我试图计算高尔夫球手在比赛中获胜的胜利顺序。
这些高尔夫比赛正在使用＆＃34; stableford＆＃34;积分评分系统，你获得最高分的每洞得分。与普通高尔夫球杆和中风比赛相比，＃34;最低分数获胜的地方（虽然这也有倒计时系统，只计算平局时的最低分数...）
规则是使用＆＃34;倒计时＆＃34;。也就是说，如果在9洞之后得分，则最佳位置是最后8洞的最佳得分。然后是7洞等。
我能想到的最好的是2个阵列。
我遍历数组1，从数组2中提取数据并使用以下公式创建一个得分最高的临时数组：
$max = array_keys($array,max($array));
如果$ max只有1个项目，则此玩家是最高得分者。通过第一个数组的循环是＆＃34;通过引用＆＃34;，所以在循环的下一次迭代中，他的playerid现在在数组中更长，因此被忽略。这种情况一直持续到第一个阵列中只剩下1个玩家。
但是，只有在每次迭代中单个玩家获胜时才有效。不起作用的场景是，如果一组玩家与任何迭代/倒计数相关联。
我认为我的问题是当前的结构我需要将原来的$ ties数组拆分，然后继续以相同的方式遍历拆分数组......
作为一个例子......
$ ties数组如下：
Array
(
[18] => Array
(
[0] => 77
[1] => 79
[2] => 76
[3] => 78
)
)
$ tie_perhole（得分数据）数组如下：
Array
(
[18] => Array
(
[77] => Array
(
[9] => 18
[8] => 16
[7] => 14
[6] => 12
[5] => 10
[4] => 8
[3] => 6
[2] => 4
[1] => 2
)
[79] => Array
(
[9] => 18
[8] => 17
[7] => 15
[6] => 14
[5] => 11
[4] => 9
[3] => 7
[2] => 5
[1] => 3
)
[76] => Array
(
[9] => 18
[8] => 16
[7] => 14
[6] => 12
[5] => 10
[4] => 8
[3] => 6
[2] => 4
[1] => 2
)
[78] => Array
(
[9] => 18
[8] => 17
[7] => 15
[6] => 13
[5] => 11
[4] => 9
[3] => 7
[2] => 5
[1] => 3
)
)
)
因此在本次比赛中，球员78和79在第8洞倒计时（17分）得分最高，因此第1和第2应该在他们之间。玩家79应该是第6洞倒数第1名（14分，相比之下13分）。对于剩下的2名其他球员，第3名和第4名也应如此。
此处可能会出现其他情况，因为在比赛中，可能会有很多不同数量的玩家（不同数量）通过排行榜处于不同的关联点。
另请注意，排行榜上会有一些玩家未被束缚并保持现状。
我的工作代码的基础是：
foreach ($ties as $comparekey => &$compareval) {
$tie_loop = 0;
for ($m = 9; $m >= 1; $m--) {
$compare = array();
foreach ($compareval as $tie) {
$compare[$tie] = $tie_perhole[$comparekey][$tie][$m];
}
$row = array_keys($compare,max($compare));
if (count($row) == 1) {
$indexties = array_search($row[0], $ties[$comparekey]);
unset($ties[$comparekey][$indexties]);
// Now update this "winners" finishing position in a sorted array
// This is a multidimensional array too, with custom function...
$indexresults = searchForId($row[0], $comp_results_arr);
$comp_results_arr[$indexresults][position] = $tie_loop;
$tie_loop++;
}
// I think I need conditions here to filter if a subset of players tie
// Other than count($row) == 1
// And possibly splitting out into multiple $ties arrays for each thread...
if (empty($ties[$comparekey])) {
break;
}
}
}
usort($comp_results_arr, 'compare_posn_asc');
foreach($comp_results_arr as $row) {
//echo an HTML table...
}
提前感谢任何有用的见解，提示，想法等......
Robert Cathay要求提供更多方案。所以这是另一个......
排行榜实际上有更多的参赛者（玩家26有一个糟糕的回合...），但我需要帮助的代码只是对排行榜内的关系感到困扰。
摘要排行榜：
Points Player
21 48
21 75
20 73
20 1
13 26
此示例生成$ tie_perhole数组：
Array
(
[21] => Array
(
[75] => Array
(
[9] => 21
[8] => 19
[7] => 16
[6] => 14
[5] => 12
[4] => 9
[3] => 7
[2] => 5
[1] => 3
)
[48] => Array
(
[9] => 21
[8] => 19
[7] => 16
[6] => 13
[5] => 11
[4] => 9
[3] => 8
[2] => 5
[1] => 3
)
)
[20] => Array
(
[73] => Array
(
[9] => 20
[8] => 18
[7] => 16
[6] => 13
[5] => 11
[4] => 8
[3] => 6
[2] => 5
[1] => 3
)
[1] => Array
(
[9] => 20
[8] => 17
[7] => 16
[6] => 14
[5] => 12
[4] => 9
[3] => 7
[2] => 4
[1] => 2
)
)
)
在这个例子中，阵列显示玩家75和48得分为21分，玩家75最终将在第6洞倒计时获胜（14分与13分相比）而玩家48则获得第2分。在下一个并列组中，玩家73和1得到20分，玩家73将在第8洞倒计时赢得这一组并获得第3名（18分与17分相比），玩家1分在第4名。然后玩家26就是第5名。
注意，$ tie_loop被添加到另一个数组以计算第1到第5位的结束位置，这样就可以了。
希望这足以帮助。
答案 0 :(得分：1)
好的，所以我根本不懂高尔夫...哈哈哈，但是！我想我得到了这个问题的要点，所以我的解决方案就是这样。
<?php
/**
* Author : Carlos Alaniz
* Email : Carlos.glvn1993@gmail.com
* Porpuse : Stackoverflow example
* Date : Aug/04/2015
**/
$golfers = [
"A" => [1,5,9,1,1,2,3,4,9],
"B" => [2,6,4,2,4,4,1,9,3],
"C" => [3,4,9,8,1,1,5,1,3],
"D" => [1,5,1,1,1,5,4,5,8]
];
//Iterate over scores.
