Unexpected round() return - php

I am trying to round down a number using PHP's round() function. Here is the code I am using:
$line_item_price = 13.775;
echo round($line_item_price, 2, PHP_ROUND_HALF_DOWN);
Now when I run the code like this I am hoping to get the output 13.77, except I am getting 0 (or nothing -- not sure which yet).
Now when I remove the PHP_ROUND_HALF_DOWN I get 13.78. Anyone see what I am doing wrong here? It seems like this should be working correctly.

The mode parameter was introduced in version 5.3, therefore it will not work for you. You'll have to find a custom function to do what you are looking for.

You are using a function that is not yet available in your current version of PHP. One way to solve this problem is using the floor function.
$line_item_price = 13.775;
echo floor($line_item_price * 100) / 100;
What I'm doing here is too first multiply the value with 100 and then floor the value. This will give you a rounded down value with the precision of 2. Then to get the correct value you need to devide with 100.
The number 100 comes from the power(10, desired precision)

can you not just do:
echo round($line_item_price, 2)
?
I'm not 100% sure but I think the ROUND_HALF_DOWN etc are for fractions such as 1.5, 2.5 and integers.

Here is a way to do it:
$num = 13.775;
$tmp = intval($num*1000);
$dec = $tmp % 10;
if ($dec > 5) {
$rounded = (1+intval($tmp/10))/100;
} else {
$rounded = intval($tmp/10)/100;
}
echo $rounded,"\n";
This gives : 13.77 for $num=13.775 and 13.78 for $num=13.776

Actually, I'm kind of surprised that it works at all, since the number 13.775 is not exactly representable in floating point:
$ php -r 'printf("%.40f\n", 13.775);'
13.7750000000000003552713678800500929355621
Indeed, it seems that round() is a bit lax about what counts as "half":
$ php -r 'echo round(13.77500000000001, 2, PHP_ROUND_HALF_DOWN) . "\n";'
13.77
Anyway, if your PHP doesn't support PHP_ROUND_HALF_DOWN, here's a simple kluge that gives approximately the same functionality:
function round_half_down ( $num, $digits ) {
$mul = pow( 10, $digits );
return ceil( $num * $mul - 0.5 ) / $mul;
}
This does turn out to work as one would naively expect, but is slightly stricter than round(): round_half_down(13.775, 2) == 13.77, but round_half_down(13.77500000000001, 2) == 13.78. Also, as a curious edge case, round_half_down(0.001, 2) returns -0. If you don't like that, you can always pass the return value through sprintf("%.{$digits}F") to format it nicely.

Related

What's the most efficient way of randomly picking a floating number within a specific range? [duplicate]

