php PHP数组组合
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PHP array combinations
提问by NVG
I have an array of 7 numbers (1,2,3,4,5,6,7) and I want to make pairs of 5 numbers like (1,2,3,4,5),(1,2,3,4,6,),(1,2,3,4,7) . (1,2,3,4,5) is equal to (4,5,3,1,2)
我有一个由 7 个数字 (1,2,3,4,5,6,7) 组成的数组，我想制作成对的 5 个数字，例如 (1,2,3,4,5),(1,2,3 ,4,6,),(1,2,3,4,7) 。(1,2,3,4,5) 等于 (4,5,3,1,2)
I would like to know if there is a function in PHP or any algorithm that can do this ? I have no idea where to start from. Can you help me ?
我想知道 PHP 中是否有函数或任何算法可以做到这一点？我不知道从哪里开始。你能帮助我吗 ？
I want all the combinations of 7 given numbers ( they are taken from an array ) put into 5 slots,disregarding order
我希望将 7 个给定数字的所有组合（它们取自一个数组）放入 5 个插槽中，不考虑顺序
回答by Galen
You can use the solution found here http://stereofrog.com/blok/on/070910.
您可以使用此处找到的解决方案http://stereofrog.com/blok/on/070910。
Incase the link goes down here's the code....
如果链接失效，这里是代码....
class Combinations implements Iterator
{
protected $c = null;
protected $s = null;
protected $n = 0;
protected $k = 0;
protected $pos = 0;
function __construct($s, $k) {
if(is_array($s)) {
$this->s = array_values($s);
$this->n = count($this->s);
} else {
$this->s = (string) $s;
$this->n = strlen($this->s);
}
$this->k = $k;
$this->rewind();
}
function key() {
return $this->pos;
}
function current() {
$r = array();
for($i = 0; $i < $this->k; $i++)
$r[] = $this->s[$this->c[$i]];
return is_array($this->s) ? $r : implode('', $r);
}
function next() {
if($this->_next())
$this->pos++;
else
$this->pos = -1;
}
function rewind() {
$this->c = range(0, $this->k);
$this->pos = 0;
}
function valid() {
return $this->pos >= 0;
}
protected function _next() {
$i = $this->k - 1;
while ($i >= 0 && $this->c[$i] == $this->n - $this->k + $i)
$i--;
if($i < 0)
return false;
$this->c[$i]++;
while($i++ < $this->k - 1)
$this->c[$i] = $this->c[$i - 1] + 1;
return true;
}
}
foreach(new Combinations("1234567", 5) as $substring)
echo $substring, ' ';
12345 12346 12347 12356 12357 12367 12456 12457 12467 12567 13456 13457 13467 13567 14567 23456 23457 23467 23567 24567 34567
12345 12346 12347 12356 12357 12367 12456 12457 12467 12567 13456 13457 13467 13567 14567 23467 23457 3 5 6 5 2 5
回答by lalu tale
<?php
echo "<pre>";
$test = array("test_1","test_2","test_3");
// Get Combination
$return = uniqueCombination($test);
//Sort
sort($return);
//Pretty Print
print_r(array_map(function($v){ return implode(",", $v); }, $return));
function uniqueCombination($in, $minLength = 1, $max = 2000) {
$count = count($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 ++) {
$b{$j} == '1' and $out[] = $in[$j];
}
count($out) >= $minLength && count($out) <= $max and $return[] = $out;
}
return $return;
}
?>
output
输出
Array
(
[0] => test_1
[1] => test_2
[2] => test_3
[3] => test_1,test_2
[4] => test_1,test_3
[5] => test_2,test_3
[6] => test_1,test_2,test_3
)
回答by Salman A
The Math_Combinatoricsin PEAR repository does exactly what you want:
将Math_Combinatorics在PEAR库不正是你想要什么：
A package that returns all the combinations and permutations, without repetition, of a given set and subset size. Associative arrays are preserved.
