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Count duplicate adjacent (Array)


Go to solution Solved by Barand,

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Hello;

I would like to do the following operation;

Count duplicate (adjacent)

my code

 

        <?php  
      
$Array = array("test", "test", "hello", "test", "world", "world", "world", "hello", "test");


 for( $i= 0 ;  $i <= 6 ;$i++ ) {
$j=1;
 if ($Array[$i] === $Array[$i+1]) {
 
 $j+=$j; 

echo  $j;
}
?>

and showing something like 2,1,1,3,1,1

Thanks

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15 minutes ago, Barand said:

if array == array[i-1]

then you have adjacent duplicates

Thank you Barand;

This my array

$Array = array("test", "test", "hello", "test", "world", "world", "world", "hello", "test");

Well, i would like to show a list with adjacent duplicates for example the word  test 2 not 4 duplicates

and finally the output will be like this 2,1,1,3,1,1 

the problem that array_count_values show the all duplicates, i would like to show the adjacent duplicates

Edited by BeHappy
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11 minutes ago, Barand said:

You've had some clues. What have you tried?

I have created this code but i think there's a problem with it

<?php   
$array = array("test", "test", "hello", "test", "world", "world", "world", "hello", "test");
             
$j=1;
for( $i= 0 ;  $i <= 7 ;$i++ ){ 
                 if ($array[$i] === $array[$i+1])
                 
	{ 
echo $j .= $j+1;
	}
	
	else
	{
      echo "1";
    }
       
}?>

output = 1211112131213121411

not 2,1,1,3,1,1  😅

Edited by BeHappy
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  • Solution

try

$arr = array("test", "test", "hello", "test", "world", "world", "world", "hello", "test");

$prev = '';
$results = [];
$j = 0;
foreach ($arr as $i => $v) {
    if ($v != $prev) {
        $j = $i;
        $results[$j] = 1;
        $prev = $v;
    }
    else {
        $results[$j]++;
    }
}

foreach ($results as $k => $v) {
    echo "$arr[$k] : $v <br>";
}

Result:

test : 2 
hello : 1 
test : 1 
world : 3 
hello : 1 
test : 1 

 

  • Great Answer 1
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