How does one generate a random float between 0 and 1 in PHP?
I'm looking for the PHP's equivalent to Java's Math.random().
You may use the standard function: lcg_value().
Here's another function given on the rand() docs:
// auxiliary function
// returns random number with flat distribution from 0 to 1
function random_0_1()
{
return (float)rand() / (float)getrandmax();
}
Example from documentation :
function random_float ($min,$max) {
return ($min+lcg_value()*(abs($max-$min)));
}
rand(0,1000)/1000 returns:
0.348 0.716 0.251 0.459 0.893 0.867 0.058 0.955 0.644 0.246 0.292
or use a bigger number if you want more digits after decimal point
class SomeHelper
{
/**
* Generate random float number.
*
* #param float|int $min
* #param float|int $max
* #return float
*/
public static function rand($min = 0, $max = 1)
{
return ($min + ($max - $min) * (mt_rand() / mt_getrandmax()));
}
}
update:
forget this answer it doesnt work wit php -v > 5.3
What about
floatVal('0.'.rand(1, 9));
?
this works perfect for me, and it´s not only for 0 - 1 for example between 1.0 - 15.0
floatVal(rand(1, 15).'.'.rand(1, 9));
function mt_rand_float($min, $max, $countZero = '0') {
$countZero = +('1'.$countZero);
$min = floor($min*$countZero);
$max = floor($max*$countZero);
$rand = mt_rand($min, $max) / $countZero;
return $rand;
}
example:
echo mt_rand_float(0, 1);
result: 0.2
echo mt_rand_float(3.2, 3.23, '000');
result: 3.219
echo mt_rand_float(1, 5, '00');
result: 4.52
echo mt_rand_float(0.56789, 1, '00');
result: 0.69
$random_number = rand(1,10).".".rand(1,9);
function frand($min, $max, $decimals = 0) {
$scale = pow(10, $decimals);
return mt_rand($min * $scale, $max * $scale) / $scale;
}
echo "frand(0, 10, 2) = " . frand(0, 10, 2) . "\n";
This question asks for a value from 0 to 1. For most mathematical purposes this is usually invalid albeit to the smallest possible degree. The standard distribution by convention is 0 >= N < 1. You should consider if you really want something inclusive of 1.
Many things that do this absent minded have a one in a couple billion result of an anomalous result. This becomes obvious if you think about performing the operation backwards.
(int)(random_float() * 10) would return a value from 0 to 9 with an equal chance of each value. If in one in a billion times it can return 1 then very rarely it will return 10 instead.
Some people would fix this after the fact (to decide that 10 should be 9). Multiplying it by 2 should give around a ~50% chance of 0 or 1 but will also have a ~0.000000000465% chance of returning a 2 like in Bender's dream.
Saying 0 to 1 as a float might be a bit like mistakenly saying 0 to 10 instead of 0 to 9 as ints when you want ten values starting at zero. In this case because of the broad range of possible float values then it's more like accidentally saying 0 to 1000000000 instead of 0 to 999999999.
With 64bit it's exceedingly rare to overflow but in this case some random functions are 32bit internally so it's not no implausible for that one in two and a half billion chance to occur.
The standard solutions would instead want to be like this:
mt_rand() / (getrandmax() + 1)
There can also be small usually insignificant differences in distribution, for example between 0 to 9 then you might find 0 is slightly more likely than 9 due to precision but this will typically be in the billionth or so and is not as severe as the above issue because the above issue can produce an invalid unexpected out of bounds figure for a calculation that would otherwise be flawless.
Java's Math.random will also never produce a value of 1. Some of this comes from that it is a mouthful to explain specifically what it does. It returns a value from 0 to less than one. It's Zeno's arrow, it never reaches 1. This isn't something someone would conventionally say. Instead people tend to say between 0 and 1 or from 0 to 1 but those are false.
This is somewhat a source of amusement in bug reports. For example, any PHP code using lcg_value without consideration for this may glitch approximately one in a couple billion times if it holds true to its documentation but that makes it painfully difficult to faithfully reproduce.
This kind of off by one error is one of the common sources of "Just turn it off and on again." issues typically encountered in embedded devices.
Solution for PHP 7. Generates random number in [0,1). i.e. includes 0 and excludes 1.
function random_float() {
return random_int(0, 2**53-1) / (2**53);
}
Thanks to Nommyde in the comments for pointing out my bug.
>>> number_format((2**53-1)/2**53,100)
=> "0.9999999999999998889776975374843459576368331909179687500000000000000000000000000000000000000000000000"
>>> number_format((2**53)/(2**53+1),100)
=> "1.0000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000"
Most answers are using mt_rand. However, mt_getrandmax() usually returns only 2147483647. That means you only have 31 bits of information, while a double has a mantissa with 52 bits, which means there is a density of at least 2^53 for the numbers between 0 and 1.
This more complicated approach will get you a finer distribution:
function rand_754_01() {
// Generate 64 random bits (8 bytes)
$entropy = openssl_random_pseudo_bytes(8);
// Create a string of 12 '0' bits and 52 '1' bits.
$x = 0x000FFFFFFFFFFFFF;
$first12 = pack("Q", $x);
// Set the first 12 bits to 0 in the random string.
$y = $entropy & $first12;
// Now set the first 12 bits to be 0[exponent], where exponent is randomly chosen between 1 and 1022.
// Here $e has a probability of 0.5 to be 1022, 0.25 to be 1021, etc.
$e = 1022;
while($e > 1) {
if(mt_rand(0,1) == 0) {
break;
} else {
--$e;
}
}
// Pack the exponent properly (add four '0' bits behind it and 49 more in front)
$z = "\0\0\0\0\0\0" . pack("S", $e << 4);
// Now convert to a double.
return unpack("d", $y | $z)[1];
}
Please note that the above code only works on 64-bit machines with a Litte-Endian byte order and Intel-style IEEE754 representation. (x64-compatible computers will have this). Unfortunately PHP does not allow bit-shifting past int32-sized boundaries, so you have to write a separate function for Big-Endian.
You should replace this line:
$z = "\0\0\0\0\0\0" . pack("S", $e << 4);
with its big-endian counterpart:
$z = pack("S", $e << 4) . "\0\0\0\0\0\0";
The difference is only notable when the function is called a large amount of times: 10^9 or more.
Testing if this works
It should be obvious that the mantissa follows a nice uniform distribution approximation, but it's less obvious that a sum of a large amount of such distributions (each with cumulatively halved chance and amplitude) is uniform.
Running:
function randomNumbers() {
$f = 0.0;
for($i = 0; $i < 1000000; ++$i) {
$f += \math::rand_754_01();
}
echo $f / 1000000;
}
Produces an output of 0.49999928273099 (or a similar number close to 0.5).
I found the answer on PHP.net
<?php
function randomFloat($min = 0, $max = 1) {
return $min + mt_rand() / mt_getrandmax() * ($max - $min);
}
var_dump(randomFloat());
var_dump(randomFloat(2, 20));
?>
float(0.91601131712832)
float(16.511210331931)
So you could do
randomFloat(0,1);
or simple
mt_rand() / mt_getrandmax() * 1;
what about:
echo (float)('0.' . rand(0,99999));
would probably work fine... hope it helps you.

