使用不同的对象数组数据更新对象数组

Update an array of objects with a different array of objects data

我有一个非常棘手的操作要做。原来如此。

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中的 id、上限值和下限值感兴趣。

这是名为 'data2'

的第二个对象数组
[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,这就是我必须要做的。我会先检查'data1'。检查 'data1' 中存在的 ID。例如对象 0 有 id="00456145"

我删除了其中的最后一个数字。所以它变成 id="0045614"。 然后我比较这个 id 是否存在于对象 'data2' 中。

如果存在,则 'data1' 中对象 0 的上限值和下限值将传递给存在 id 的对象 'data2'。在这种情况下,'data2' 的对象 0 具有 id='0045614'。

因此上限值和下限值将分别更新为 0.67 和 -0.34。

其他数组也一样。所以最终的输出应该是这样的 '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

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,您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;
    }
  }
}

这是一个使用 Array.prototype.forEach 的解决方案。遍历 data2 个对象,如果 id 与 data1 中的任何对象匹配,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);

您可以使用

  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;
          };
      });
  });

做这个。

也许这就是你想要的。

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);

我喜欢这种情况下的 array#some,这样您就可以跳出循环,从而避免不必要的迭代:

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);

我建议使用对象作为散列 table 或地图。那么只需要两个循环,一个用于获取引用,一个用于分配。

大O:O(n+m)

虽然我其实不知道哪个更适合(Map vs Object),你可以得到你自己的照片:

  • Map vs Object in JavaScript

提案 Object

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);

提案 Map

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);