PHP Array check the count of same values

Question

I'm having troubles comparing two arrays with array_inersect(), tried lot of suggestion from SO but none worked. So here is my problem:

I have two arrays:

$base_array = array(2,2,2,1);

Array
(
    [0] => 2
    [1] => 2
    [2] => 2
    [3] => 1
)

And another one $risks that prints like this:

Array
(
    [0] => 2
    [1] => 3
    [2] => 4
    [3] => 1
    [4] => 2
    [5] => 8
    [6] => 6
)

What I want is to see if the same number and type of values from $base_array fits into $risks array. Right now it should return false cause I only have two occurences of number 2 and base array has 3. But if I set $base_array to:

$base_array = array(2,2,1);

This should return true.

I can't make any code to give me what I need, everything I tried was array_intersect and array_diff but didn't work. Always returning true cause 2 exists in both but no same occurence. — rgomez
– rgomez, Commented Oct 2, 2016 at 15:28
Is it supposed to return true when there are two 2's in $base_array? i.e we are checking if there are two 2's and one 1, right? — Indrasis Datta
– Indrasis Datta, Commented Oct 2, 2016 at 15:28
Add the attempts with array_intersect and array_diff in your question, and in which sense they didn't work. Then we can/will help. — trincot
– trincot, Commented Oct 2, 2016 at 15:29
@ObjectManipulator exatcly. This is just an example, but yes in this case it should return false cause there are three 2's in $base_array but if we set $base_array = array(2,2,1); it should return True. — rgomez
– rgomez, Commented Oct 2, 2016 at 15:30

Risul Islam · Accepted Answer · 2016-10-05 02:46:44Z

1

One approach can be counting occurrence of each number then manually comparing them. For this you need two array to keep the occurrence count.

$base_array_count = [];
$risks_count = [];

In your case $base_array_count will look like

$base_array_count[1] = 1
$base_array_count[2] = 2

And the $risks_array_count will look like

$risks_count[1] = 1
$risks_count[2] = 2
$risks_count[3] = 1
$risks_count[4] = 1
$risks_count[6] = 1
$risks_count[8] = 1

Now loop through $base_array and check if each numbers $base_array_count value is same as in $risks_array_count

Complete code after edit:

$base_array = array(1,2,2,7);
$risks_array = array(2,3,4,5,6,7);

$base_array_count = array();
$risks_array_count = [];

for($i = 0; $i < sizeof($base_array); $i++){
    if( !isset($base_array_count[$base_array[$i]] ))
        $base_array_count[$base_array[$i]] = 0;
    $base_array_count[$base_array[$i]]++;
}

for($i = 0; $i < sizeof($risks_array); $i++){
    if( !isset($risks_array_count[$risks_array[$i]] ))
        $risks_array_count[$risks_array[$i]] = 0;
    $risks_array_count[$risks_array[$i]]++;
}

$same = true;

if (sizeof($base_array) <= sizeof($risks_array)) {
    for ($i = 0; $i < sizeof($base_array); $i++) {
        if(!isset($risks_array_count[$base_array[$i]]) || $base_array_count[$base_array[$i]] != $risks_array_count[$base_array[$i]]){
            $same = false;
            break;
        }
    }
} else {
    $same = false;
}
if($same)
    echo "TRUE";
else
    echo "FALSE";

edited Oct 5, 2016 at 2:46

answered Oct 2, 2016 at 15:48

Risul Islam

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7 Comments

rgomez Over a year ago

thanks for your suggestion, I tried with array_diff but it didn't worked very well. Can you please explain how should I loop for each $base_array_count and compare?

Risul Islam Over a year ago

@rgomez Added the complete code, check if it solves your problem.

rgomez Over a year ago

almost perfect and sorry for being such meticulous but this will be very important on my code. It works pefectly even tried adding other numbers into base_array like 8 or 4 ($base_array = array(1,2,2,4,8);) and of course If I set 2,2,2 it returns false. But the problem comes if I set for instance $base_array = array(1,2,2,7); I get Undefined offset: 7

rgomez Over a year ago

@ Risul Islam, after last change according to your edit the Undefined offset Notice goes away but setting $base_array = array(1,2,2,7); should return False and returns True.

Risul Islam Over a year ago

@rgomez It shows me False. I don't know why it shows you True. Did you update your code properly? I am changing complete code according to edit try this.

|

Imran Hossain · Accepted Answer · 2016-10-08 07:03:25Z

1

It can be done by array_diff with some extra check. See here

$base_array = array(1,2,2,7);
$risks_array = array(1,2,2,3,4,5,6);

if(array_diff($base_array, $risks_array)) // false if $base_array has type of elemenes which $risk_array does not contain 
 echo "false"; //
else{
 // $risk_array has all type of elements of $base_array. now check for similar frequency
 $freq_cnt_base = array_count_values($base_array);
 $freq_cnt_risks = array_count_values($risks_array);
 $is_similar = "true";

 foreach($freq_cnt_base as $key => $val){
    if($freq_cnt_risks[$key] !== $freq_cnt_base[$key]){
        $is_similar = "false";
        break;
    }
 }
 echo $is_similar;
}

edited Oct 8, 2016 at 7:03

answered Oct 2, 2016 at 19:02

Imran Hossain

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