How to get log() of a very big number (PHP)?

I've looked at php-big numbers, BC Math, and GMP for dealing with very big numbers in php. But none seem to have a function equivilent to php's log(). For example I want to do this:
$result = log($bigNumber, 2);
Would anyone know of an alternate way to get the log base 2 of a arbitray precision point number in php? Maybe Ive missed a function, or library, or formula.
edit: php-bignumbers seems to have a log base 10 function only log10()
In general if you want to implement your high precision log own calculation, I'd suggest 1st use the basic features of logarithm:
log_a(x) = log_b(x) / log_b(a) |=> thus you can recalulate logarith to any base
log(x*y) = log(x) + log(y)
log(a**n) = n*log(a)
where log_a(x) - meaning logarithm to the base a of x; log means natural logarithm
So log(1000000000000000000000.123) = 21*log(1.000000000000000000000123)
and for high precision of log(1+x)
use algorithm referenced at
http://en.wikipedia.org/wiki/Natural_logarithm#High_precision
One solution combining the suggestions so far would be to use this formula:
log2($num) = log10($num) / log10(2)
in conjunction with php-big numbers since it has a pre-made log10 function.
eg, after installing the php-big numbers library, use:
$log2 = log10($bigNum) / log10(2);
Personally I've decided to use different math/logic so as to not need the log function, and just using bcmath for the big numbers.
One of the great things about base 2 is that counting and shifting become part of the tool set.
So one way to get a 'log2' of a number is to convert it to a binary string and count the bits.
You can accomplish this equivalently by dividing by 2 in a loop. But it seems to me that counting would be more efficient.
gmp_scan0 and gmp_scan1 can be used if you are counting from the right. But you'd have to somehow convert the mixed bits to all ones and zeroes.
But using gmp_strval(num, 2), you can produce a string and do a strpos on it.
if the whole value is being converted, you can do a (strlen - 1) on it.
Obviously this only works when you want an integer log.
I've had a very similar problem just recently.. and so I just scaled the number considerably in order to use the inbuild log to find the fractional part.. (I prefere the log10 for some reason.. don't ask... people are strange, me too)
I hope this is selfexplanatory enough..
it returns a float value (since that's what I needed)
function gmp_log($num, $base=10, $full=true)
{
if($base == 10)
$string = gmp_strval($num);
else
$string = gmp_strval($num,$base);
$intpart = strlen($string)-1;
if(!$full)
return $intpart;
if($base ==10)
{
$string = substr_replace($string, ".", 1, 0);
$number = floatval($string);
$lg = $intpart + log10($number);
return $lg;
}
else
{
$string = gmp_strval($num);
$intpart = strlen($string)-1;
$string = substr_replace($string, ".", 1, 0);
$number = floatval($string);
$lg = $intpart + log10($number);
$lb = $lg / log10($base);
return $lb;
}
}
it's quick, it's dirty... but it works well enough to get the log of some RSA sized integers ;)
usage is straight forward as well
$N = gmp_init("11002930366353704069");
echo gmp_log($N,10)."\n";
echo gmp_log($N,10, false)."\n";
echo gmp_log($N,2)."\n";
echo gmp_log($N,16)."\n";
returns
19.041508364472
19
63.254521604973
15.813630401243

Random Float between 0 and 1 in PHP

How does one generate a random float between 0 and 1 in PHP?
I'm looking for the PHP's equivalent to Java's Math.random().
You may use the standard function: lcg_value().
Here's another function given on the rand() docs:
// auxiliary function
// returns random number with flat distribution from 0 to 1
function random_0_1()
{
return (float)rand() / (float)getrandmax();
}
Example from documentation :
function random_float ($min,$max) {
return ($min+lcg_value()*(abs($max-$min)));
}
rand(0,1000)/1000 returns:
0.348 0.716 0.251 0.459 0.893 0.867 0.058 0.955 0.644 0.246 0.292
or use a bigger number if you want more digits after decimal point
class SomeHelper
{
/**
* Generate random float number.
*
* #param float|int $min
* #param float|int $max
* #return float
*/
public static function rand($min = 0, $max = 1)
{
return ($min + ($max - $min) * (mt_rand() / mt_getrandmax()));
}
}
update:
forget this answer it doesnt work wit php -v > 5.3
What about
floatVal('0.'.rand(1, 9));
?
this works perfect for me, and it´s not only for 0 - 1 for example between 1.0 - 15.0
floatVal(rand(1, 15).'.'.rand(1, 9));
function mt_rand_float($min, $max, $countZero = '0') {
$countZero = +('1'.$countZero);
$min = floor($min*$countZero);
$max = floor($max*$countZero);
$rand = mt_rand($min, $max) / $countZero;
return $rand;
}
example:
echo mt_rand_float(0, 1);
result: 0.2
echo mt_rand_float(3.2, 3.23, '000');
result: 3.219
echo mt_rand_float(1, 5, '00');
result: 4.52
echo mt_rand_float(0.56789, 1, '00');
result: 0.69
$random_number = rand(1,10).".".rand(1,9);
function frand($min, $max, $decimals = 0) {
$scale = pow(10, $decimals);
return mt_rand($min * $scale, $max * $scale) / $scale;
}
echo "frand(0, 10, 2) = " . frand(0, 10, 2) . "\n";
This question asks for a value from 0 to 1. For most mathematical purposes this is usually invalid albeit to the smallest possible degree. The standard distribution by convention is 0 >= N < 1. You should consider if you really want something inclusive of 1.
Many things that do this absent minded have a one in a couple billion result of an anomalous result. This becomes obvious if you think about performing the operation backwards.
(int)(random_float() * 10) would return a value from 0 to 9 with an equal chance of each value. If in one in a billion times it can return 1 then very rarely it will return 10 instead.
Some people would fix this after the fact (to decide that 10 should be 9). Multiplying it by 2 should give around a ~50% chance of 0 or 1 but will also have a ~0.000000000465% chance of returning a 2 like in Bender's dream.
Saying 0 to 1 as a float might be a bit like mistakenly saying 0 to 10 instead of 0 to 9 as ints when you want ten values starting at zero. In this case because of the broad range of possible float values then it's more like accidentally saying 0 to 1000000000 instead of 0 to 999999999.
With 64bit it's exceedingly rare to overflow but in this case some random functions are 32bit internally so it's not no implausible for that one in two and a half billion chance to occur.
The standard solutions would instead want to be like this:
mt_rand() / (getrandmax() + 1)
There can also be small usually insignificant differences in distribution, for example between 0 to 9 then you might find 0 is slightly more likely than 9 due to precision but this will typically be in the billionth or so and is not as severe as the above issue because the above issue can produce an invalid unexpected out of bounds figure for a calculation that would otherwise be flawless.
