Php 将重复数组与计数合并
我正在处理大量的数据,当数组重复时,我需要合并它们。如果它们被合并,我需要在数组中添加一个计数Php 将重复数组与计数合并,php,arrays,laravel-5.1,array-merge,Php,Arrays,Laravel 5.1,Array Merge,我正在处理大量的数据,当数组重复时,我需要合并它们。如果它们被合并,我需要在数组中添加一个计数 array:3721 [▼ 0 => array:3 [▼ "subscriber" => "gmail.com." "code" => 554 "status" => 50 ] 1 => array:3 [▼ "subscriber" => "apied.be" "code" => 550 "sta
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
array_unique($hardbounces, SORT_REGULAR);
我只剩下534个结果,而不是3721个。这太棒了,只是我还需要知道数量,而且它必须有一定的效率,因为结果集可能非常大(更大)
之后,它需要被排序以及域和金额
我正在使用laravel 5.1,如果有必要,我可以将数组转换为集合,这样就可以使用辅助函数了我自己设法用foreach循环修复了它,它的运行速度比我预期的快(0.3179秒) 我在反弹中循环,如果合并为空(第一次迭代),它只添加硬反弹。 从那时起,我将循环新数组并检查是否存在重复。当这种情况发生时,我们只需在数量上增加一个,然后从foreach中分离出来。否则,它最终仍将添加重复项。 在我检查它是否进入最后一次迭代之后,这意味着如果它仍然没有中断,我们应该添加硬反弹,因为它还不存在
为了说明这一点,我确保在循环运行之前,数量1已经存在,而不是在循环过程中添加它。我自己设法用foreach循环修复它,它的运行速度比我预期的快(0.3179秒) 我在反弹中循环,如果合并为空(第一次迭代),它只添加硬反弹。 从那时起,我将循环新数组并检查是否存在重复。当这种情况发生时,我们只需在数量上增加一个,然后从foreach中分离出来。否则,它最终仍将添加重复项。 在我检查它是否进入最后一次迭代之后,这意味着如果它仍然没有中断,我们应该添加硬反弹,因为它还不存在
为了明确起见,我确保在循环运行之前数量1已经存在,而不是在循环期间添加它。我不确定这段代码对很多项的效率有多高(顺便说一句,您没有提到数字的数量级),但我认为它比将数组转换为Laravel集合更有效
$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);
我不确定这段代码在处理很多项时有多高效(顺便说一句,您没有提到数字的数量级),但我认为它比将数组转换为Laravel集合更高效
$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);
试一试
试试看
<?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";
?>
有趣的问题。您是否尝试过array\u count\u values()
?您当前是否正在从MySql之类的东西收集这些数据集?是的,请使用雄辩的方式获取这些值。为什么不使用然后聚合SQL函数进行计数?我不太擅长子查询和所有有趣的问题。您是否尝试过array\u count\u values()
?您当前是否正在从MySql之类的工具收集这些数据集?是的,请使用雄辩的方法获取这些值。为什么不使用然后聚合SQL函数进行计数?我不太擅长子查询和所有这些东西这仍然会给我3721个结果。除非我以某种方式拥有完整的数据集,否则很难进行调试。你看到你提供的样本是如何工作的了吗?是的,我看到了,我只是用8个结果运行了它。这仍然不是它应该看起来的样子。它将域名设置为一个键,我认为这是不必要的,在该数组中,它为计数添加了另一个数组,该数组应该是一个带值的键。我自己解决了这个问题,看看我给出的答案,看看你在做什么,为什么要将域名添加为键,看看它是否已经存在,这并不意味着它是独一无二的。还应考虑状态和代码。我应该把我的案子做得更好。SorryThis仍然给我3721个结果。除非我有完整的数据集,否则很难调试。你看到你提供的样本是如何工作的了吗?是的,我看到了,我只是用8个结果运行了它。这仍然不是它应该看起来的样子。它将域名设置为一个键,我认为这是不必要的,在该数组中,它为计数添加了另一个数组,该数组应该是一个带值的键。我自己解决了这个问题,看看我给出的答案,看看你在做什么,为什么要将域名添加为键,看看它是否已经存在,这并不意味着它是独一无二的。还应考虑状态和代码。我应该把我的案子做得更好。抱歉,这仍然输出完整的结果集吗?仍然得到3721个结果,看看我自己提供的答案。这仍然输出完整的结果集?仍然得到3721个结果,看看我自己提供的答案。