java 从文本文件中读取整数并将它们放入已排序的数组中
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Reading integers from a text file and putting them into a sorted array
提问by Oran
I've been scratching my head on this all day trying to figure out how to do this. I have tried multiple different ways, but I feel I've been going about this all wrong. 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.
我整天都在挠头，试图弄清楚如何做到这一点。我尝试了多种不同的方法，但我觉得我一直在做这一切都是错误的。我的任务是从 .txt 文件中读取 50 个整数并将内容放入一个排序数组中，然后列出最高/最低/平均数，但我几乎无法通过第一步。
These are the 50 numbers in the text file
这些是文本文件中的 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.
在这个文件中，我设法让“proj8”文件至少打印出来。
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”文件并对其进行了排序，但非常疯狂。
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. I need to be able to input the .txt file into a sorted array in integer and then list the highest/lowest/average number. Any help is good help, but the more simple the code the better.
我知道我做这一切都是错的，但我不知道如何正确地做到这一点。我需要能够将 .txt 文件输入到整数排序数组中，然后列出最高/最低/平均数。任何帮助都是好的帮助，但代码越简单越好。
回答by chathux
you are trying to sort lines of the file which contains several numbers separated by space. you have to split each line into numbers and for each number you have to parse it as an Integerand store those in the ArrayList. finally you can sort the ArrayList.
您正在尝试对包含多个由空格分隔的数字的文件行进行排序。您必须将每一行拆分为数字，对于每个数字，您必须将其解析为 anInteger并将它们存储在ArrayList. 最后你可以对ArrayList.
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);
回答by Raniz
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. You also need to convert them to numbers or they will be sorted as strings (meaning that "10" comes before "2").
您走在正确的轨道上，但查看您的文件，它似乎每行包含多个数字，因此您需要处理此问题。您还需要将它们转换为数字，否则它们将作为字符串排序（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:
我不会给你完整的答案，但这里有一些方法/类你可以用来解决这个问题：
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，但它们应该能够帮助您找到解决方案。
回答by wassim sabra
You dont have to go with complex methods all you have to do is to open the file create a scanner object. 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. then you compute the average and you output all results.
您不必使用复杂的方法，您只需打开文件创建一个扫描仪对象。1 for 循环从文本文件中构造数组 1 for 循环查找最低/最高值和所有值的总和。然后计算平均值并输出所有结果。
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);
}
}