Java's Math.random will also never produce a value of 1. Some of this comes from that it is a mouthful to explain specifically what it does. It returns a value from 0 to less than one. It's Zeno's arrow, it never reaches 1. This isn't something someone would conventionally say. Instead people tend to say between 0 and 1 or from 0 to 1 but those are false.
This is somewhat a source of amusement in bug reports. For example, any PHP code using lcg_value without consideration for this may glitch approximately one in a couple billion times if it holds true to its documentation but that makes it painfully difficult to faithfully reproduce.
This kind of off by one error is one of the common sources of "Just turn it off and on again." issues typically encountered in embedded devices.
Solution for PHP 7. Generates random number in [0,1). i.e. includes 0 and excludes 1.
function random_float() {
return random_int(0, 2**53-1) / (2**53);
}
Thanks to Nommyde in the comments for pointing out my bug.
>>> number_format((2**53-1)/2**53,100)
=> "0.9999999999999998889776975374843459576368331909179687500000000000000000000000000000000000000000000000"
>>> number_format((2**53)/(2**53+1),100)
=> "1.0000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000"
Most answers are using mt_rand. However, mt_getrandmax() usually returns only 2147483647. That means you only have 31 bits of information, while a double has a mantissa with 52 bits, which means there is a density of at least 2^53 for the numbers between 0 and 1.
This more complicated approach will get you a finer distribution:
function rand_754_01() {
// Generate 64 random bits (8 bytes)
$entropy = openssl_random_pseudo_bytes(8);
// Create a string of 12 '0' bits and 52 '1' bits.
$x = 0x000FFFFFFFFFFFFF;
$first12 = pack("Q", $x);
// Set the first 12 bits to 0 in the random string.
$y = $entropy & $first12;
// Now set the first 12 bits to be 0[exponent], where exponent is randomly chosen between 1 and 1022.
// Here $e has a probability of 0.5 to be 1022, 0.25 to be 1021, etc.
$e = 1022;
while($e > 1) {
if(mt_rand(0,1) == 0) {
break;
} else {
--$e;
}
}
// Pack the exponent properly (add four '0' bits behind it and 49 more in front)
$z = "\0\0\0\0\0\0" . pack("S", $e << 4);
// Now convert to a double.
return unpack("d", $y | $z)[1];
}
Please note that the above code only works on 64-bit machines with a Litte-Endian byte order and Intel-style IEEE754 representation. (x64-compatible computers will have this). Unfortunately PHP does not allow bit-shifting past int32-sized boundaries, so you have to write a separate function for Big-Endian.
You should replace this line:
$z = "\0\0\0\0\0\0" . pack("S", $e << 4);
with its big-endian counterpart:
$z = pack("S", $e << 4) . "\0\0\0\0\0\0";
The difference is only notable when the function is called a large amount of times: 10^9 or more.
Testing if this works
It should be obvious that the mantissa follows a nice uniform distribution approximation, but it's less obvious that a sum of a large amount of such distributions (each with cumulatively halved chance and amplitude) is uniform.
Running:
function randomNumbers() {
$f = 0.0;
for($i = 0; $i < 1000000; ++$i) {
$f += \math::rand_754_01();
}
echo $f / 1000000;
}
Produces an output of 0.49999928273099 (or a similar number close to 0.5).
I found the answer on PHP.net
<?php
function randomFloat($min = 0, $max = 1) {
return $min + mt_rand() / mt_getrandmax() * ($max - $min);
}
var_dump(randomFloat());
var_dump(randomFloat(2, 20));
?>
float(0.91601131712832)
float(16.511210331931)
So you could do
randomFloat(0,1);
or simple
mt_rand() / mt_getrandmax() * 1;
what about:
echo (float)('0.' . rand(0,99999));
would probably work fine... hope it helps you.

