多维数组中的数组排列保留键 PHP
这两天我一直在疯狂地尝试完成这个任务,也许你可以启发我。这是针对赛马投注排列的。每次用户玩游戏时,我都会得到一个多维数组(2 个级别)。第一级包含比赛 ID,第二级包含用户为该比赛选择的马匹。它看起来像这样:
$play = array
(
'4' => array(7, 32),
'8' => array(4),
'2' => array(9),
'12' => array('5'),
'83' => array('10', '11', '12', ''),
'9' => array('3'),
);
我需要知道该游戏的所有可能组合是什么。使用此函数可以轻松完成此操作:
function permutations(array $array)
{
switch (count($array)) {
case 1:
return $array[0];
break;
case 0:
throw new InvalidArgumentException('Requires at least one array');
break;
}
$a = array_shift($array);
$b = permutations($array);
$return = array();
foreach ($a as $key => $v) {
if(is_numeric($v))
{
foreach ($b as $key2 => $v2) {
$return[] = array_merge(array($v), (array) $v2);
}
}
}
return $return;
}
这将返回一个包含所有可能组合的数组。到目前为止一切顺利,结果如下所示:
Array
(
[0] => Array
(
[0] => 7
[1] => 4
[2] => 9
[3] => 5
[4] => 10
[5] => 3
)
[1] => Array
(
[0] => 7
[1] => 4
[2] => 9
[3] => 5
[4] => 11
[5] => 3
)
[2] => Array
(
[0] => 7
[1] => 4
[2] => 9
[3] => 5
[4] => 12
[5] => 3
)
[3] => Array
(
[0] => 32
[1] => 4
[2] => 9
[3] => 5
[4] => 10
[5] => 3
)
[4] => Array
(
[0] => 32
[1] => 4
[2] => 9
[3] => 5
[4] => 11
[5] => 3
)
[5] => Array
(
[0] => 32
[1] => 4
[2] => 9
[3] => 5
[4] => 12
[5] => 3
)
)
我的问题: 我需要每匹马的数组“key”作为“比赛 ID”,而不是 0,1,2,3。 我需要这样的结果:
Array
(
[0] => Array
(
[4] => 7
[8] => 4
[2] => 9
[12] => 5
[83] => 10
[9] => 3
)
[1] => Array
(
[4] => 7
[8] => 4
[2] => 9
[12] => 5
[83] => 11
[9] => 3
)
[2] => Array
(
[4] => 7
[8] => 4
[2] => 9
[12] => 5
[83] => 12
[9] => 3
)
[3] => Array
(
[4] => 32
[8] => 4
[2] => 9
[12] => 5
[83] => 10
[9] => 3
)
[4] => Array
(
[4] => 32
[8] => 4
[2] => 9
[12] => 5
[83] => 11
[9] => 3
)
[5] => Array
(
[4] => 32
[8] => 4
[2] => 9
[12] => 5
[83] => 12
[9] => 3
)
)
我怎样才能做到这一点?我知道这是一篇很长的文章,但我需要用图表来表示。我在思考函数递归时遇到了问题,并且我完全迷失在每个循环中。
For two days I've been running crazy trying to accomplish this, maybe you can enlighten me. This is for a horse betting permutation. Every time a user plays, I get a multidimensional array (2 levels). The first level contains the race ID, the the second level contains thee horses selected by the user for that race. It looks like this:
$play = array
(
'4' => array(7, 32),
'8' => array(4),
'2' => array(9),
'12' => array('5'),
'83' => array('10', '11', '12', ''),
'9' => array('3'),
);
I need to know what are all the possible combinations for that play. Which is easily done with this function:
function permutations(array $array)
{
switch (count($array)) {
case 1:
return $array[0];
break;
case 0:
throw new InvalidArgumentException('Requires at least one array');
break;
}
$a = array_shift($array);
$b = permutations($array);
$return = array();
foreach ($a as $key => $v) {
if(is_numeric($v))
{
foreach ($b as $key2 => $v2) {
$return[] = array_merge(array($v), (array) $v2);
}
}
}
return $return;
}
This returns an array with all the possible combinations beautifully. So far so good, and the result looks like this:
Array
(
[0] => Array
(
[0] => 7
[1] => 4
[2] => 9
[3] => 5
[4] => 10
[5] => 3
)
[1] => Array
(
[0] => 7
[1] => 4
[2] => 9
[3] => 5
[4] => 11
[5] => 3
)
[2] => Array
(
[0] => 7
[1] => 4
[2] => 9
[3] => 5
[4] => 12
[5] => 3
)
[3] => Array
(
[0] => 32
[1] => 4
[2] => 9
[3] => 5
[4] => 10
[5] => 3
)
[4] => Array
(
[0] => 32
[1] => 4
[2] => 9
[3] => 5
[4] => 11
[5] => 3
)
[5] => Array
(
[0] => 32
[1] => 4
[2] => 9
[3] => 5
[4] => 12
[5] => 3
)
)
My problem: I need the array "key" for every horse to be the "race ID", not 0,1,2,3. I need the result to be like this:
Array
(
[0] => Array
(
[4] => 7
[8] => 4
[2] => 9
[12] => 5
[83] => 10
[9] => 3
)
[1] => Array
(
[4] => 7
[8] => 4
[2] => 9
[12] => 5
[83] => 11
[9] => 3
)
[2] => Array
(
[4] => 7
[8] => 4
[2] => 9
[12] => 5
[83] => 12
[9] => 3
)
[3] => Array
(
[4] => 32
[8] => 4
[2] => 9
[12] => 5
[83] => 10
[9] => 3
)
[4] => Array
(
[4] => 32
[8] => 4
[2] => 9
[12] => 5
[83] => 11
[9] => 3
)
[5] => Array
(
[4] => 32
[8] => 4
[2] => 9
[12] => 5
[83] => 12
[9] => 3
)
)
How can I accomplish this? I know its a long post but I needed to graph this. I am having problems to wrap my head around the function recursion and I get totally lost in each loop.
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我也遇到了同样的问题,丹尼的解决方案对我来说并不好。
我管理数千个排列并将它们存储在内存中非常昂贵。
这是我的解决方案:
如果有人有更好的主意,我就在这里!
I've got the same problem and Danny's solution wasn't good for me.
I manage thousand of permutation and store them in memory is damn expensive.
Here my solution:
If someone has a better idea i'm here!
这就是您所需要的。我已根据需要发表了评论:
查看实际操作。
顺便说一句,递归计算所有排列很简洁,但您可能不想在生产环境中执行此操作。您绝对应该考虑进行健全性检查,计算有多少排列,并且至少在它们超过某个限制时不允许继续处理。
Here's what you need. I have commented as necessary:
See it in action.
By the way, calculating all the permutations recursively is neat, but you might not want to do it in a production environment. You should definitely consider a sanity check that calculates how many permutations there are and doesn't allow processing to continue if they are over some limit, at the very least.
我通过将乔恩的算法与我最初的算法合并来改进乔恩的函数。我所做的是检查函数是否正在执行递归,如果是,我使用原始的 array_merge() (它正在工作),否则我使用 Jon 的 array_combine() (以保留数组键)。
我将乔恩的答案标记为正确,因为他提出了一个巧妙的解决方案来保持数组键完整。
已成功使用多种阵列组合进行测试。
I improved Jon's function by merging his algorithm with the one I had initially. What I did, was check if the function was doing a recursion, if so, I use the original array_merge() (which was working), else I use Jon's array_combine() (to keep the arrays keys).
I'm marking Jon's answer as correct since he proposed a slick solution to keep the array keys intact.
Tested successfully with several array combinations.