function get_winners(&$golfers, $hole = 9){
$positions = array(); // The score numer is the key!
foreach ($golfers as $golfer=>$score ) { // Get key and value
$score_sub = array_slice($score,0,$hole); // Get the scores subset, first iteration is always all holes
$total_score = (string)array_sum($score_sub); // Get the key
if(!isset($positions[$total_score])){
$positions[$total_score] = array(); // Make array
}
$positions[$total_score][] = $golfer; // Add Golpher to score.
}
ksort($positions, SORT_NUMERIC); // Sort based on key, low -> high
return array(end($positions), key($positions)); // The last shall be first
}
//Recursion is Awsome
function getWinner(&$golfers, $hole = 9){
if ($hole == 0) return;
$winner = get_winners($golfers,$hole); // Get all ties, if any.
if(count($winner[0]) > 1){ // If theirs ties, filter again!
$sub_golfers =
array_intersect_key($golfers,
array_flip($winner[0])); // Only the Worthy Shall Pass.
$winner = getWinner($sub_golfers,$hole - 1); // And again...
}
return $winner; // We got a winner, unless they really tie...
}
echo "<pre>";
print_R(getWinner($golfers));
echo "</pre>";
好的......现在生病解释我的方法...... 既然我们需要知道得分最高并且它可能是关系，那么在单独的数组中维护所有内容是没有意义的，而我只是颠倒了
golfer =＆gt;分数 至 Tota_score =＆gt;高尔夫球手
这样我们可以按键对数组进行排序，并获得所有得分最高的高尔夫球手。
现在total_score是孔得分数组子集的总和。所以...这个功能第一次运行时，它会添加所有9个洞，在这种情况下，3个高尔夫球手最终得到相同的分数。
Array
(
[0] => Array
(
[0] => A
[1] => B
[2] => C
)
[1] => 35
)
由于高尔夫球手的总数不是1而且我们仍然在第9洞，我们再次运行，但这次只针对那3个高尔夫球手和当前的洞 - 1，所以我们只加到第8洞这次打洞。
Array
(
[0] => Array
(
[0] => B
[1] => C
)
[1] => 32
)
我们还有另一个平局....这个过程将一直持续到我们到达最后一洞或者胜利者为止。
Array
(
[0] => Array
(
[0] => C
)
[1] => 31
)
修改
<?php
/**
* Author : Carlos Alaniz
* Email : Carlos.glvn1993@gmail.com
* Porpuse : Stackoverflow example
**/
$golfers = [
"77" => [2,4,6,8,10,12,14,16,18],
"79" => [3,5,7,9,11,14,15,17,18],
"76" => [2,4,6,8,10,12,14,16,18],
"78" => [3,5,7,9,11,13,15,17,18]
];
//Iterate over scores.
function get_winners(&$golfers, $hole = 9){
$positions = array(); // The score numer is the key!
foreach ($golfers as $golfer => $score) { // Get key and value
//$score_sub = array_slice($score,0,$hole); // Get the scores subset, first iteration is always all holes
$total_score = (string)$score[$hole-1]; // Get the key
if(!isset($positions[$total_score])){
$positions[$total_score] = array(); // Make array
}
$positions[$total_score][] = $golfer; // Add Golpher to score.
}
ksort($positions, SORT_NUMERIC); // Sort based on key, low -> high
return [
"winner"=> end($positions),
"score" => key($positions),
"tiebreaker_hole" => [
"hole"=>$hole,
"score"=> key($positions)],
]; // The last shall be first
}
//Recursion is Awsome
function getWinner(&$golfers, $hole = 9){
if ($hole == 0) return;
$highest = get_winners($golfers,$hole); // Get all ties, if any.
$winner = $highest;
if(count($winner["winner"]) > 1){ // If theirs ties, filter again!
$sub_golfers =
array_intersect_key($golfers,
array_flip($winner["winner"])); // Only the Worthy Shall Pass.
$winner = getWinner($sub_golfers,$hole - 1); // And again...
}
$winner["score"] = $highest["score"];
return $winner; // We got a winner, unless they really tie...
}
echo "<pre>";
print_R(getWinner($golfers));
echo "</pre>";
结果：
Array
(
[winner] => Array
(
[0] => 79
)
[score] => 18
[tiebreaker_hole] => Array
(
[hole] => 6
[score] => 14
)
)