Java 如何将小数转换为分数？
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How to convert decimal to fractions?
提问by Search Results Web results Pi
What I need to convert decimal to fractions. It is easy to convert to 10's feet.
我需要将十进制转换为分数。很容易转换为 10 英尺。
1.5 => 15/10
This can do via this code:
这可以通过以下代码完成：
public class Rational {
private int num, denom;
public Rational(double d) {
String s = String.valueOf(d);
int digitsDec = s.length() - 1 - s.indexOf('.');
int denom = 1;
for (int i = 0; i < digitsDec; i++) {
d *= 10;
denom *= 10;
}
int num = (int) Math.round(d);
this.num = num;
this.denom = denom;
}
public Rational(int num, int denom) {
this.num = num;
this.denom = denom;
}
public String toString() {
return String.valueOf(num) + "/" + String.valueOf(denom);
}
public static void main(String[] args) {
System.out.println(new Rational(1.5));
}
}
But what I want is
但我想要的是
1.5 => 3/2
and I don't get how to proceed. My question is not a duplication. Because other related question is C#. This is java.
我不知道如何继续。我的问题不是重复。因为其他相关问题是 C#。这是爪哇。
采纳答案by Hristo93
You should find the greatest common divisor of the resulted numbers and divide the numerator and denominator by it.
您应该找到结果数的最大公约数，并将分子和分母除以它。
Here is one way to do it:
这是一种方法：
public class Rational {
private int num, denom;
public Rational(double d) {
String s = String.valueOf(d);
int digitsDec = s.length() - 1 - s.indexOf('.');
int denom = 1;
for (int i = 0; i < digitsDec; i++) {
d *= 10;
denom *= 10;
}
int num = (int) Math.round(d);
int g = gcd(num, denom);
this.num = num / g;
this.denom = denom /g;
}
public Rational(int num, int denom) {
this.num = num;
this.denom = denom;
}
public String toString() {
return String.valueOf(num) + "/" + String.valueOf(denom);
}
public static int gcd(int num, int denom) {
....
}
public static void main(String[] args) {
System.out.println(new Rational(1.5));
}
}
回答by Rémi Becheras
Here is a simple algorythme :
这是一个简单的算法：
numerato = 1.5
denominator = 1;
while (!isInterger(numerator*denominator))
do
denominator++;
done
return numerator*denominator + '/' + denominator
// => 3/2
You just have to implement it in java + implement the isInteger(i)where iis a float.
你只需要在java中实现它+实现isInteger(i)where iis a float。
回答by Joop Eggen
Given double x >= 0, int p, int q, find p/q as closest approximation:
给定 double x >= 0, int p, int q，找到 p/q 作为最接近的近似值：
- iterate on q from 1 upwards, determine p above and below; check deviations
- 从 1 向上迭代 q，确定上下 p；检查偏差
So (not tested):
所以（未测试）：
public static Rational toFraction(double x) {
// Approximate x with p/q.
final double eps = 0.000_001;
int pfound = (int) Math.round(x);
int qfound = 1;
double errorfound = Math.abs(x - pfound);
for (int q = 2; q < 100 && error > eps; ++q) {
int p = (int) (x * q);
for (int i = 0; i < 2; ++i) { // below and above x
double error = Math.abs(x - ((double) p / q));
if (error < errorfound) {
pfound = p;
qfound = q;
errorfound = error;
}
++p;
}
}
return new Rational(pfound, qfound);
}
You could try it for Math.PI and E.
你可以试试 Math.PI 和 E。
回答by Abhiram Kulkarni
Including the method to find highest common factor and modifying toString method, solves your question i suppose.
