Javascript按值合并数组中的对象

Question

在Javascript中，我有一个包含对象的数组，如下所示：

var arr = [
    {
        value_1: 0,
        value_2: 4,
        encounter_bits: {
            min_level: 8,
            max_level: 12,
            rarity: 20,
            conditions: [1, 5]
        }
    },
    {
        value_1: 1,
        value_2: 4,
        encounter_bits: {
            min_level: 5,
            max_level: 5,
            rarity: 20,
            conditions: [2, 9]
        }
    },
    {
        value_1: 0,
        value_2: 4,
        encounter_bits: {
            min_level: 8,
            max_level: 12,
            rarity: 5,
            conditions: [1, 5]
        }
    },
];

我需要合并具有相同min_level、max_level和条件的对象。合并后的对象将会将它们的稀有度相加。我还需要保留数组顺序。

因此arr和arr2将变成：

    arr[0] = {
        value_1: 0,
        value_2: 4,
        encounter_bits: {
            min_level: 8,
            max_level: 12,
            rarity: 25,
            conditions: [1, 5]
        }
    }

在大致相同的数据集中，这是在Python中完成的：

# Combine "level 3-4, 50%" and "level 3-4, 20%" into "level 3-4, 70%".
existing_encounter = filter(lambda enc: enc['min_level'] == encounter.min_level
                                    and enc['max_level'] == encounter.max_level,
                            encounter_bits)
if existing_encounter:
    existing_encounter[0]['rarity'] += encounter.slot.rarity
else:
    encounter_bits.append({
        'min_level': encounter.min_level,
        'max_level': encounter.max_level,
        'rarity': encounter.slot.rarity,
    })

我知道我可能需要对array.sort和array.splice做点什么，但是我想不通。实现这一目标的最有效方法是什么？

Answer 1

这是我刚刚想出来的一些实验代码和它的原生javascript。

首先定义一个比较两个数组的函数。

function compareArrays( val_one, val_two ) {
  var arr_one = val_one, arr_two = val_two;

   if ( arr_one.length !== arr_two.length ) return false;

   for ( var i = 0; i < arr_one.length; i++ ) {
     if( arr_one[i] !== arr_two[i] ) return false;
   }  
  return true;
}

**还定义了一个合并两个对象的函数

function mergeObjects( obj_1, obj_2 ) {
  var a = obj_1, b = obj_2;

  for ( prop in b ) { 
    if ( !a[prop] ) {
      a[prop] = b[prop];
    }
    else if ( prop === 'encounter_bits' ) {
      a[prop]['rarity'] += b[prop]['rarity'];
    }
  }
  return a;
}

让我们遍历数组

for ( var i=0; i<arr.length; i++ ) {  // loop over the array that holds objects.
 if(arr[i] !== null){    // if object has not been used
  var _min = arr[i].encounter_bits.min_level,  
      _max = arr[i].encounter_bits.max_level,
      _con = arr[i].encounter_bits.conditions; //create variables that hold values you want to compare

for ( var c = 0; c < arr.length; c++ ) { /*  create an inner loop that will also traverse the array */

  if ( c !== i && arr[c] !== null ) {   // if inner loop is not on current loop index 
    if ( _min === arr[c].encounter_bits.min_level &&
        _max === arr[c].encounter_bits.max_level &&
        compareArrays(_con, arr[c].encounter_bits.conditions)  
      ) 
        {
         var newObject = mergeObjects(arr[i], arr[c]);
         arr[i] = newObject;     
         arr[c] = null; 
         }
        }

    }
  }       
}

如果可以的话，请告诉我。

Answer 2

我还没有测试过它，但它应该给出了基本的想法。不使用拼接或排序。

// Collect members that have same min-max levels in one array.
// this array is saved in a map keyed by a string made up of min-max levels
var m = {};
for(var i=0; i<arr.length; i++)
{
    var a = arr[i];
    tag = ''+a.encounter_bits.min_level+'-'+a.encounter_bits.max_level);
    if(m[tag]){
        m[tag].push(i);
    }else {
        m[tag] = [i]
    }
}

// for each element of map, calculate sum of rarities
// set the rarity of all member in that element of map to sum of rarities
for(var i in m){
    var candidates = m[i];
    var r = 0;
    for(var j=0; j<candidates.length; j++){
      r += candidates[j].encounter_bits.rarity;
    }
    for(var j=0; j<candidates.length; j++){
       candidates[j].encounter_bits.rarity = r;
    }            
 }

Answer 3

下面是用代码编写的名称空间良好且自包含的解决方案。它没有使用像forEach和filter这样的“现代”(可能是不可用的)特性。它在说完和做完之后清理无效的记录，而不是在过程进行时删除活动数组中的项。如果您想要更改识别重复项的方式或合并对象的方式，也可以非常容易地进行修改。

在pastie，here上有一个注释版本(带有示例和精简版本)。

(function (global, ns) {
function mergeRecordsAndCleanUp(arr) {
  var rec, dupeIndices, foundDupes;
  for(var idx=0, len=arr.length; idx<len-1; ++idx) {
    rec = arr[idx];
    if (rec===null) continue;
    dupeIndices = findDupeIndices(rec, arr.slice(idx+1), idx+1);
    if (dupeIndices.length===0) continue;
    foundDupes = true;
    processDupes(rec, dupeIndices, arr);
  }
  if (foundDupes) cleanUpArray(arr);
}
function cleanUpArray(arr) {
  for (var idx=0, len=arr.length; idx<len; ++idx) {
    if (arr[idx]===null) arr.splice(idx--, 1);
  }
}
function processDupes(rec, dupeIndices, arr) {
  var dupeRealIdx, dupeRec;
  for (var dupeIdx=0, dupesLen=dupeIndices.length; dupeIdx<dupesLen; ++dupeIdx) {
    dupeRealIdx = dupeIndices[dupeIdx];
    dupeRec = arr[dupeRealIdx];
    updateRecord(rec, dupeRec);
    arr[dupeRealIdx] = null;
  }
}
function findDupeIndices(rec, arr, offset) {
  var other, result = [];
  for (var idx=0, len=arr.length; idx<len; ++idx) {
    other = arr[idx];
    if (other===null) continue;
    if (isDupe(rec, other)) result.push(idx+offset);
  }
  return result;
}
function identicalArrays(arr0, arr1) {
  if (arr0.length!==arr1.length) return false;
  for (var idx=0, len=arr0.length; idx<len; ++idx) {
    if (arr0[idx]!==arr1[idx]) return false;
  }
  return true;
}
function isDupe(original_record, candidate_record) {
  var orec_bits = original_record.encounter_bits
    , crec_bits = candidate_record.encounter_bits;
  return (crec_bits.min_level===orec_bits.min_level && crec_bits.max_level===orec_bits.max_level) 
    && (identicalArrays(crec_bits.conditions, orec_bits.conditions));
}
function updateRecord(rec, dupe) {rec.encounter_bits.rarity += dupe.encounter_bits.rarity}
global[ns] = {
  npup: mergeRecordsAndCleanUp
};
})(this, 'npup'/* sweet namespace eh */);
// npup.npup(arr);