使用不同的对象数组数据更新对象数组

Question

我有一个非常棘手的操作。 就是这样

我有一个名为“ data1”的对象数组

[Object, Object, Object]
       0:Object
               id="00456145"     //check this 
               name: "Rick"
               upper:"0.67"
               lower:"-0.34"
      1:Object
               id="00379321"     
               name:"Anjie"
               upper:"0.46"
               lower:"-0.56"
      2:Object
               id="00323113"      
               name:"dillan"
               upper:"0.23"
               lower:"-0.11"

我只对这些对象数组的id，上限值和下限值感兴趣。

这是名为“ data2”的对象的第二个数组

[Object, Object]
     0:Object
             id="0045614"   //present here if we remove last element of '00456145'
             cast="Rick"
             Contact: "Yes"
             upper:"0.11"    //need to be updated to '0.67'
             lower:"-0.11"   //need to be updated to '-0.34'
     1:Object
             id="0032311"     //present here if we remove last element of '00323113'
             cast:"dillan"
             Contact:"Maybe"
             upper:"0.11"
             lower:"-0.11"

所以，这就是我要做的。 我将首先检查“ data1”。 检查“ data1”中存在的ID。 例如，对象0具有id =“ 00456145”

我删除其中的最后一个数字。 因此它变为id =“ 0045614”。 然后，我比较此id是否存在于对象“ data2”中。

如果存在，则在“ data1”中为该对象0显示的上限值和下限值传递到存在id的对象“ data2”中。 在这种情况下，“ data2”的对象0具有id =“ 0045614”。

因此上限和下限值将分别更新为0.67和-0.34。

对于其他阵列也是如此。 因此最终输出应类似于“ data2”

[Object, Object]
     0:Object
             id="0045614"   
             cast="Rick"
             Contact: "Yes"
             upper:"0.67"    //updated
             lower:"-0.34"   //updated
     1:Object
             id="0032311"  
             cast:"dillan"
             Contact:"Maybe"
             upper:"0.23"      //updated
             lower:"-0.11"     //updated

Answer 1

在这种情况下，我喜欢array＃some ，这样您就可以摆脱循环，因此不会出现不必要的迭代：

 var data1 = [{ id: "00456145", name: "Rick", upper: "0.67", lower: "-0.34" }, { id: "00379321", name: "Anjie", upper: "0.46", lower: "-0.56" }, { id: "00323113", name: "dillan", upper: "0.23", lower: "-0.11" }]; var data2 = [{ id: "0045614", cast: "Rick", Contact: "Yes", upper: "0.11", lower: "-0.11" }, { id: "0032311", cast: "dillan", Contact: "Maybe", upper: "0.11", lower: "-0.11" }]; data2 = data2.map(function(item) { data1.some(function(a) { if (item.id == a.id.slice(0, -1)) { item.upper = a.upper; item.lower = a.lower; return true; } }); return item; }); console.log(data2); 

Answer 2

我建议使用一个对象作为哈希表或映射。 然后，仅需要两个循环，一个循环用于获取引用，一个循环用于分配。

大O： O（n + m）

虽然我实际上不知道哪一种（地图与对象）更适合，但您可能会得到自己的照片：

• JavaScript中的地图与对象
• ES6中的地图与对象，何时使用？

带Object提案

 var data1 = [{ id: "00456145", name: "Rick", upper: "0.67", lower: "-0.34", }, { id: "00379321", name: "Anjie", upper: "0.46", lower: "-0.56", }, { id: "00323113", name: "dillan", upper: "0.23", lower: "-0.11" }], data2 = [{ id: "0045614", cast: "Rick", Contact: "Yes", upper: "0.11", lower: "-0.11", }, { id: "0032311", cast: "dillan", Contact: "Maybe", upper: "0.11", lower: "-0.11" }], hash = Object.create(null); data1.forEach(function (a) { hash[a.id.slice(0, -1)] = a; }); data2.forEach(function (a) { var o = hash[a.id]; o && Object.keys(o).forEach(function (k) { if (k !== 'id' && a[k] !== o[k]) { a[k] = o[k]; } }); }); console.log(data2); 

