计算多维数组中特定值的出现次数

Question

假设我有一个多维数组:

[
    ["Thing1", "OtherThing1"],
    ["Thing1", "OtherThing2"],
    ["Thing2", "OtherThing3"]
]

我怎样才能计算出"Thing1"发生在多维数组中?

Answer 1

您可以使用array_search获取更多信息，请参阅此http://www.php.net/manual/en/function.array-search.php

这段代码是PHP文档sample

的样本

<?php 
function recursiveArraySearchAll($haystack, $needle, $index = null) 
{ 
 $aIt     = new RecursiveArrayIterator($haystack); 
 $it    = new RecursiveIteratorIterator($aIt); 
 $resultkeys; 
 while($it->valid()) {        
 if (((isset($index) AND ($it->key() == $index)) OR (!isset($index))) AND (strpos($it->current(), $needle)!==false)) { //$it->current() == $needle 
 $resultkeys[]=$aIt->key(); //return $aIt->key(); 
 } 
 $it->next(); 
 } 
 return $resultkeys;  // return all finding in an array 
} ; 
?> 

如果在干草堆中不止一次找到针，则返回第一个匹配的键。若要返回所有匹配值的键，请使用带可选search_value参数的array_keys()。

http://www.php.net/manual/en/function.array-keys.php

Answer 2

试试这个:

$arr =array(
array("Thing1","OtherThing1"),
array("Thing1","OtherThing2"),
array("Thing2","OtherThing3")
);
echo "<pre>";
$res  = array_count_values(call_user_func_array('array_merge', $arr));
echo $res['Thing1'];

输出:

Array
(
    [Thing1] => 2
    [OtherThing1] => 1
    [OtherThing2] => 1
    [Thing2] => 1
    [OtherThing3] => 1
)

给出每个值的出现次数。即:Thing1出现2次。

EDIT:根据OP的评论:"您指的是哪个数组?"—输入阵列。例如，这将是输入数组:array(array(1,1)，array(2,1)，array(3,2))，我只希望它计算第一个值(1,2,3)而不是第二个值(1,1,2)- gdscei 7 mins ago

$arr =array(
array("Thing1","OtherThing1"),
array("Thing1","OtherThing2"),
array("Thing2","OtherThing3")
);
$res  = array_count_values(array_map(function($a){return $a[0];}, $arr));
echo $res['Thing1'];

Answer 3

function showCount($arr, $needle, $count=0)
{
    // Check if $arr is array. Thx to Waygood
    if(!is_array($arr)) return false;
    foreach($arr as $k=>$v)
    {
        // if item is array do recursion
        if(is_array($v))
        {
            $count = showCount($v, $needle, $count);
        }
        elseif($v == $needle){
            $count++;
        }
    }
    return $count;  
}

Answer 4

使用in_array可以帮助:

$cont = 0;
//for each array inside the multidimensional one
foreach($multidimensional as $m){
    if(in_array('Thing1', $m)){
        $cont++;
    }
}
echo $cont;

更多信息:http://php.net/manual/en/function.in-array.php

Answer 5

try this

$arr =array(
array("Thing1","OtherThing1"),
 array("Thing1","OtherThing2"),
 array("Thing2","OtherThing3")
 );
   $abc=array_count_values(call_user_func_array('array_merge', $arr));
  echo $abc[Thing1];

Answer 6

$count = 0;
foreach($array as $key => $value)
{
if(in_array("Thing1", $value)) $count++;
}

Answer 7

如果您希望代码简洁，零全局作用域污染，您可以计算每个值并访问您想要的一个计数:

echo array_count_values(array_merge(...$array))['Thing1'] ?? 0;

---

如果您不想在永远不需要计数的地方计数值，那么您可以在每次遇到目制值时使用array_walk_recursive()和+1访问叶节点。

$thing1Count = 0;
array_walk_recursive($array, function($v) use(&$thing1Count) { $thing1Count += ($v === 'Thing1'); });
echo $thing1Count;

两个代码段都返回2。这是一个Demo