Question

我有一个带有属性的关系模式（A B C D）。我也有一套功能依赖。

现在我需要确定R属性的所有可能子集的闭包。这就是我被困的地方。我需要学习如何在PHP中找到子集（非重复）。

My Array存储方式如下。

$ATTRIBUTES = ('A', 'B', 'C', 'D').

所以我的子集应该是

$SUBSET = ('A', 'B', 'C', 'D', 'AB', 'AC', AD', 'BC', 'BD', 'CD', 'ABC', 'ABD', 'BCD', 'ABCD')

代码不应该是大的东西，但出于某种原因，我无法理解它。

Answer 1

您希望$attributes的功率设定？这就是你的问题所暗示的。

可以找到一个示例here（引用完整性）

<?php 
/** 
* Returns the power set of a one dimensional array, a 2-D array. 
* [a,b,c] -> [ [a], [b], [c], [a, b], [a, c], [b, c], [a, b, c] ]
*/ 
function powerSet($in,$minLength = 1) { 
   $count = count($in); 
   $members = pow(2,$count); 
   $return = array(); 
   for ($i = 0; $i < $members; $i++) { 
      $b = sprintf("%0".$count."b",$i); 
      $out = array(); 
      for ($j = 0; $j < $count; $j++) { 
         if ($b{$j} == '1') $out[] = $in[$j]; 
      } 
      if (count($out) >= $minLength) { 
         $return[] = $out; 
      } 
   } 
   return $return; 
}

Answer 2

使用php array_merge，我们可以有一个很好的短powerSet函数

function powerSet($array) {
    // add the empty set
    $results = array(array());

    foreach ($array as $element) {
        foreach ($results as $combination) {
            $results[] = array_merge(array($element), $combination);
        }
    }

    return $results;
}

Answer 3

这是一个回溯解决方案。

给定一个返回输入集的所有L-lenght子集的函数，找到从L = 2到数据集输入长度的所有L-lenght子集

<?php

function subsets($S,$L) {
    $a = $b = 0;
    $subset = [];
    $result = [];
    while ($a < count($S)) {
        $current = $S[$a++];
        $subset[] = $current;
        if (count($subset) == $L) {
            $result[] = json_encode($subset);
            array_pop($subset);
        }
        if ($a == count($S)) {
            $a = ++$b;
            $subset = [];
        }
    }
    return $result;
}



$S = [ 'A', 'B', 'C', 'D'];
$L = 2;


// L = 1 -> no need to do anything
print_r($S);

for ($i = 2; $i <= count($S); $i++)
    print_r(subsets($S,$i));

Answer 4

基于@Yada's answer，这将生成数组的幂集，但会在每个子集中保留原始数组的键（返回值仍会按数字顺序索引）。如果您需要关联数组的子集，这将非常有用。

子集还保留原始数组的元素顺序。我向$results添加了稳定的排序，因为我需要它，但是您可以忽略它。

function power_set($array) {
    $results = [[]];
    foreach ($array as $key => $value) {
        foreach ($results as $combination) {
            $results[] = $combination + [$key => $value];
        }
    }

    # array_shift($results); # uncomment if you don't want the empty set in your results
    $order = array_map('count', $results);
    uksort($results, function($key_a, $key_b) use ($order) {
        $comp = $order[$key_a] - $order[$key_b]; # change only this to $order[$key_b] - $order[$key_a] for descending size
        if ($comp == 0) {
            $comp = $key_a - $key_b;
        }
        return $comp;
    });
    return array_values($results);
}

给出OP的输入，var_dump(power_set(['A', 'B', 'C', 'D']));提供：

array(16) {
  [0] =>
  array(0) {
  }
  [1] =>
  array(1) {
    [0] =>
    string(1) "A"
  }
  [2] =>
  array(1) {
    [1] =>
    string(1) "B"
  }
  [3] =>
  array(1) {
    [2] =>
    string(1) "C"
  }
  [4] =>
  array(1) {
    [3] =>
    string(1) "D"
  }
  [5] =>
  array(2) {
    [0] =>
    string(1) "A"
    [1] =>
    string(1) "B"
  }
  [6] =>
  array(2) {
    [0] =>
    string(1) "A"
    [2] =>
    string(1) "C"
  }
  [7] =>
  array(2) {
    [1] =>
    string(1) "B"
    [2] =>
    string(1) "C"
  }
  [8] =>
  array(2) {
    [0] =>
    string(1) "A"
    [3] =>
    string(1) "D"
  }
  [9] =>
  array(2) {
    [1] =>
    string(1) "B"
    [3] =>
    string(1) "D"
  }
  [10] =>
  array(2) {
    [2] =>
    string(1) "C"
    [3] =>
    string(1) "D"
  }
  [11] =>
  array(3) {
    [0] =>
    string(1) "A"
    [1] =>
    string(1) "B"
    [2] =>
    string(1) "C"
  }
  [12] =>
  array(3) {
    [0] =>
    string(1) "A"
    [1] =>
    string(1) "B"
    [3] =>
    string(1) "D"
  }
  [13] =>
  array(3) {
    [0] =>
    string(1) "A"
    [2] =>
    string(1) "C"
    [3] =>
    string(1) "D"
  }
  [14] =>
  array(3) {
    [1] =>
    string(1) "B"
    [2] =>
    string(1) "C"
    [3] =>
    string(1) "D"
  }
  [15] =>
  array(4) {
    [0] =>
    string(1) "A"
    [1] =>
    string(1) "B"
    [2] =>
    string(1) "C"
    [3] =>
    string(1) "D"
  }
}

Answer 5

在 @fbstj answer 之后，我更新函数：

使用 bitwize operators 而不是 sprintf (@Titus comments)
处理空集 (@James Stott & @fbstj comments)
将语法更新为 PHP 7+

function powerSet(array $in, int $minLength = 0): array
{
    $return = [];
    
    if ($minLength === 0) {
        $return[] = [];
    }

    for ($i = 1 << count($in); --$i;) {
        $out = [];

        foreach ($in as $j => $u) {
            if ($i >> $j & 1) {
                $out[] = $u;
            }
        }

        if (count($out) >= $minLength) {
            $return[] = $out;
        }
    }
    
    return $return;
}

由于幂集函数会大量增加内存负载（2^count($in) 次迭代），考虑使用Generator：

function powerSet(array $in, int $minLength = 0): \Generator
{
    if ($minLength === 0) {
        yield [];
    }

    for ($i = 1 << count($in); --$i;) {
        $out = [];

        foreach ($in as $j => $u) {
            if ($i >> $j & 1) {
                $out[] = $u;
            }
        }

        if (count($out) >= $minLength) {
            yield $out;
        }
    }
}

用法：

foreach (powerSet(range(1, 10)) as $value) {
    echo implode(', ', $value) . "\n";
}

在PHP中查找数组的子集

5 个答案: