如何在数组中搜索并找到最接近给定目标值的值?
假设我有以下示例数组:
array(0, 5, 10, 11, 12, 20)
例如,当我使用目标值0进行搜索时,该函数应返回0;当我使用3进行搜索时,它应返回5;当我使用14进行搜索时,它应返回12。将您搜索的数字作为第一个参数传入,将数字数组作为第二个参数传入:
function getClosest($search, $arr) {
$closest = null;
foreach ($arr as $item) {
if ($closest === null || abs($search - $closest) > abs($item - $search)) {
$closest = $item;
}
}
return $closest;
}
一种常见的懒惰方法是让PHP根据与搜索数字的距离对数组进行排序:
$num = 3;
$array = array(0, 5, 10, 11, 12, 20);
$smallest = [];
foreach ($array as $i) {
$smallest[$i] = abs($i - $num);
}
asort($smallest);
print key($smallest);
我为大数组排序编写了高性能函数。
测试表明,对于一个包含20000个元素的数组,主循环只需要约20次迭代。
请注意数组必须是按升序排列的!
define('ARRAY_NEAREST_DEFAULT', 0);
define('ARRAY_NEAREST_LOWER', 1);
define('ARRAY_NEAREST_HIGHER', 2);
/**
* Finds nearest value in numeric array. Can be used in loops.
* Array needs to be non-assocative and sorted.
*
* @param array $array
* @param int $value
* @param int $method ARRAY_NEAREST_DEFAULT|ARRAY_NEAREST_LOWER|ARRAY_NEAREST_HIGHER
* @return int
*/
function array_numeric_sorted_nearest($array, $value, $method = ARRAY_NEAREST_DEFAULT) {
$count = count($array);
if($count == 0) {
return null;
}
$div_step = 2;
$index = ceil($count / $div_step);
$best_index = null;
$best_score = null;
$direction = null;
$indexes_checked = Array();
while(true) {
if(isset($indexes_checked[$index])) {
break ;
}
$curr_key = $array[$index];
if($curr_key === null) {
break ;
}
$indexes_checked[$index] = true;
// perfect match, nothing else to do
if($curr_key == $value) {
return $curr_key;
}
$prev_key = $array[$index - 1];
$next_key = $array[$index + 1];
switch($method) {
default:
case ARRAY_NEAREST_DEFAULT:
$curr_score = abs($curr_key - $value);
$prev_score = $prev_key !== null ? abs($prev_key - $value) : null;
$next_score = $next_key !== null ? abs($next_key - $value) : null;
if($prev_score === null) {
$direction = 1;
}else if ($next_score === null) {
break 2;
}else{
$direction = $next_score < $prev_score ? 1 : -1;
}
break;
case ARRAY_NEAREST_LOWER:
$curr_score = $curr_key - $value;
if($curr_score > 0) {
$curr_score = null;
}else{
$curr_score = abs($curr_score);
}
if($curr_score === null) {
$direction = -1;
}else{
$direction = 1;
}
break;
case ARRAY_NEAREST_HIGHER:
$curr_score = $curr_key - $value;
if($curr_score < 0) {
$curr_score = null;
}
if($curr_score === null) {
$direction = 1;
}else{
$direction = -1;
}
break;
}
if(($curr_score !== null) && ($curr_score < $best_score) || ($best_score === null)) {
$best_index = $index;
$best_score = $curr_score;
}
$div_step *= 2;
$index += $direction * ceil($count / $div_step);
}
return $array[$best_index];
}
ARRAY_NEAREST_DEFAULT
:查找最近的元素。ARRAY_NEAREST_LOWER
:查找最近且小于给定值的元素。ARRAY_NEAREST_HIGHER
:查找最近且大于给定值的元素。用法:
$test = Array(5,2,8,3,9,12,20,...,52100,52460,62000);
// sort an array and use array_numeric_sorted_nearest
// for multiple searches.