Map提案

 var data1 = [{ id: "00456145", name: "Rick", upper: "0.67", lower: "-0.34", }, { id: "00379321", name: "Anjie", upper: "0.46", lower: "-0.56", }, { id: "00323113", name: "dillan", upper: "0.23", lower: "-0.11" }], data2 = [{ id: "0045614", cast: "Rick", Contact: "Yes", upper: "0.11", lower: "-0.11", }, { id: "0032311", cast: "dillan", Contact: "Maybe", upper: "0.11", lower: "-0.11" }], map = new Map; data1.forEach(function (a) { map.set(a.id.slice(0, -1), a); }); data2.forEach(function (a) { var o = map.get(a.id); o && Object.keys(o).forEach(function (k) { if (k !== 'id' && a[k] !== o[k]) { a[k] = o[k]; } }); }); console.log(data2); 

Answer 3

您可以尝试类似的方法：

for (var i = 0; i < data1.length; i++) {
  var item = data1[i];
  var trim = item.id.substr(0, item.id.length - 1);
  for (var j = 0; j < data2.length; j++) {
    var item2 = data2[j];
    if (item2.id === trim) {
      item2.upper = item.upper;
      item2.lower = item.lower;
    }
  }
}

如果您使用的是es6，则可以通过以下方式简化代码：

for (let item of data1) {
  let trim = item.id.substr(0, item.id.length - 1);
  for (let item2 of data2) {
    if (item2.id === trim) {
      item2.upper = item.upper;
      item2.lower = item.lower;
    }
  }
}

Answer 4

这是使用Array.prototype.forEach的解决方案。 遍历data2对象，如果id与data1任何对象匹配，则用匹配的对象更新值。

var data1 = [{
    id: "00456145",
    name: "Rick",
    upper: "0.67",
    lower: "-0.34"
  }, {
    id: "00379321",
    name: "Anjie",
    upper: "0.46",
    lower: "-0.56"
  }, {
    id: "00323113",
    name: "dillan",
    upper: "0.23",
    lower: "-0.11"
  }];

  var data2 = [{
    id: "0045610", //present here if we remove last element of '00456145'
    cast: "Rick",
    Contact: "Yes",
    upper: "0.11", //need to be updated to '0.67'
    lower: "-0.11" //need to be updated to '-0.34'
  }, {
    id: "0032311", //present here if we remove last element of '00323113'
    cast: "dillan",
    Contact: "Maybe",
    upper: "0.11",
    lower: "-0.11"
  }];

data2.forEach(function(item2){
   data1.forEach(function(item1){
   if (item1.id.substring(0,7) === item2.id){ // can be `item1.id.indexOf(item2.id) == 0
     item2.upper = item1.upper;
     item2.lower = item1.lower;
   }
  });
});

console.log(data2);

Answer 5

您可以使用类似

  data1.forEach(function(obj) {
      var search = obj.id.slice(0, -1);
      data2.forEach(function(d) {
          if (d.id === search) {
              d.upper = obj.upper;
              d.lower = obj.lower;
          };
      });
  });

去做这个。

Answer 6

也许这就是您想要的。

 var data1 = [{ id: "00456145", name: "Rick", upper: "0.67", lower: "-0.34" }, { id: "00379321", name: "Anjie", upper: "0.46", lower: "-0.56" }, { id: "00323113", name: "dillan", upper: "0.23", lower: "-0.11" }]; var data2 = [{ id:"0045614", cast:"Rick", Contact: "Yes", upper:"0.11", lower:"-0.11", }, { id:"0032311", cast:"dillan", Contact:"Maybe", upper:"0.11", lower:"-0.11", }]; data2.forEach(function(data2Value) { data1.forEach(function(data1Value) { if(data1Value.id.substr(0, data1Value.id.length - 1) === data2Value.id) { data2Value.upper = data1Value.upper; data2Value.lower = data1Value.lower; } }); }); console.log(data2);