包括找到最高公因数的方法和修改 toString 方法，我想解决了你的问题。
public String toString() {
int hcf = findHighestCommonFactor(num, denom);
return (String.valueOf(num/hcf) + "/" + String.valueOf(denom/hcf));
}
private int findHighestCommonFactor(int num, int denom) {
if (denom == 0) {
return num;
}
return findHighestCommonFactor(denom, num % denom);
}
回答by Krishanthy Mohanachandran
Not only for the decimal number 1.5, for all you can use the following steps:
不仅对于十进制数1.5，对于所有您都可以使用以下步骤：
Find Number of decimal digits:
double d = 1.5050;//Example I useddouble d1 = 1;String text = Double.toString(Math.abs(d));int integerPlaces = text.indexOf('.');int decimalPlaces = text.length() - integerPlaces - 1;System.out.println(decimalPlaces);//4Then convert to integer:
static int ipower(int base, int exp) {int result = 1; for (int i = 1; i <= exp; i++) { result *= base; } return result; }//using the methodint i = (int) (d*ipower(10, decimalPlaces));int i1 = (int) (d1*ipower(10, decimalPlaces));System.out.println("i=" + i + " i1 =" +i1);//i=1505 i1 =1000Then find highest common factor
private static int commonFactor(int num, int divisor) {if (divisor == 0) { return num; } return commonFactor(divisor, num % divisor); }
查找小数位数：
double d = 1.5050;//Example I useddouble d1 = 1;String text = Double.toString(Math.abs(d));int integerPlaces = text.indexOf('.');int decimalPlaces = text.length() - integerPlaces - 1;System.out.println(decimalPlaces);//4然后转换为整数：
static int ipower(int base, int exp) {int result = 1; for (int i = 1; i <= exp; i++) { result *= base; } return result; }//using the methodint i = (int) (d*ipower(10, decimalPlaces));int i1 = (int) (d1*ipower(10, decimalPlaces));System.out.println("i=" + i + " i1 =" +i1);//i=1505 i1 =1000然后求最大公因数
private static int commonFactor(int num, int divisor) {if (divisor == 0) { return num; } return commonFactor(divisor, num % divisor); }
//using common factor
//using common factor
int commonfactor = commonFactor(i, i1);
int commonfactor = commonFactor(i, i1);
System.out.println(commonfactor);//5
System.out.println(commonfactor);//5
Finally print results:
System.out.println(i/commonfactor + "/" + i1/commonfactor);//301/200
最后打印结果：
System.out.println(i/commonfactor + "/" + i1/commonfactor);//301/200
Here you can find:
您可以在这里找到：
public static void main(String[] args) {
double d = 1.5050;
double d1 = 1;
String text = Double.toString(Math.abs(d));
int integerPlaces = text.indexOf('.');
int decimalPlaces = text.length() - integerPlaces - 1;
System.out.println(decimalPlaces);
System.out.println(ipower(10, decimalPlaces));
int i = (int) (d*ipower(10, decimalPlaces));
int i1 = (int) (d1*ipower(10, decimalPlaces));
System.out.println("i=" + i + " i1 =" +i1);
int commonfactor = commonFactor(i, i1);
System.out.println(commonfactor);
System.out.println(i/commonfactor + "/" + i1/commonfactor);
}
static int ipower(int base, int exp) {
int result = 1;
for (int i = 1; i <= exp; i++) {
result *= base;
}
return result;
}
private static int commonFactor(int num, int divisor) {
if (divisor == 0) {
return num;
}
return commonFactor(divisor, num % divisor);
}
回答by Matthew556
static private String convertDecimalToFraction(double x){
if (x < 0){
return "-" + convertDecimalToFraction(-x);
}
double tolerance = 1.0E-6;
double h1=1; double h2=0;
double k1=0; double k2=1;
double b = x;
do {
double a = Math.floor(b);
double aux = h1; h1 = a*h1+h2; h2 = aux;
aux = k1; k1 = a*k1+k2; k2 = aux;
b = 1/(b-a);
} while (Math.abs(x-h1/k1) > x*tolerance);
return h1+"/"+k1;
}
I got this answer from here. All I had to do is convert his answer to java.
