我在java中有一个赋值,按升序对数字进行排序,找出最大值,最小值,平均值和标准差
我已经这样做但我想改变程序以使用双值但是有一个异常显示,我无法解决问题,请帮助可以有人解决它。
import java.io.*;
import java.util.Scanner;
import java.io.FileReader;
import java.io.FileWriter;
import java.io.BufferedReader;
import java.io.PrintWriter;
import java.io.IOException;
public class FileJava2 {
static double min,max,sum,k;
static double PS1,PS2;
static double stdev=0;
static double aa, x[]=new double[1000], no;
static String source= "",source2= "";
public static void main (String args[]) throws Exception
{
FileJava2.fileinput();
FileJava2.fileoutput();
FileJava2.sort();
FileJava2.display();
}
public static void sort() throws Exception
{
sum=0;
for(int j=0;j<k;j++){
sum+=x[j];
System.out.print(x[j]+" ");
}
double t;
for(int dd=0; dd<k; dd++){
for(int in=0;in<k-1;in++){
if(x[dd]<x[in])
{t=x[dd];
x[dd]=x[in];
x[in]=t;
}
}
}
min=x[0];max=0;
System.out.print("\nSorted Elements: ");
for(int j=0;j<k;j++){
if(x[j]<min)
min=x[j];
if(x[j]>max)
max=x[j];
source2+=x[j];
System.out.print(x[j]+" ");
source+=x[j];
}
PS1=0;PS2=0;
for( int i=0; i<k;i++) {
PS1 += x[i];
PS2 += Math.pow(x[i], 2);
stdev = Math.sqrt(i*PS2 - Math.pow(PS1, 2))/i;
}
byte buf1[]=source2.getBytes();
OutputStream fo1=new FileOutputStream("SortData.txt");
for (int i=0;i<buf1.length;i++)
{
fo1.write(buf1[i]);
}
}
public static void fileinput() throws Exception{
Scanner s = new Scanner(System.in);
do{
System.out.print("Enter Numbers: ");
aa=s.nextDouble();
if(aa==0)
break;
else
source+=(aa+" ");
} while(aa!=0);
System.out.println("YOUR INPUT: "+source);
k=0;
byte buf[]=source.getBytes();
OutputStream fo=new FileOutputStream("waitingtime");
for (int i=0;i<buf.length;i++)
{
fo.write(buf[i]);
}
System.out.println("\nElements successfuly saved into waitingtime.dat ");
}
public static void fileoutput() throws Exception{
BufferedReader inputStream = null;
try {
inputStream =
new BufferedReader(new FileReader("waitingtime"));
String l;
while ((l = inputStream.readLine()) != null) {
// System.out.println(l);
for ( int i = 0; i < l.length(); i++ ) {
String cc=" "+l.charAt( i );
x[(int)k++]=Integer.parseInt(cc);
// System.out.println(no);
}
}
}
finally {
if (inputStream != null) {
inputStream.close();
}
}
}
public static void display(){
System.out.println("\nElements successfuly saved into SortData.dat ");
System.out.print("\nMinimum: "+min);
System.out.print("\nMaximun:"+max);
System.out.print("\nMean:"+sum/k);
System.out.print("\nMidrange:"+(min+max)/2);
System.out.println("\nStandard Deviation:"+stdev);
}
}
和异常消息
Enter Numbers: 1
Enter Numbers: 1
Enter Numbers: 2
Enter Numbers: 5
Enter Numbers: 0
YOUR INPUT: 1.0 1.0 2.0 5.0
Elements successfuly saved into waitingtime.dat
Exception in thread "main" java.lang.NumberFormatException: For input string: "."
at java.lang.NumberFormatException.forInputString(NumberFormatException.java:65)
at java.lang.Integer.parseInt(Integer.java:481)
at java.lang.Integer.parseInt(Integer.java:527)
at FileJava2.fileoutput(FileJava2.java:110)
at FileJava2.main(FileJava2.java:21)
有人可以告诉我为什么当我输入1时它会显示1.0?