一个包，返回给定集合和子集大小的所有组合和排列，没有重复。关联数组被保留。
require_once 'Math/Combinatorics.php';
$combinatorics = new Math_Combinatorics;
$input = array(1, 2, 3, 4, 5, 6, 7);
$output = $combinatorics->combinations($input, 5); // 5 is the subset size
// 1,2,3,4,5
// 1,2,3,4,6
// 1,2,3,4,7
// 1,2,3,5,6
// 1,2,3,5,7
// 1,2,3,6,7
// 1,2,4,5,6
// 1,2,4,5,7
// 1,2,4,6,7
// 1,2,5,6,7
// 1,3,4,5,6
// 1,3,4,5,7
// 1,3,4,6,7
// 1,3,5,6,7
// 1,4,5,6,7
// 2,3,4,5,6
// 2,3,4,5,7
// 2,3,4,6,7
// 2,3,5,6,7
// 2,4,5,6,7
// 3,4,5,6,7
回答by Nguyen Van Vinh
Another solution that bases on stack. It's quit fast but eats much memory.
另一种基于堆栈的解决方案。它退出很快，但占用了很多内存。
Hope that helps someone.
希望能帮助某人。
In detail:
详细：
function _combine($numbers, $length)
{
$combinations = array();
$stack = array();
// every combinations can be ordered
sort($numbers);
// startup
array_push($stack, array(
'store' => array(),
'options' => $numbers,
));
while (true) {
// pop a item
$item = array_pop($stack);
// end of stack
if (!$item) {
break;
}
// valid store
if ($length <= count($item['store'])) {
$combinations[] = $item['store'];
continue;
}
// bypass when options are not enough
if (count($item['store']) + count($item['options']) < $length) {
continue;
}
foreach ($item['options'] as $index => $n) {
$newStore = $item['store'];
$newStore[] = $n;
// every combine can be ordered
// so accept only options which is greater than store numbers
$newOptions = array_slice($item['options'], $index + 1);
// push new items
array_push($stack, array(
'store' => $newStore,
'options' => $newOptions,
));
}
}
return $combinations;
}
回答by Nguyen Van Vinh
New solution which optimizes speed and memory for combining algorithm
优化组合算法的速度和内存的新解决方案
Mindset: generate combinations K numbers of Array of numbers. New solution will use K 'for' statements. One 'for' One number. Such as: $K = 5 mean that 5 of 'for' statements is used
心态：生成 K 个数字数组的组合。新的解决方案将使用 K 'for' 语句。一个“为”一个数字。如：$K = 5 表示使用了 5 个 'for' 语句
$total = count($array);
$i0 = -1;
for ($i1 = $i0 + 1; $i1 < $total; $i1++) {
for ($i2 = $i1 + 1; $i2 < $total; $i2++) {
for ($i3 = $i2 + 1; $i3 < $total; $i3++) {
for ($i4 = $i3 + 1; $i4 < $total; $i4++) {
for ($i5 = $i4 + 1; $i5 < $total; $i5++) {
$record = array();
for ($i = 1; $i <= $k; $i++) {
$t = "i$i";
$record[] = $array[$$t];
}
$callback($record);
}
}
}
}
}
And detail of code that generated the real code that will be execute by eval() function
以及生成将由 eval() 函数执行的真实代码的代码的详细信息
function combine($array, $k, $callback)
{
$total = count($array);
$init = '
$i0 = -1;
';
$sample = '
for($i{current} = $i{previous} + 1; $i{current} < $total; $i{current}++ ) {
{body}
}
';
$do = '
$record = array();
for ($i = 1; $i <= $k; $i++) {
$t = "i$i";
$record[] = $array[$$t];
}
$callback($record);
';
$for = '';
for ($i = $k; $i >= 1; $i--) {
switch ($i) {
case $k:
$for = str_replace(['{current}', '{previous}', '{body}'], [$i, $i - 1, $do], $sample);
break;
case 1:
$for = $init . str_replace(['{current}', '{previous}', '{body}'], [$i, $i - 1, $for], $sample);
break;
default:
$for = str_replace(['{current}', '{previous}', '{body}'], [$i, $i - 1, $for], $sample);
break;
}
}
// execute
eval($for);
}
How to combine K numbers of Array
如何组合 K 个数组
$k = 5;
$array = array(1, 2, 3, 4, 5, 6, 7);
$callback = function ($record) {
echo implode($record) . "\n";
};
combine($array, $k, $callback);
