在数组中查找匹配或最接近的值

Question

对于给定的目标值，如何搜索和查找数组中最接近的值？

假设我有这个示例数组：

array(0, 5, 10, 11, 12, 20)

例如，当我用目标值0搜索时，函数应该返回0； 当我用 3 搜索时，它将返回 5； 当我用 14 搜索时，它会返回 12。

Answer 1

将您要搜索的数字作为第一个参数传递，并将数字数组传递给第二个参数：

function getClosest($search, $arr) {
   $closest = null;
   foreach ($arr as $item) {
      if ($closest === null || abs($search - $closest) > abs($item - $search)) {
         $closest = $item;
      }
   }
   return $closest;
}

Answer 2

一种特殊的懒惰方法是让 PHP 按照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);

foreach ($array as $i) {
    $smallest[$i] = abs($i - $num);
}
asort($smallest);
print key($smallest);

Answer 3

这是我为排序大数组编写的高性能函数

经测试，对于具有 20000 个元素的数组，主循环只需要约 20 次迭代。

请注意数组必须排序（升序）！

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查找最近的 LOWER 元素
• ARRAY_NEAREST_HIGHER找到最近的 HIGHER 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);

Answer 4

<?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);
}
?>

Answer 5

大多数时候， Tim 的实现都会削减它。 尽管如此，为了性能谨慎，您可以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中断搜索。

<?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] );
}

Answer 6

要在对象数组中搜索最近的值，您可以使用Tim Cooper's answer 中的改编代码。

<?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://stackoverflow.com/a/5464961/496176
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);
?>

Answer 7

function closestnumber($number, $candidates) {
$last = null;
foreach ($candidates as $cand) {
    if ($cand < $number) {
        $last = $cand;
    } else if ($cand == $number) {
        return $number;
    } else if ($cand > $number) {
        return $last;
    }
}
return $last;
}

这应该可以满足您的需求。

Answer 8

您可以简单地使用array_search ，它返回一个键，如果在数组中找到了许多搜索实例，它将返回它找到的第一个。

引用自 PHP ：

如果在 haystack 中多次找到 Needle，则返回第一个匹配的键。 要返回所有匹配值的键，请改用array_keys()和可选的 search_value 参数。

示例用法：

if(false !== ($index = array_search(12,array(0, 5, 10, 11, 12, 20))))
{
    echo $index; //5
}

---

更新：

function findNearest($number,$Array)
{
    //First check if we have an exact number
    if(false !== ($exact = array_search($number,$Array)))
    {
         return $Array[$exact];
    }

    //Sort the array
    sort($Array);

   //make sure our search is greater then the smallest value
   if ($number < $Array[0] ) 
   { 
       return $Array[0];
   }

    $closest = $Array[0]; //Set the closest to the lowest number to start

    foreach($Array as $value)
    {
        if(abs($number - $closest) > abs($value - $number))
        {
            $closest = $value;
        }
    }

    return $closest;
}

Answer 9

我根据Piyush Dholariya 的回答找到的最佳方法：

$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

Answer 10

例如，考虑到输入数组按升序排序asort() ，使用二分搜索进行搜索会快得多。

这是我用来在按 DateTime 对象排序的Iterable事件列表中插入新事件的一些代码的快速和肮脏的改编......

因此，此代码将返回左侧最近的点（之前/较小）。

如果您想找到数学上最近的点：考虑将搜索值与返回值的距离以及返回值右侧（下一个）的点（如果存在）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);        
}

Answer 11

试试这些……这不仅给出了最接近的匹配，而且还提供了与给定数字最接近的较高值和最接近的较低值。

 getNearestValue(lookup_array, lookup_value) {

        if (lookup_array.length > 0) {
          let nearestHighValue = lookup_array[0];
          let nearestLowValue = lookup_array[0];
          let nearestValue=0;
          let diff, diffPositive = Infinity;
          let diffNeg = -Infinity;
          lookup_array.forEach(num => {
            diff = lookup_value - num;
            if (diff >= 0 && diff <= diffPositive) {
              nearestLowValue = num;
              diffPositive = diff;
            }
            if (diff <= 0 && diff >= diffNeg) {
              nearestHighValue = num;
              diffNeg = diff;
            }

          })
          //If no value is higher than GivenNumber then  keep nearest low value as clossets
          if (diffNeg == -Infinity) {
            nearestHighValue = nearestLowValue;
          }
              //If no value is Lower than Givennumber then  keep nearest High value as clossets
          if (diffPositive == Infinity) {
            nearestLowValue = nearestHighValue;
          }
          if((lookup_value-nearestLowValue)<=(nearestHighValue-lookup_value))
            {
              nearestValue=nearestLowValue;
             }
         else
            {
            nearestValue=nearestHighValue;
            }
          return { NearHighest: nearestHighValue, NearLowest: nearestLowValue,NearestValue:nearestValue };
        }
        else {
          return null;
        }
      }