答案 0 :(得分:1)
在for
方法的fileoutput()
循环
for (int i = 0; i < l.length(); i++) {
if (l.charAt(i) != '.' && l.charAt(i) != ' ') {
String cc = (" " + l.charAt(i)).trim();
int result = Integer.parseInt(cc);
if (result != 0) {
x[(int) k++] = result;
}
}
}
答案 1 :(得分:0)
此例外:
java.lang.NumberFormatException: For input string: "."
at java.lang.NumberFormatException.forInputString(NumberFormatException.java:65)
at java.lang.Integer.parseInt(Integer.java:481)
告诉您,无法将字符串.
解析为int
。
并且有道理,因为int可以由.
表示?
答案 2 :(得分:0)
您正在尝试将字符串中的每个字符解析为整数。在读取行"1.0 1.0 2.0 5.0"
之后,您需要拆分数字并将子串传递给解析int / double。您可以使用三个空格字符进行拆分,如下所示:
while ((l = inputStream.readLine()) != null) {
for(String ss:l.split(" ") {
x[(int)k++] = Double.parseDouble(ss);
}
}
答案 3 :(得分:0)
import java.io.*;
import java.util.Scanner;
import java.io.FileReader;
import java.io.FileWriter;
import java.io.BufferedReader;
import java.io.PrintWriter;
import java.io.IOException;
public class FileJava2 {
static double min,max,sum,k;
static double PS1,PS2;
static double stdev=0;
static double aa, x[]=new double[1000], no;
static String source= "",source2= "";
public static void main (String args[]) throws Exception
{
FileJava2.fileinput();
FileJava2.fileoutput();
FileJava2.sort();
FileJava2.display();
}
public static void sort() throws Exception
{
sum=0;
for(int j=0;j<k;j++){
sum+=x[j];
System.out.print(x[j]+" ");
}
double t;
for(int dd=0; dd<k; dd++){
for(int in=0;in<k-1;in++){
if(x[dd]<x[in])
{t=x[dd];
x[dd]=x[in];
x[in]=t;
}
}
}
min=x[0];max=0;
System.out.print("\nSorted Elements: ");
for(int j=0;j<k;j++){
if(x[j]<min)
min=x[j];
if(x[j]>max)
max=x[j];
source2+=x[j];
System.out.print(x[j]+" ");
source+=x[j];
}
PS1=0;PS2=0;
for( int i=0; i<k;i++) {
PS1 += x[i];
PS2 += Math.pow(x[i], 2);
stdev = Math.sqrt(i*PS2 - Math.pow(PS1, 2))/i;
}
byte buf1[]=source2.getBytes();
OutputStream fo1=new FileOutputStream("SortData.txt");
for (int i=0;i<buf1.length;i++)
{
fo1.write(buf1[i]);
}
}
public static void fileinput() throws Exception{
Scanner s = new Scanner(System.in);
do{
System.out.print("Enter Numbers: ");
aa=s.nextDouble();
if(aa==0)
break;
else
source+=(aa+" ");
} while(aa!=0);
System.out.println("YOUR INPUT: "+source);
k=0;
byte buf[]=source.getBytes();
OutputStream fo=new FileOutputStream("waitingtime");
for (int i=0;i<buf.length;i++)
{
fo.write(buf[i]);
}
System.out.println("\nElements successfuly saved into waitingtime.dat ");
}
public static void fileoutput() throws Exception{
BufferedReader inputStream = null;
try {
inputStream =
new BufferedReader(new FileReader("waitingtime"));
String l;
while ((l = inputStream.readLine()) != null) {
// System.out.println(l);
String numbers[] = l.split(" ");
for (String cc : numbers) {
x[(int)k++]=Double.parseDouble(cc);
// System.out.println(no);
}
}
}
finally {
if (inputStream != null) {
inputStream.close();
}
}
}
public static void display(){
System.out.println("\nElements successfuly saved into SortData.dat ");
System.out.print("\nMinimum: "+min);
System.out.print("\nMaximun:"+max);
System.out.print("\nMean:"+sum/k);
System.out.print("\nMidrange:"+(min+max)/2);
System.out.println("\nStandard Deviation:"+stdev);
}
}
您的FileJava2.output()已更改解决您的问题。使用split获取值,然后将string转换为double而不是int。