner;

public class textToArray {

        public static void main(String[] args) throws FileNotFoundException{
        //Open the file 
        File f = new File("C:\\Users\\YOUR COMPUTER\\Desktop\\java.txt");//when the file directory contain \ you need to use escape ex "c:\\user\\desktop"
        //Create a new scanner to read the file
        Scanner s = new Scanner(f);
        //Declare an array with length of 100
        int[] array = new int[100];
        int i = 0, min, max, sum = 0;
        //Loop through the file if there is a next int to process it will continue

        for(i = 0; s.hasNextInt(); i++){
            //We store every int we read in the array
            array[i] = s.nextInt();                             
        }

        //Close the scanner object
        s.close();
        //As the for loop will stop when there is no new int to read so the i++ will stop at the number of int in the file
        //We use this number to calculate the average
        int nbInArray = i;
        //initialize min and max to the first array value which is for now empty
        min = max = array[0];
        for(int j = 0; j<nbInArray ; j++){
            //Now we test if the new value if bigger than the initial value (0 or empty) the new value will become the max
            if (array[j]>max){
                max = array[j];
                }
            //Otherwise if the the value is smaller than the initial value the new value will become the minimum
            //And so on every time we test if we have a smaller value the new one will become the min
            if (array[j]<min){
                min = array[j];
                }
            //The sum was initialised to 0 so we add every new int to it
            sum = sum + array[j];
        }


        //We compute the avg , i declared it double in case we have decimal results
        double avg = sum / nbInArray;
        //finally we print out the min, max and the average
        System.out.println("The minimum is: "+min);
        System.out.println("The maximum is: "+max);
        System.out.println("The averange is: "+avg);

}
}

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