Count repeated null values in array

Question

I have an array like this:

$items = ['a', 'b', null, null]; 

I want count [nulls] in this array which is 2 .

But: Without replacing them before working with. Like this:

$items = array_replace($items,array_fill_keys(array_keys($items, null),''));

array_count_values($items);

And: Without removing them from the array and put them out of the total. Like this:

count($items) - count(array_filter($items));

I'm trying to know more better solutions.

Answer 1

You could array_filter on the null values and count once on the result:

count(array_filter($items, function ($item) { return $item === null; }))

or in PHP 7.4:

count(array_filter($items, fn($item) => $item === null));

Demo: https://3v4l.org/TOHdP

Answer 2

You can use array_count_values method. you can use it like this:

- Solution 1

$items = ['a', 'b', null, null]; 

$counts = 0;
foreach(array_count_values($items) as $item => $counter){
    $counts += $counter;
}
echo count($items) - $counts;

PS: array_count_values() just count string and integers.

- Solution 2

Or you can use array_filter to create null values array and count it.

echo count(array_filter($items, function ($item) { 
    return $item === null; 
}))

Answer 3

Usually the best is to be explicit, so simply iterate over the input array and count:

<?php
$items = ['a', 'b', null, null]; 
array_walk($items, function($value) use (&$count) {
    if ($value === null)
      $count++;
});
var_dump($count);

The output obviously is:

int(2)

Answer 4

No need for array_filter

$items = ['a', 'b', null, null]; 

echo(count(array_keys($items, null, true))); // Output: 2

Note the third argument must be true if you only want null to be counted, as true means do strict comparison (===)