// for every iteration it start from half of chunk where
// first chunk is whole array
// function doesn't work with unosrted arrays, and it's much
// faster than other solutions here for sorted arrays
sort($test);
$nearest = array_numeric_sorted_nearest($test, 8256);
$nearest = array_numeric_sorted_nearest($test, 3433);
$nearest = array_numeric_sorted_nearest($test, 1100);
$nearest = array_numeric_sorted_nearest($test, 700);
<?php
$arr = array(0, 5, 10, 11, 12, 20);
function getNearest($arr,$var){
usort($arr, function($a,$b) use ($var){
return abs($a - $var) - abs($b - $var);
});
return array_shift($arr);
}
?>
基于Piyush Dholariya的回答,我找到的最佳方法如下:
$array = [4, 9, 15, 6, 2];
$goal = 7;
$closest = array_reduce($array, function($carry, $item) use($goal) {
return (abs($item - $goal) < abs($carry - $goal) ? $item : $carry);
}, reset($array)); // Returns 6
要在对象数组中搜索最近的值,您可以使用从Tim Cooper's answer改编的代码。
<?php
// create array of ten objects with random values
$images = array();
for ($i = 0; $i < 10; $i++)
$images[ $i ] = (object)array(
'width' => rand(100, 1000)
);
// print array
print_r($images);
// adapted function from Tim Copper's solution
// https://dev59.com/im035IYBdhLWcg3wW-z1#5464961
function closest($array, $member, $number) {
$arr = array();
foreach ($array as $key => $value)
$arr[$key] = $value->$member;
$closest = null;
foreach ($arr as $item)
if ($closest === null || abs($number - $closest) > abs($item - $number))
$closest = $item;
$key = array_search($closest, $arr);
return $array[$key];
}
// object needed
$needed_object = closest($images, 'width', 320);
// print result
print_r($needed_object);
?>
Tim的实现在大多数情况下都可以胜任。然而,对于那些注重性能的人来说,在迭代之前对列表进行排序,并在下一个差异大于上一个差异时停止搜索,是一种更好的选择。
<?php
function getIndexOfClosestValue ($needle, $haystack) {
if (count($haystack) === 1) {
return $haystack[0];
}
sort($haystack);
$closest_value_index = 0;
$last_closest_value_index = null;
foreach ($haystack as $i => $item) {
if (abs($needle - $haystack[$closest_value_index]) > abs($item - $needle)) {
$closest_value_index = $i;
}
if ($closest_value_index === $last_closest_value_index) {
break;
}
}
return $closest_value_index;
}
function getClosestValue ($needle, $haystack) {
return $haystack[getIndexOfClosestValue($needle, $haystack)];
}
// Test
$needles = [0, 2, 3, 4, 5, 11, 19, 20];
$haystack = [0, 5, 10, 11, 12, 20];
$expectation = [0, 0, 1, 1, 1, 3, 5, 5];
foreach ($needles as $i => $needle) {
var_dump( getIndexOfClosestValue($needle, $haystack) === $expectation[$i] );
}
这与Mario的答案相同,但我使用array_search()
和min()
而不是排序。性能相同,所以只是个人偏好的问题。
function findClosest(array $values, $match)
{
$map = [];
foreach ($values as $v) {
$map[$v] = abs($match - $v);
}
return array_search(min($map), $map);
}
考虑到输入数组是按升序排序的,例如使用asort()
,使用二分搜索将更快。
这是我正在使用的一些代码的快速适应,用于在按DateTime对象排序的Iterable事件列表中插入新事件...
因此,此代码将返回左侧(之前/较小)最近的点。
如果您想找到数学上最接近的点:请考虑将搜索值的距离与返回值及其右侧(下一个)点(如果存在)的距离进行比较。
function dichotomicSearch($search, $haystack, $position=false)
{
// Set a cursor between two values
if($position === false)
{ $position=(object) array(
'min' => 0,
'cur' => round(count($haystack)/2, 0, PHP_ROUND_HALF_ODD),
'max' => count($haystack)
);
}
// Return insertion point (to push using array_splice something at the right spot in a sorted array)
if(is_numeric($position)){return $position;}
// Return the index of the value when found
if($search == $haystack[$position->cur]){return $position->cur;}
// Searched value is smaller (go left)
if($search <= $haystack[$position->cur])
{
// Not found (closest value would be $position->min || $position->min+1)
if($position->cur == $position->min){return $position->min;}
// Resetting the interval from [min,max[ to [min,cur[
$position->max=$position->cur;
// Resetting cursor to the new middle of the interval
$position->cur=round($position->cur/2, 0, PHP_ROUND_HALF_DOWN);
return dichotomicSearch($search, $haystack, $position);
}
// Search value is greater (go right)
// Not found (closest value would be $position->max-1 || $position->max)
if($position->cur < $position->min or $position->cur >= $position->max){return $position->max;}
// Resetting the interval from [min,max[ to [cur,max[
$position->min = $position->cur;
// Resetting cursor to the new middle of the interval
$position->cur = $position->min + round(($position->max-$position->min)/2, 0, PHP_ROUND_HALF_UP);
if($position->cur >= $position->max){return $position->max;}
return dichotomicSearch($search, $haystack, $position);
}
function closestnumber($number, $candidates) {
$last = null;
foreach ($candidates as $cand) {
if ($cand < $number) {
$last = $cand;
} elseif ($cand == $number) {
return $number;
} elseif ($cand > $number) {
return $last;
}
}
return $last;
}
abs($search - $closest) >= abs($item - $search)
来将数字四舍五入到最接近的值。 - Sos.$closest === null
是用于第一次迭代之前,因为此时 closest 还没有被设置为任何值。 - Andy