Question

我们有一阵列填充类别名称(其中许多重复),我们需要建立两个层面阵列,消除阵列一面的重复,而阵列另一边的类别名称在原阵列中出现的次数多。这里有两幅照片,使您能更好地了解我描述的情况:http://postimage.org/image/ptms64cl9/a>和。现在我确信,这样做有不止一种方式,但我想理解该书是如何做到的。这里是该法典,它指出,舱面值-类别具有重复类别的最初阵列:

 $category_totals = array(array($mismatch_categories[0], 0));
    foreach ($mismatch_categories as $category) {
      if ($category_totals[count($category_totals) - 1][0] != $category) {
        array_push($category_totals, array($category, 1));
      }
      else {
        $category_totals[count($category_totals) - 1][1]++;
      }
    }

我对这个例子不了解的一个主要内容是阵列中的阵列。这里有3个阵列:

 $category_totals = array(array($mismatch_categories[0], 0));

如果有3个阵列,我如何使用它们的指数? 可能的话:

 $category_totals[0][0][0];

Answer 1

希望将有助于你们理解。

<?php
     echo  <pre> ;
     $mismatch_categories = array( cat ,  cat ,  cow ,  book ,  box ,  box ,  box );
     echo  Input Mismatch Category::<br /> ;
     print_r($mismatch_categories);
     echo  <br /> ;
     $category_totals = array(array($mismatch_categories[0], 0));
     echo  categroy totals that holds final data  .  <br /> ;
     $counter = 0;
     print_r($category_totals);
     foreach ($mismatch_categories as $category) {
         echo  Iteration   . $counter++ .  <br /><br /> ;
         echo  Current category value::  . $category . "<br /><br />";
         echo  Value of category_totals[  . count($category_totals) .  - 1][0] ::   .         $category_totals[count($category_totals) - 1][0] .  <br/><br /> ;
         echo  Are they equal  .  <br /> ;
         if ($category_totals[count($category_totals) - 1][0] != $category) {
              echo  Not matched so pushed into array with occurence of one<br /> ;
              array_push($category_totals, array($category, 1));
         } else {
              echo  matches so count is increased by 1  . "<br />";
              $category_totals[count($category_totals) - 1][1]++;
         }
         echo  category totals:  .  <br /> ;
         print_r($category_totals);
   }
   echo  Final value of category_totals:: ;
   print_r($category_totals);
?>

Answer 2

核对阵列_count_ Values()功能()

资料来源:

array(
    0 =>  Cat 1 ,
    1 =>  Cat 1 ,
    2 =>  Cat 1 ,
    3 =>  Cat 2 ,
    4 =>  Cat 2 ,
    5 =>  Cat 3 ,
    6 =>  Cat 4 ,
)

成果阵列——面值:

array(
     Cat 1  => 3,
     Cat 2  => 2,
     Cat 3  => 1,
     Cat 4  => 1,
)

Answer 3

落实评论意见中的建议:

<?php
$mismatch_categories = array( cat ,  cat ,  cow ,  book ,  box ,  box ,  box );

$cat_counts = array_count_values($mismatch_categories);
// = array( cat  => 2,  cow  => 1,  book  => 1,  box  => 3)

$categories = array_unique($mismatch_categories);
// = array( cat ,  cow ,  book ,  box );
?>

换言之,这正是你所期望的。

在这里写你自己处理文件的唯一原因是编码做法。

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