Answer 12

    $closest = 0;
    while ($rowDecontos = mysql_fetch_array($pvDescontos)) {
        if ($rowDecontos['n_dias'] > $closest && $rowDecontos['n_dias'] <= $numDias) {
            $closest = $rowDecontos['n_dias'] ;
        }
    };

Answer 13

我通过一些修改将接受的答案转换为 Kotlin。 请随时根据您的要求进行修改。

private fun getClosestTo1(array: List<Float>): Float? {
            var closest: Float? = null
            for (item in array) {
                if (item == 1.0f) {
                    return item
                } else if (closest == null || abs(1.minus(closest)) > abs(item - 1)) {
                    closest = item
                }
            }

            return closest
        }

Answer 14

二进制搜索以找到最接近的值（数组必须排序）：

function findClosest($sortedArr, $val)
{
    $low = 0;
    $high = count($sortedArr) - 1;
    while ($low <= $high) {
        if ($high - $low <= 1) {
            if (abs($sortedArr[$low] - $val) < abs($sortedArr[$high] - $val)) {
                return $sortedArr[$low];
            } else {
                return $sortedArr[$high];
            }
        }

        $mid = (int)(($high + $low) / 2);
        if ($val < $sortedArr[$mid]) {
            $high = $mid;
        } else {
            $low = $mid;
        }
    }

    // Empty array
    return false;
}

Answer 15

我将提供一个迟到的答案，通过维护两个临时变量和实现提前返回来努力避免不必要的迭代和过多的函数调用。

一个优雅的解决方案不应该需要大于n的时间复杂度——换句话说，大O应该是O(n)而小o应该是o(1) 。 通过预先对干草堆进行排序，然后再次迭代干草堆，大O只会变得更糟。 要实现o(1) ，您将需要在遇到相同匹配项时提前返回——无需进一步搜索。

我的代码段将任意返回距离最小的第一个出现的值（以防多个值具有相同的距离）。 OP 未指定任何其他行为。

与其他一些答案相比，一个微不足道的性能改进是abs()是循环内的唯一函数调用，每次迭代最多调用 1 次。 以前的一些答案会重新计算当前值的距离以及每次迭代中当前最接近的匹配——这是不必要的工作。

代码：（演示）

$haystack = [-6, 0, 5, 10, 11, 12, 20];

$needles = [0, 3, 14, -3];

function getNearest($needle, $haystack) {
    if (!$haystack) {
        throw new Exception('empty haystack');
    }
    $bestDistance = PHP_INT_MAX;
    foreach ($haystack as $value) {
        if ($value === $needle) {
            return $needle;
        }
        $distance = abs($value - $needle);
        if ($distance < $bestDistance) {
            $bestDistance = $distance;
            $keep = $value;
        }
    }
    return $keep ?? $value; // coalesce to silence potential IDE complaint
}

foreach ($needles as $needle) { // each test case
    echo "$needle -> " . getNearest($needle, $haystack) . "\n";
}

输出：

0 -> 0
3 -> 5
14 -> 12
-3 -> -6

Answer 16

这与Mario's answer 的方法相同，但我使用array_search()和min()而不是排序。 性能是一样的，所以它只是归结为偏好问题。

function findClosest(array $values, $match)
{
    $map = [];
    foreach ($values as $v) {
        $map[$v] = abs($match - $v);
    }
    return array_search(min($map), $map);
}

Answer 17

试试这个：（尚未测试）

function searchArray($needle, $haystack){
    $return = $haystack[0];
    $prevReturn = $return;
    foreach($haystack as $key=>$val){
         if($needle > $val) {
            $prevReturn = $return;
            $return = $val;
         }
         if($val >= $needle) {
            $prevReturn = $return;
            $return = $val;
            break;
         }
    }
    if((($return+$needle)/2) > (($prevReturn+$needle)/2)){
         //means that the needle is closer to $prevReturn
         return $prevReturn;
    }
    else return $return;
}

Answer 18

给定的代码可用于获取您的解决方案：

var counts = [4, 9, 15, 6, 2],
  goal = 5;

var closest = counts.reduce(function(prev, curr) {
  return (Math.abs(curr - goal) < Math.abs(prev - goal) ? curr : prev);
});

console.log(closest);