How can I make sure a float will always be rounded up with PHP?

I want to make sure a float in PHP is rounded up if any decimal is present, without worrying about mathematical rounding rules. This function would work as follows:
1.1 to 2
1.2 to 2
1.9 to 2
2.3 to 3
2.8 to 3
I know the round() function exists but I don't see any function for rounding up if any decimal is found. Is there any easy way to do this?
Use the ceil function:
$number = ceil(1.1); //2
I know this is an old topic, however it appears in Google. I will extend Blake Plumb's answer regarding precision.
ceil(1024.321 * 100) / 100;
Multiplying by 100 and dividing by 100 only works with one-hundredths. This isn't accurate on tenths, one-thousandths, one-hundred thousandths, etc.
function round_up($number, $precision = 2)
{
$fig = pow(10, $precision);
return (ceil($number * $fig) / $fig);
}
Results:
var_dump(round_up(1024.654321, 0)); // Output: float(1025)
var_dump(round_up(1024.654321, 1)); // Output: float(1024.7)
var_dump(round_up(1024.654321, 2)); // Output: float(1024.66)
var_dump(round_up(1024.654321, 3)); // Output: float(1024.655)
var_dump(round_up(1024.654321, 4)); // Output: float(1024.6544)
var_dump(round_up(1024.654321, 5)); // Output: float(1024.65433)
var_dump(round_up(1024.654321, 6)); // Output: float(1024.654321)
Notes:
Thanks for the contributions from Joseph McDermott and brandom for improving my original snippet.
The official Ceil function will do that for you.
Taken from the example:
<?php
echo ceil(4.3); // 5
echo ceil(9.999); // 10
echo ceil(-3.14); // -3
?>
I know this question has long since been answered, but it came up when I did a google search on the topic. If you want to round up with precision, then a good method would be to use the ceil function and times the number by how many decimal points you want to represent and then divide by that number.
ceil(1024.321*100)/100
would produce 1024.33
I like Ash's response, although I would have:
$fig = (int) str_pad('1', $precision + 1, '0');
Makes sense that if I provide precision '2', I would expect it rounded to 2 decimal places. Matter of choice though I suppose. Thanks for the answer Ash, works well.

What's the best way to get the fractional part of a float in PHP?

How would you find the fractional part of a floating point number in PHP?
For example, if I have the value 1.25, I want to return 0.25.
$x = $x - floor($x)
$x = fmod($x, 1);
Here's a demo:
<?php
$x = 25.3333;
$x = fmod($x, 1);
var_dump($x);
Should ouptut
double(0.3333)
Credit.
Don't forget that you can't trust floating point arithmetic to be 100% accurate. If you're concerned about this, you'll want to look into the BCMath Arbitrary Precision Mathematics functions.
$x = 22.732423423423432;
$x = bcsub(abs($x),floor(abs($x)),20);
You could also hack on the string yourself
$x = 22.732423423423432;
$x = strstr ( $x, '.' );
The answer provided by nlucaroni will only work for positive numbers. A possible solution that works for both positive as well as negative numbers is:
$x = $x - intval($x)
If if the number is negative, you'll have to do this:
$x = abs($x) - floor(abs($x));
My PHP skills are lacking but you could minus the result of a floor from the original number
However, if you are dealing with something like perlin noise or another graphical representation, the solution which was accepted is correct. It will give you the fractional part from the lower number.
i.e:
.25 : 0 is integer below, fractional part is .25
-.25 : -1 is integer below, fractional part is .75
With the other solutions, you will repeat 0 as integer below, and worse, you will get reversed fractional values for all negative numbers.
Some of the preceding answers are partial. This, I believe, is what you need to handle all situations:
function getDecimalPart($floatNum) {
return abs($floatNum - intval($floatNum));
}
$decimalPart = getDecimalPart($floatNum);
You can use fmod function:
$y = fmod($x, 1); //$x = 1.25 $y = 0.25
To stop the confusion on this page actually this is the best answer, which is fast and works for both positive and negative values of $x:
$frac=($x<0) ? $x-ceil($x) : $x-floor($x);
I ran speed tests of 10 million computations on PHP 7.2.15 and even though both solutions give the same results, fmod is slower than floor/ceil.
$frac=($x<0) ? $x-ceil($x) : $x-floor($x);
-> 490-510 ms (depending on the sign of $x)
$frac=fmod($x, 1);
-> 590 - 1000 ms (depending on the value of $x)
Whereas the actual empty loop itself takes 80 ms (which is included in above timings).
Test script:
$x=sqrt(2)-0.41421356237;
$time_start = microtime(true);
for ($i=0;$i<=9999999;$i++) {
//$frac=fmod($x, 1); // version a
$frac=($x<0) ? $x-ceil($x) : $x-floor($x); // version b
}
$time_end = microtime(true);
$time = $time_end - $time_start;

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