我正在尝试将用户键入的具有小数结果的计算转换为分数。例如；66.6666666667 成 66 2/3。任何指针？提前感谢
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7 回答
32
连分数可用于找到严格意义上的“最佳”实数的有理逼近。这是一个 PHP 函数，它找到给定（正）浮点数的有理逼近，其相对误差小于$tolerance：
<?php
function float2rat($n, $tolerance = 1.e-6) {
$h1=1; $h2=0;
$k1=0; $k2=1;
$b = 1/$n;
do {
$b = 1/$b;
$a = floor($b);
$aux = $h1; $h1 = $a*$h1+$h2; $h2 = $aux;
$aux = $k1; $k1 = $a*$k1+$k2; $k2 = $aux;
$b = $b-$a;
} while (abs($n-$h1/$k1) > $n*$tolerance);
return "$h1/$k1";
}
printf("%s\n", float2rat(66.66667)); # 200/3
printf("%s\n", float2rat(sqrt(2))); # 1393/985
printf("%s\n", float2rat(0.43212)); # 748/1731
我已经写了更多关于这个算法及其工作原理的内容，甚至还有一个 JavaScript 演示：https ://web.archive.org/web/20180731235708/http://jonisalonen.com/2012/converting-decimal-numbers-to -比率/
于 2013-01-16T11:19:46.000 回答
7
在这种情况下，法雷分数可能非常有用。
它们可用于将任何小数转换为分母最小的分数。
抱歉 - 我没有 PHP 原型，所以这里有一个 Python 原型：
def farey(v, lim):
"""No error checking on args. lim = maximum denominator.
Results are (numerator, denominator); (1, 0) is 'infinity'."""
if v < 0:
n, d = farey(-v, lim)
return (-n, d)
z = lim - lim # Get a "zero of the right type" for the denominator
lower, upper = (z, z+1), (z+1, z)
while True:
mediant = (lower[0] + upper[0]), (lower[1] + upper[1])
if v * mediant[1] > mediant[0]:
if lim < mediant[1]:
return upper
lower = mediant
elif v * mediant[1] == mediant[0]:
if lim >= mediant[1]:
return mediant
if lower[1] < upper[1]:
return lower
return upper
else:
if lim < mediant[1]:
return lower
upper = mediant
于 2013-01-15T03:40:27.067 回答
7
根据@Joni 的回答，这是我用来提取整数的内容。
function convert_decimal_to_fraction($decimal){
$big_fraction = float2rat($decimal);
$num_array = explode('/', $big_fraction);
$numerator = $num_array[0];
$denominator = $num_array[1];
$whole_number = floor( $numerator / $denominator );
$numerator = $numerator % $denominator;
if($numerator == 0){
return $whole_number;
}else if ($whole_number == 0){
return $numerator . '/' . $denominator;
}else{
return $whole_number . ' ' . $numerator . '/' . $denominator;
}
}
function float2rat($n, $tolerance = 1.e-6) {
$h1=1; $h2=0;
$k1=0; $k2=1;
$b = 1/$n;
do {
$b = 1/$b;
$a = floor($b);
$aux = $h1; $h1 = $a*$h1+$h2; $h2 = $aux;
$aux = $k1; $k1 = $a*$k1+$k2; $k2 = $aux;
$b = $b-$a;
} while (abs($n-$h1/$k1) > $n*$tolerance);
return "$h1/$k1";
}
于 2014-10-31T23:44:41.703 回答
7
将答案中的 Python 代码从 @APerson241 转换为 PHP
<?php
function farey($v, $lim) {
// No error checking on args. lim = maximum denominator.
// Results are array(numerator, denominator); array(1, 0) is 'infinity'.
if($v < 0) {
list($n, $d) = farey(-$v, $lim);
return array(-$n, $d);
}
$z = $lim - $lim; // Get a "zero of the right type" for the denominator
list($lower, $upper) = array(array($z, $z+1), array($z+1, $z));
while(true) {
$mediant = array(($lower[0] + $upper[0]), ($lower[1] + $upper[1]));
if($v * $mediant[1] > $mediant[0]) {
if($lim < $mediant[1])
return $upper;
$lower = $mediant;
}
else if($v * $mediant[1] == $mediant[0]) {
if($lim >= $mediant[1])
return $mediant;
if($lower[1] < $upper[1])
return $lower;
return $upper;
}
else {
if($lim < $mediant[1])
return $lower;
$upper = $mediant;
}
}
}
// Example use:
$f = farey(66.66667, 10);
echo $f[0], '/', $f[1], "\n"; # 200/3
$f = farey(sqrt(2), 1000);
echo $f[0], '/', $f[1], "\n"; # 1393/985
$f = farey(0.43212, 2000);
echo $f[0], '/', $f[1], "\n"; # 748/1731
于 2013-02-18T16:22:33.383 回答
1
基于@APerson 和@Jeff Monteiro 的回答，我创建了Farey 分数的PHP 版本，它将被简化为具有最低分母的分数的整数：
<?php
class QuantityTransform
{
/**
* @see https://stackoverflow.com/questions/14330713/converting-float-decimal-to-fraction
*/
public static function decimalToFraction(float $decimal, $glue = ' ', int $limes = 10): string
{
if (null === $decimal || $decimal < 0.001) {
return '';
}
$wholeNumber = (int) floor($decimal);
$remainingDecimal = $decimal - $wholeNumber;
[$numerator, $denominator] = self::fareyFraction($remainingDecimal, $limes);
