我正在处理大量数据，我需要在重复时合并数组。如果它们合并，我需要在阵列中添加一个计数。
array:3721 [▼
0 => array:3 [▼
"subscriber" => "gmail.com."
"code" => 554
"status" => 50
]
1 => array:3 [▼
"subscriber" => "apied.be"
"code" => 550
"status" => 51
]
2 => array:3 [▼
"subscriber" => "beton-dobbelaere.be"
"code" => 550
"status" => 50
]
3 => array:3 [▼
"subscriber" => "live.be"
"code" => 550
"status" => 51
]
4 => array:3 [▼
"subscriber" => "hotmail.be"
"code" => 550
"status" => 51
]
5 => array:3 [▼
"subscriber" => "telenet.be"
"code" => 550
"status" => 50
]
6 => array:3 [▼
"subscriber" => "telenet.be"
"code" => 550
"status" => 55
]
7 => array:3 [▼
"subscriber" => "telenet.be"
"code" => 550
"status" => 51
]
8 => array:3 [▼
"subscriber" => "telenet.be"
"code" => 550
"status" => 51
]
这应该类似于：
array:3721 [▼
0 => array:3 [▼
"subscriber" => "gmail.com."
"code" => 554
"status" => 50
"amount" => 1
]
1 => array:3 [▼
"subscriber" => "apied.be"
"code" => 550
"status" => 51
"amount" => 1
]
2 => array:3 [▼
"subscriber" => "beton-dobbelaere.be"
"code" => 550
"status" => 50
"amount" => 1
]
3 => array:3 [▼
"subscriber" => "live.be"
"code" => 550
"status" => 51
"amount" => 1
]
4 => array:3 [▼
"subscriber" => "hotmail.be"
"code" => 550
"status" => 51
"amount" => 1
]
5 => array:3 [▼
"subscriber" => "telenet.be"
"code" => 550
"status" => 50
"amount" => 1
]
6 => array:3 [▼
"subscriber" => "telenet.be"
"code" => 550
"status" => 55
"amount" => 1
]
7 => array:3 [▼
"subscriber" => "telenet.be"
"code" => 550
"status" => 51
"amount" => 2
]
当我使用
合并此示例时array_unique($hardbounces, SORT_REGULAR);
我留下了大约534个结果，而不是3721，这很好，只是我需要知道数量，它必须有点效率，因为结果集可能非常大（更大）。
在需要对域和金额进行排序之后。
我正在使用laravel 5.1，如果有必要，我可以将数组转换为集合，因此可以使用辅助函数
答案 0 :(得分：0)
我设法使用foreach循环自行修复它比我预期的更快（0.3179秒）
$merged = [];
foreach($hardbounces as &$hardbounce){
if(empty($merged)){
$merged[] = $hardbounce;
}else{
$i = 0;
foreach($merged as $key => $merge){
$i++;
if($hardbounce['subscriber'] == $merge['subscriber'] && $hardbounce['code'] == $merge['code'] && $hardbounce['status'] == $merge['status']){
$merged[$key]['amount']++;
break;
}
if(count($merged) == $i){
$merged[] = $hardbounce;
}
}
}
}
我循环跳过，如果合并为空（第一次迭代），它只是添加硬弹跳。 从那时起，我将遍历新数组并检查是否存在重复。当发生这种情况时，我们只需添加一个金额并打破foreach。否则它最终仍会添加重复项。 在我检查我们是否进入最后一次迭代后，这意味着如果它仍然没有破坏，我们应该添加硬弹跳，因为它还不存在。
要清楚，我确保在此循环运行之前已存在金额1，而不是在此循环期间添加它。
答案 1 :(得分：0)
我不确定这段代码对于很多项目有多高效（BTW，你没有提到数字的数量级），但我认为它比将数组转换成Laravel集合更有效率< / p>
$arr = [
0 => [
"subscriber" => "gmail.com.",
"code" => 554,
"status" => 50,
],
1 => [
"subscriber" => "apied.be",
"code" => 550,
"status" => 51,
],
2 => [
"subscriber" => "beton-dobbelaere.be",
"code" => 550,
"status" => 50,
],
3 => [
"subscriber" => "live.be",
"code" => 550,
"status" => 51,
],
4 => [
"subscriber" => "hotmail.be",
"code" => 550,
"status" => 51,
],
5 => [
"subscriber" => "telenet.be",
"code" => 550,
"status" => 50,
],
6 => [
"subscriber" => "telenet.be",
"code" => 550,
"status" => 55,
],
7 => [
"subscriber" => "telenet.be",
"code" => 550,
"status" => 51,
],
8 => [
"subscriber" => "telenet.be",
"code" => 550,
"status" => 51,
],
];
$res = [];
foreach($arr as $element) {
if(empty($res[$element['subscriber']])) {
$res[$element['subscriber']] = [$element, 'count' => 1];
} else {
$res[$element['subscriber']]['count']++;
}
}
var_dump($res);
答案 2 :(得分：0)
尝试
<?php
$input = array(
0 => array(
"subscriber" => "gmail.com.",
"code" => 554,
"status" => 50),
1 => array(
"subscriber" => "apied.be",
"code" => 550,
"status" => 51),
2 => array(
"subscriber" => "beton-dobbelaere.be",
"code" => 550,
"status" => 50),
3 => array(
"subscriber" => "live.be",
"code" => 550,
"status" => 51),
4 => array(
"subscriber" => "hotmail.be",
"code" => 550,
"status" => 51),
5 => array(
"subscriber" => "telenet.be",
"code" => 550,
"status" => 50),
6 => array(
"subscriber" => "telenet.be",
"code" => 550,
"status" => 55),
7 => array(
"subscriber" => "telenet.be",
"code" => 550,
"status" => 51),
8 => array(
"subscriber" => "telenet.be",
"code" => 550,
"status" => 51)
);
/**
*@param array $counted The array already counted or NULL
*@param array $new The array to count or to merge with the counted $counted
*/
function merge_xor_count(array $counted = NULL, array $new){
if($counted === NULL){
$counted = array();
}
foreach($new as $keyNew => $valueNew){
$matches = false;
foreach($counted as $keyOut => $valueOut){
if ($valueOut['subscriber'] == $valueNew['subscriber'] && $valueOut['code'] == $valueNew['code'] &&
$valueOut['status'] == $valueNew['status']){
$matches = $keyOut;
}
}
if($matches !== false){
$counted[$matches]['amount']++;
}
else{
if(!isset($valueNew['amount'])) $valueNew['amount'] = 1;
$counted[] = $valueNew;
}
}
return $counted;
}
$output = merge_xor_count(NULL, $input);
print_r ($output)."\n";
$output = merge_xor_count($output, $input);
print_r ($output)."\n";
?>