我从这里得到了这个答案。我所要做的就是将他的答案转换为 java。
回答by John R Perry
I tried adding this as an edit, but it was denied. This answer builds off of @Hristo93's answerbut finishes the gcd method:
我尝试将其添加为编辑，但被拒绝。这个答案建立客@ Hristo93的答案，但完成GCD的方法：
public class DecimalToFraction {
private int numerator, denominator;
public Rational(double decimal) {
String string = String.valueOf(decimal);
int digitsDec = string.length() - 1 - s.indexOf('.');
int denominator = 1;
for (int i = 0; i < digitsDec; i++) {
decimal *= 10;
denominator *= 10;
}
int numerator = (int) Math.round(decimal);
int gcd = gcd(numerator, denominator);
this.numerator = numerator / gcd;
this.denominator = denominator /gcd;
}
public static int gcd(int numerator, int denom) {
return denominator == 0 ? numerator : gcm(denominator, numerator % denominator);
}
public String toString() {
return String.valueOf(numerator) + "/" + String.valueOf(denominator);
}
public static void main(String[] args) {
System.out.println(new Rational(1.5));
}
}
回答by Orhan
I prepared a solution for this question. Maybe it's look like primitive but working. I tested many decimal number. At least it can converting 1.5 to 3/2 :)
我为这个问题准备了一个解决方案。也许它看起来像原始但有效。我测试了很多十进制数。至少它可以将 1.5 转换为 3/2 :)
public String kesirliYap(Double sayi){
String[] a=payPaydaVer(sayi);
return a[0]+"/"+a[1];
}
public String[] payPaydaVer(Double sayi){
long pay;
long payda;
DecimalFormat df=new DecimalFormat("#");
df.setRoundingMode(RoundingMode.FLOOR);
String metin=sayi.toString();
int virguldenSonra=(metin.length() -metin.indexOf("."))-1;
double payyda=Math.pow(10,virguldenSonra);
double payy=payyda*sayi;
String pays=df.format(payy);
String paydas=df.format(payyda);
pay=Long.valueOf(pays);
payda=Long.valueOf(paydas);
String[] kesir=sadelestir(pay,payda).split(",");
return kesir;
}
private String sadelestir(Long pay,Long payda){
DecimalFormat df=new DecimalFormat("#");
df.setRoundingMode(RoundingMode.FLOOR);
Long a=pay<payda ? pay : payda;
String b = "",c = "";
int sayac=0;
for(double i = a;i>1;i--){
double payy=pay/i;
double paydaa=payda/i;
String spay=df.format(payy);
String spayda=df.format(paydaa);
Long lpay=Long.valueOf(spay);
Long lpayda=Long.valueOf(spayda);
if((payy-lpay)==0&&(paydaa-lpayda)==0){
b=df.format(pay/i);
c=df.format(payda/i);
sayac++;
break;
}
}
return sayac>0 ? b+","+c:pay+","+payda;
}
回答by Hasan Shadi
First of all, if you want to convert a decimal number, you need to know the state of the situation before you convert it, let us say you have 0.333333, number 3 is being repeated infinitely. We all know that 0.333333 is 1/3 . Some people think that multiplying by the number of digits after the decimal point will convert it. That is completely false. It is something related mathematics, another situation is 0.25, take the numbers after the decimal point and divide them by 100 and simplify them, which equals to 1/4, completely true!
首先，如果你想转换一个十进制数，你需要在转换之前知道情况的状态，假设你有0.333333，数字3被无限重复。我们都知道 0.333333 是 1/3 。有人认为乘以小数点后的位数就可以换算了。那是完全错误的。是数学相关的东西，另一种情况是0.25，取小数点后的数除以100化简，等于1/4，完全正确！
However, in mathematics we have 3 situations for converting a decimal number into a fraction, I am not going to explain them because It will take a lot of space and time, I have already written a program for this problem. This is the code :
但是，在数学中我们有3种情况将十进制数转换为分数，我不打算解释它们，因为这会占用大量空间和时间，我已经为这个问题编写了一个程序。这是代码：
import java.math.BigDecimal;
import java.math.BigInteger;
public class Main {
static BigDecimal finalResult = new BigDecimal("0");
static boolean check(short[] checks) {
boolean isContinues = true;
int index = -1;
for (short ind : checks) {
index++;
if (ind==1) {
}
else if (ind==0) {
isContinues = false;
break;
}
else if (ind==-1) {
if (index==0) {
isContinues = false;
}
break;
}
}
return isContinues;
}
static int[] analyzeDecimal() { // will return int[3]
int[] analysis = new int[3];
int dot = finalResult.toString().indexOf(".");
String num = finalResult.toString();
int state = -1;
int firstPart = 0; // first part will be compared with each secondPart!
int secondPart = 0;
String part = ""; // without the dot
int index = 0; // index for every loop!
int loop = 6;
int originalLoop = loop;
int size = 0; // until six!
int ps = -1;
short[] checks = new short[] {-1, -1, -1, -1, -1, -1, -1, -1, -1, -1}; // 10 compares for each part!
