查找数组中频繁的数字序列

Question

排列 (3, 5, 1, 3, 5, 48, 4, 7, 13, 55, 65, 4, 7, 13, 32)

频繁的数字序列将是 (3, 5) f=2 + (4, 7, 13) f=2

任何算法或伪代码来找到？

更新(1):

如果(7, 13)也发生它将通过更新其频率而被包括在最长的那个中

(4, 7, 13) f=3 等等...

更新(2):

如果(1,2,3,4,1,2,3,4,1,2,7,8,7,8,3,4,3,4,1,2)输出应该是(1,2,3,4)&(3,4,1,2)

&(7,8),明确地将每个数字视为一个单词,并且您希望找到最常用的短语

所以在很多短语中看到相同的单词是很常见的,但是如果任何短语是任何其他短语的子串

短语不应被视为短语,而是将更新每个短语的频率包括它

Answer 1

**编辑**:稍微好一点的实现,现在也返回频率,并有一个更好的序列过滤器.

function getFrequences($input, $minimalSequenceSize = 2) {
   $sequences = array();
   $frequences = array();

   $len = count($input);
   for ($i=0; $i<$len; $i++) {
      $offset = $i;

      for ($j=$i+$minimalSequenceSize; $j<$len; $j++) {
         if ($input[$offset] == $input[$j]) {
            $sequenceSize = 1;
            $sequence = array($input[$offset]);
            while (($offset + $sequenceSize < $j) 
               && ($input[$offset+$sequenceSize] == $input[$j+$sequenceSize])) {

               if (false !== ($seqIndex = array_search($sequence, $frequences))) {
                  // we already have this sequence, since we found a bigger one, remove the old one
                  array_splice($sequences, $seqIndex, 1);
                  array_splice($frequences, $seqIndex, 1);
               }              

               $sequence[] = $input[$offset+$sequenceSize];
               $sequenceSize++;
            }

            if ($sequenceSize >= $minimalSequenceSize) {
               if (false !== ($seqIndex = array_search($sequence, $sequences))) {
                  $frequences[$seqIndex]++;
               } else {
                  $sequences[] = $sequence;
                  $frequences[] = 2;  // we have two occurances already
               }
               // $i += $sequenceSize;  // move $i so we don't reuse the same sub-sequence
               break;
            }
         }
      }
   }

   // remove sequences that are sub-sequence of another frequence
   // ** comment this to keep all sequences regardless ** 
   $len = count($sequences);
   for ($i=0; $i<$len; $i++) {
      $freq_i = $sequences[$i];
      for ($j=$i+1; $j<$len; $j++) {
         $freq_j = $sequences[$j];
         $freq_inter = array_intersect($freq_i, $freq_j);
         if (count($freq_inter) != 0) {
            $len--;
            if (count($freq_i) > count($freq_j)) {
               array_splice($sequences, $j, 1);
               array_splice($frequences, $j, 1);
               $j--;
            } else {
               array_splice($sequences, $i, 1);
               array_splice($frequences, $i, 1); 
               $i--;
               break;
            }
         }
      }
   }

   return array($sequences, $frequences);
};

测试用例

header('Content-type: text/plain');

$input = array(3, 5, 1, 3, 5, 48, 4, 7, 13, 55, 3, 5, 65, 4, 7, 13, 32, 5, 48, 4, 7, 13);

list($sequences, $frequences) = getFrequences($input);
foreach ($sequences as $i => $s) {
   echo "(" . implode(',', $s) . ') f=' . $frequences[$i] . "\n";
}

**编辑**:这是对该功能的更新.它几乎完全重写了......告诉我这是不是你要找的东西.我还添加了冗余检查,以防止计算相同的序列或子序列两次.

function getFrequences2($input, $minSequenceSize = 2) {
  $sequences = array();

  $last_offset = 0;
  $last_offset_len = 0;

  $len = count($input);
  for ($i=0; $i<$len; $i++) {
     for ($j=$i+$minSequenceSize; $j<$len; $j++) {
        if ($input[$i] == $input[$j]) {
           $offset = 1;
           $sub = array($input[$i]);
           while ($i + $offset < $j && $j + $offset < $len) {
              if ($input[$i + $offset] == $input[$j + $offset]) {
                 array_push($sub, $input[$i + $offset]);
              } else {
                 break;
              }
              $offset++;
           }

           $sub_len = count($sub);
           if ($sub_len >= $minSequenceSize) {
              // $sub must contain more elements than the last sequence found
              // otherwise we will count the same sequence twice
              if ($last_offset + $last_offset_len >= $i + $sub_len) {
                 // we already saw this sequence... ignore
                 continue;
              } else {
                 // save offset and sub_len for future check
                 $last_offset = $i;
                 $last_offset_len = $sub_len;
              }

              foreach ($sequences as & $sequence) {
                 $sequence_len = count($sequence['values']);
                 if ($sequence_len == $sub_len && $sequence['values'] == $sub) {
                    //echo "Found add-full ".var_export($sub, true)." at $i and $j...\n";
                    $sequence['frequence']++;
                    break 2;
                 } else {
                    if ($sequence_len > $sub_len) {
                       $end = $sequence_len - $sub_len;
                       $values = $sequence['values'];
                       $slice_len = $sub_len;
                       $test = $sub;
                    } else {
                       $end = $sub_len - $sequence_len;
                       $values = $sub;
                       $slice_len = $sequence_len;
                       $test = $sequence['values'];
                    }
                    for ($k=0; $k<=$end; $k++) {
                       if (array_slice($values, $k, $slice_len) == $test) {
                          //echo "Found add-part ".implode(',',$sub)." which is part of ".implode(',',$values)." at $i and $j...\n";
                          $sequence['values'] = $values;
                          $sequence['frequence']++;
                          break 3;
                       }
                    }
                 }
              }

              //echo "Found new ".implode(',',$sub)." at $i and $j...\n";
              array_push($sequences, array('values' => $sub, 'frequence' => 2));
              break;
           }
        }
     }
  }

  return $sequences;
};