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up vote 2 down vote favorite
1

Suppose I have the following array $diff.

a, a, a, a, b, b, b, a, a, b, b, b, a, a, a, b

A represents a value inside $diff.
B represents an Array inside $diff.

Now I have to count A if it occurred more than two times in its sequence and is not an Array (instead a value). Otherwise, disregard it.

For the above input, the code should function as follows

[a] = not an array; 0
[a,a] = not an array; 0
[a,a,a] = not an array; 3
[a,a,a,a] = not an array; 4
[b] = array;
[b,b] = array;
[b,b,b] = array;
[a] = not an array; 0
[a,a] = not an array; 0
[b] = array;
[b,b] = array;
[b,b,b] = array;
[a] = not an array; 0
[a,a] = not an array; 0
[a,a,a] = not an array; 3
[b] = array;

Here is my attempt, but it doesn't work!, the value gets changed because the values get replaced. 

<?php

foreach($diff as $key => $val)  {

    if (!is_array($diff[$key])) { // THIS MEANS THAT THE CURRENT ELEMENT IS NOT AN ARRAY. 
       if(is_array($diff[$key-1]) ) {   //START OF SEQ. IF THE PREVIOUS ELEMENT IS AN ARRAY AND CURRENT ELEMENT IS NOT AN ARRAY.

        $SEQ_START=$key;
        $n=1;

            for($i=0; $i<=count($diff); $i+=1) { // I AM CHECKING HERE IF THE NEXT 3 ELEMENTS are NOT ARRAY, HENCE I CAN INCREMENT IT

            if(!is_array($diff[$SEQ_START+$i])) $n+=1;
            else $n=0;
            }
        }
    }
}

?>
share|improve this question
    
I understand the logic of the count, but what is the end result you are looking for? The number of contiguous segments (> 2), or the number of elements in the contiguous segments (> 2)? (e.x. if the former, the answer would be 2, if the latter, the answer would be 7) –  Grexis Dec 22 '11 at 10:06
    
I need the answer to be 7 in the above example. –  SupaOden Dec 22 '11 at 10:12

3 Answers 3

up vote 3 down vote accepted

UPDATED as per @Grexis' comment below

$diff = array(array(),'a', 'a', 'a', 'a', array(), array(), array(), 'a', 'a', array(), array(), array(), 'a', 'a', 'a', array());

// Counter to hold current sequence total
$count = 0;
// Array to hold results
$counted = array();

// Loop array
foreach ($diff as $key => $val) {
  if (is_array($val)) { // If it is an array
    if ($count > 2) { // If the counter is more than 2
      $counted[(isset($seq)) ? $seq + 1 : 0] = $count; // add it to the array
    }
    // Reset the counter
    $count = 0;
    $seq = $key;
  } else {
    // Increment the counter
    $count++;
  }
} 
// If there is a >2 counter at the end, add one more result
if ($count > 2) {
  $counted[(isset($seq)) ? $seq + 1 : 0] = $count;
}

print_r($counted);
// Outputs:
// Array
// (
//     [0] => 4
//     [12] => 3
// )

// or if you want the total count
$total = array_sum($counted);
echo $total; // 7

See it working

share|improve this answer
    
It's unnecessary, but if you include the index in the foreach, you could store the index of the start of the sequence as the index in $counted. Something like $counted[$index-$count] = $count; on line 10 and $counted[count($diff)-$count] = $count; on line 21 –  Grexis Dec 22 '11 at 10:20
    
@Grexis I updated the code with how I would do that :-D –  DaveRandom Dec 22 '11 at 10:26
    
Not gonna lie, your implementation is definitely neater... showoff :p –  Grexis Dec 22 '11 at 10:35
up vote 1 down vote

All you need is a counter that counts the consecutive non-array values. Increase it with every array value and reset it with every non-array value:

$seqLength = 0;
foreach ($arr as $index => $value) {
    if (is_array($value)) {
        $seqLength++;
        echo 'array';
    } else {
        $seqLength = 0;
        echo 'not an array';
    }
    if ($seqLength > 2) {
        echo '; '.$seqLength;
    } else {
        echo '; 0';
    }
}
share|improve this answer
up vote 0 down vote

maybe:

$a = array(1,1,1,1,1,array(1,1),1,1,1,1,1,array(1,1));
$i = 0;
$j = 0;
foreach ($a as $item){
    if ($i>=2){
        $j=$i + 1;
    }
    if (!is_array($item)){
        $i++;
        echo $j.'<br>';
    }else{
        $i=0;
        $j=0;
        echo '<br>';
    }   
}

output:

0
0
3
4
5

0
0
3
4
5
share|improve this answer

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