// Values rounded to 1 should be added to base value and returned without fraction part
if (is_int($simplifiedFraction = $numerator / $denominator)) {
$wholeNumber += $simplifiedFraction;
$numerator = 0;
}
return (0 === $wholeNumber && 0 === $numerator)
// Too small values will be returned in original format
? (string) $decimal
// Otherwise let's format value - only non-0 whole value / fractions will be returned
: trim(sprintf(
'%s%s%s',
(string) $wholeNumber ?: '',
$wholeNumber > 0 ? $glue : '',
0 === $numerator ? '' : ($numerator . '/' . $denominator)
));
}
/**
* @see https://stackoverflow.com/a/14330799/842480
*
* @return int[] Numerator and Denominator values
*/
private static function fareyFraction(float $value, int $limes): array
{
if ($value < 0) {
[$numerator, $denominator] = self::fareyFraction(-$value, $limes);
return [-$numerator, $denominator];
}
$zero = $limes - $limes;
$lower = [$zero, $zero + 1];
$upper = [$zero + 1, $zero];
while (true) {
$mediant = [$lower[0] + $upper[0], $lower[1] + $upper[1]];
if ($value * $mediant[1] > $mediant[0]) {
if ($limes < $mediant[1]) {
return $upper;
}
$lower = $mediant;
} elseif ($value * $mediant[1] === $mediant[0]) {
if ($limes >= $mediant[1]) {
return $mediant;
}
if ($lower[1] < $upper[1]) {
return $lower;
}
return $upper;
} else {
if ($limes < $mediant[1]) {
return $lower;
}
$upper = $mediant;
}
}
}
}
然后你像这样使用它：
QuantityTransform::decimalToFraction(0.06); // 0.06
QuantityTransform::decimalToFraction(0.75); // 3/4
QuantityTransform::decimalToFraction(1.75, ' and '); // 1 and 3/4
QuantityTransform::decimalToFraction(2.33, ' and '); // 2 and 1/3
QuantityTransform::decimalToFraction(2.58, ' ', 5); // 2 3/5
QuantityTransform::decimalToFraction(2.58, ' & ', 10); // 2 & 4/7
QuantityTransform::decimalToFraction(1.97); // 2
于 2021-04-14T08:43:52.200 回答
0
这是我解决这个问题的方法。适用于有理数。
function dec2fracso($dec){
//Negative number flag.
$num=$dec;
if($num<0){
$neg=true;
}else{
$neg=false;
}
//Extracts 2 strings from input number
$decarr=explode('.',(string)$dec);
//Checks for divided by zero input.
if($decarr[1]==0){
$decarr[1]=1;
$fraccion[0]=$decarr[0];
$fraccion[1]=$decarr[1];
return $fraccion;
}
//Calculates the divisor before simplification.
$long=strlen($decarr[1]);
$div="1";
for($x=0;$x<$long;$x++){
$div.="0";
}
//Gets the greatest common divisor.
$x=(int)$decarr[1];
$y=(int)$div;
$gcd=gmp_strval(gmp_gcd($x,$y));
//Calculates the result and fills the array with the correct sign.
if($neg){
$fraccion[0]=((abs($decarr[0])*($y/$gcd))+($x/$gcd))*(-1);
}else{
$fraccion[0]=(abs($decarr[0])*($y/$gcd))+($x/$gcd);
}
$fraccion[1]=($y/$gcd);
return $fraccion;
}
于 2014-03-26T04:29:10.950 回答
0
有时有必要只处理浮点数的小数。所以我创建了一个代码，它使用@Joni 创建的函数来呈现一种在烹饪食谱中很常见的格式，至少在巴西是这样。
因此，不是使用 3/2（即 1.5 的结果），而是使用我创建的函数，可以显示值 1 1/2，如果需要，还可以添加一个字符串来连接这些值，创建类似“1 和 1/2”。
function float2rat($n, $tolerance = 1.e-6) {
$h1=1; $h2=0;
$k1=0; $k2=1;
$b = 1/$n;
do {
$b = 1/$b;
$a = floor($b);
$aux = $h1; $h1 = $a*$h1+$h2; $h2 = $aux;
$aux = $k1; $k1 = $a*$k1+$k2; $k2 = $aux;
$b = $b-$a;
} while (abs($n-$h1/$k1) > $n*$tolerance);
return "$h1/$k1";
}
function float2fraction($float, $concat = ' '){
// ensures that the number is float,
// even when the parameter is a string
$float = (float)$float;
if($float == 0 ){
return $float;
}
// when float between -1 and 1
if( $float > -1 && $float < 0 || $float < 1 && $float > 0 ){
$fraction = float2rat($float);
return $fraction;
}
else{
// get the minor integer
if( $float < 0 ){
$integer = ceil($float);
}
else{
$integer = floor($float);
}
// get the decimal
$decimal = $float - $integer;
if( $decimal != 0 ){
$fraction = float2rat(abs($decimal));
$fraction = $integer . $concat . $fraction;
return $fraction;
}
else{
return $float;
}
}
}
用法例如：
echo float2fraction(1.5);
will return "1 1/2"
于 2020-07-15T14:53:00.953 回答