重命名Javascript数组中的重复项

时间:2012-05-17 21:42:24

标签: jquery arrays duplicates

我正在寻找一种最有效的方法来重命名(append-1,-2等)一个变量,如果它已经存在于一个字符串中。

所以我保留了一个数组“

dupeCheck = [];

一旦我看到变量:

var UID;

已经在我的dupeCheck数组中,我想立即用-1附加UID的值,

另外,我需要防止第三个副本变成string-1-1,而不是string-2 ..

我之前看过这个:Appending the count to duplicates in a javascript string array,但它正是我想要的......

任何聪明的想法?我更喜欢jQuery ..

/编辑:

例如:

var dupeUIDCheck = [];  

$.each(data[IDS].UIDs[keys], function(keys, val)
     {
     var currString = val;
     switch (key)
 {
      case "UID":

       UID = unquote(currString);

   //TODO:
   //Detect if multiple UIDs are loaded from a single source, and
   //rename them:

   dupeUIDCheck.push(UID); //Push current ID onto existing array

       //Check if ID exists
       ?
       //If exists rename value of currString, save it in currString
       newName = currSting;
      break;

      case "otherstuff":
           //Other vars to parse
      break;
     }

因此,当我们退出“UID”案例时,我想确保它具有唯一值

3 个答案:

答案 0 :(得分:6)

您可以将功能包装在函数中以便能够重用它。下面的函数采用字符串列表并返回-1-2等后缀字符串列表。

function suffixDuplicates( list )
{
    // Containers

    var count = { };
    var firstOccurences = { };

    // Loop through the list

    var item, itemCount;
    for( var i = 0, c = list.length; i < c; i ++ )
    {
        item = list[ i ];
        itemCount = count[ item ];
        itemCount = count[ item ] = ( itemCount == null ? 1 : itemCount + 1 );

        if( itemCount == 2 )
            list[ firstOccurences[ item ] ] = list[ firstOccurences[ item ] ] + "-1";
        if( count[ item ] > 1 )
            list[ i ] = list[ i ] + "-" + count[ item ]
        else
            firstOccurences[ item ] = i;       
    }

    // Return
    return list;
}

例如,输入

[ "Barry", "Henk", "Jaap", "Peter", "Jaap", "Jaap", "Peter", "Henk", "Adam" ]

返回

的输出
[ "Barry", "Henk-1", "Jaap-1", "Peter-1", "Jaap-2", "Jaap-3", "Peter-2", "Henk-2", "Adam" ]

要查看它的实际效果,here是指向jsFiddle示例的链接。

答案 1 :(得分:1)

保留要检查重复项的列表的最佳方法是将它们放在对象中,而不是数组中,以便您可以快速查找它们,然后生成一个尚未使用的唯一后缀每一次。此函数允许您将id传递给函数,并让函数返回尚未使用的该id的唯一版本。如果传入的内容未被使用,则只返回该内容。如果传入的内容正在使用中,它会删除任何后缀并生成一个尚未使用的新后缀并返回新创建的id。然后,新创建的id将存储在数据结构中,以便将来不会重复。

var idList = {};

makeIdUnique(id) {
    if (id in idList) {
        // remove any existing suffix
        var base = id.replace(/-\d+$/, "");
        // generate a new suffix
        var cnt = idList[base] || 1;
        // while new suffix is in the list, keep making a different suffix
        do {
            id = base + "-" + cnt++;
        } while (id in idList);
        // save cnt for more efficient generation next time
        idList[base] = cnt;
    }
    // put the final id in the list so it won't get used again in the future
    idList[id] = true;
    // return the newly generated unique id
    return(id);
}

答案 2 :(得分:1)

理解你的问题有点困难,但如果你指的是这个,请告诉我。假设我们有一串单词,有些单词重复。我们希望使用新的后缀修改这些重复的字词,例如-1-2,具体取决于它的实例。

// Start by creating our string, word array, and result array
var string = "one two one one two one three three one three",
    values = string.split(" "), result = [];

// For every word in the values array
for ( var i = 0; i < values.length; i++ ) {

  // Set a word variable, and an integer suffix
  var word = values[i], int = 1;

  // While our current word (with/without suffix) exists in results
  while ( strArr(word, result) >= 0 ) 
    // Increment the suffix on our word
    word = values[i] + "-" + int++;

  // Push word into result array
  result.push(word);
}

// Function for determining if a string is in an array
function strArr(s,a){
  for ( var j = 0; j < a.length; j++ )
    if ( a[ j ] == s ) return j;
  return -1;
}

// Compare the before and after
console.log( string );
console.log( result.join(" ") );

我们的结果是

one two one one two one three three one three
one two one-1 one-2 two-1 one-3 three three-1 one-4 three-2