// length of checks is 10!
int continues = -1; // -1 means there is no continues part!
boolean stop = false;
while (true) { // while for size!
if (size!=6) {
while (true) { // we need to compare a part with a part!
// while for loop
// 6 loops, every loop will increase the compared part by 1!
if (loop!=-1) { // TODO : check every part with the increasing pos
firstPart = dot+1+(originalLoop-loop); // changed
try {
part = num.substring(firstPart, firstPart+(size+1));
}
catch (StringIndexOutOfBoundsException ex) {
break;
}
int partSize = part.length();
int afterDecimal = num.length()-(dot+1);
while (index!=checks.length &&
firstPart+partSize+index*partSize-(dot+1)<=afterDecimal) { // while for index!
secondPart = firstPart+partSize+index*partSize;
String comparedPart;
try {
comparedPart = num.substring(secondPart, secondPart+partSize);
}
catch (StringIndexOutOfBoundsException ex) {
break;
}
if (part.equals(comparedPart)) {
checks[index] = 1;
}
else {
checks[index] = 0;
}
index++;
}
index = 0;
if (check(checks)) {
stop = true;
continues = firstPart;
ps = partSize;
}
for (int i = 0 ; i!=10 ; i++) {
checks[i] = -1;
}
}
else { // finished!
break;
}
loop--;
if (stop) {
break;
}
}
loop = originalLoop;
size++;
if (stop) {
break;
}
}
else {
break;
}
}
if (continues==-1) {
state = 2;
}
else {
if (dot+1==continues) {
state = 1;
}
else {
state = 0;
}
}
analysis[0] = state;
analysis[1] = continues;
analysis[2] = ps;
return analysis;
}
static String convertToStandard() {
// determine the state first :
int[] analysis = analyzeDecimal();
int dot = finalResult.toString().indexOf('.')+1;
int continues = analysis[1];
int partSize = analysis[2]; // how many steps after the continues part
if (analysis[0]==0) { // constant + continues
String number = finalResult.toString().substring(0, continues+partSize);
int numOfConst = continues-dot;
int numOfDecimals = continues+partSize-dot;
int den = (int)(Math.pow(10, numOfDecimals)-Math.pow(10, numOfConst)); // (10^numOfDecimals)-(10^numOfConst);
int num;
int toSubtract = Integer.parseInt(number.substring(0, dot-1)+number.substring(dot, dot+numOfConst));
if (number.charAt(0)==0) {
num = Integer.parseInt(number.substring(dot));
}
else {
num = Integer.parseInt(number.replace(".", ""));
}
num -= toSubtract;
return simplify(num, den);
}
else if (analysis[0]==1) { // continues
int num, den;
// we always have to subtract by only one x!
String n = finalResult.toString().substring(0, dot+partSize).replace(".", "");
num = Integer.parseInt(n);
den = nines(partSize);
int toSubtract = Integer.parseInt(finalResult.toString().substring(0, dot-1));
num -= toSubtract;
return simplify(num, den);
}
else if (analysis[0]==2) { // constant
partSize = finalResult.toString().length()-dot;
int num = Integer.parseInt(finalResult.toString().replace(".", ""));
int den = (int)Math.pow(10, partSize);
return simplify(num, den);
}
else {
System.out.println("[Error] State is not determined!");
}
return "STATE NOT DETERMINED!";
}
static String simplify(int num, int den) {
BigInteger n1 = new BigInteger(Integer.toString(num));
BigInteger n2 = new BigInteger(Integer.toString(den));
BigInteger GCD = n1.gcd(n2);
String number = Integer.toString(num/GCD.intValue())+"/"+Integer.toString(den/GCD.intValue());
return number;
}
static int nines(int n) {
StringBuilder result = new StringBuilder();
while (n!=0) {
n--;
result.append("9");
}
return Integer.parseInt(result.toString());
}
public static void main(String[] args) {
finalResult = new BigDecimal("1.222222");
System.out.println(convertToStandard());
}
}
The program above will give you an optimal result with high precision. All you have to do is change the finalResult variable in the main function.
上面的程序将为您提供高精度的最佳结果。您所要做的就是更改主函数中的 finalResult 变量。
