PHP : array_intersect not giving expected result [duplicate]

Question

I am trying to count the matches between expected and actual in a PHP array, I have this...

$array = array(
    "item" => array(
        'expected' => array(
            '1' => 25,
            '2' => 4,
            '3' => 4,
        ),
        'color' => 'red',
        'actual' => array(
            '1' => 25,
            '2' => 4,
            '3' => 3,
        ),
    ),
);

foreach ($array as $key => $arrayItem) {

    $matches = array (
        'matches'  => count ( array_intersect ( $arrayItem['expected'], $arrayItem['actual'] ) ),
    );

}

echo "Matches = " . $matches['matches'];

I am expecting this to return 2 but it is actually returning 3. If I change the values like in the example below then it does work...

$array = array(
    "item" => array(
        'expected' => array(
            '1' => 25,
            '2' => 84,
            '3' => 4,
        ),
        'color' => 'red',
        'actual' => array(
            '1' => 25,
            '2' => 84,
            '3' => 3,
        ),
    ),
);

foreach ($array as $key => $arrayItem) {

    $matches = array (
        'matches'  => count ( array_intersect ( $arrayItem['expected'], $arrayItem['actual'] ) ),
    );

}

echo "Matches = " . $matches['matches'];

Anyone any ideas why the top version is not giving me the expected result?

In first approach, it matches 2nd 4 of expected with 2nd 4 of actual. — mighty_dev
– mighty_dev, Commented Feb 13, 2020 at 10:21
From the manual: array_intersect: "Returns an array containing all of the values in array1 whose values exist in all of the parameters. " — Progrock
– Progrock, Commented Feb 13, 2020 at 10:27
I think array_intersect_assoc might be the answer to this one. Reading up on it now — fightstarr20
– fightstarr20, Commented Feb 13, 2020 at 10:31

Chemaclass · Accepted Answer · 2020-02-13 12:06:29Z

3

Because it returns an array containing all values in array1 whose values exist in all of the parameters.

array_intersect(array $array1, array $array2[, array $... ]): array

https://www.php.net/manual/en/function.array-intersect.php

Maybe you can see it clearly from this perspective:

var_dump(array_intersect([25, 4, 4, 4], [25, 4, 3])); // [25, 4, 4, 4] 
// because the number `4` is in the second array!

var_dump(array_intersect([25, 4, 3], [25, 4, 4, 4])); // [25, 4]

edited Feb 13, 2020 at 12:06

answered Feb 13, 2020 at 10:16

Chemaclass

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Comments

Mark · Accepted Answer · 2020-02-13 10:30:28Z

2

The count is actually correct.

It doesn't happen in your second example because you use the numbers 84 and 4, but essentially here are the matches:

$arrayItem['expected'][1] matches with $arrayItem['actual'][1] (25 and 25)

$arrayItem['expected'][2] matches with $arrayItem['actual'][2] (4 and 4)

$arrayItem['expected'][3] matches with $arrayItem['actual'][2] (4 and 4)

The count of 3 is correct.

You can test this by changing your code to the following:

$matches = array(
    'matches' => array_intersect ($arrayItem['expected'], $arrayItem['actual'])
);

var_dump($matches);

Here you'll see this output:

array(1) {
    ["matches"] => array(3) {
        [1]=> int(25) 
        [2]=> int(4) 
        [3]=> int(4)
    }
}

edited Feb 13, 2020 at 10:30

answered Feb 13, 2020 at 10:16

Mark

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2 Comments

mighty_dev Over a year ago

$arrayItem['expected'][2] and $arrayItem['expected'][3] are counting as a match Not true. See demo

u_mulder Over a year ago

Number 4 is met twice in expected, that's why 4 is checked twice in actual and gets a match of course. Nothing is matched with itself.

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