将 Json 对象的所有对象中的小数值四舍五入

Question

这是我的 JSON 字符串

[{"$id":"1","GeoName":"EAST","Value2":87.88221970554928,"Value4":90.71955219607294,"Value1":18.44377295716579,"Value3":16.732108234801206},{"$id":"2","GeoName":"WEST","Value2":87.88221970554928,"Value4":90.71955219607294,"Value1":18.44377295716579,"Value3":16.732108234801206}]"

这是我在 Stringyfying 之前的 JSON 对象

**MyObj:**


0:Object
    $id:"1"
    Value1:18.44377295716579
    GeoName:"EAST"
    Value2:87.88221970554928
    Value3:16.732108234801206
    Value4:90.71955219607294
1:Object
    $id:"2"
    Value1:18.44377295716579
    GeoName:"WEST"
    Value2:87.88221970554928
    Value3:16.732108234801206
    Value4:90.71955219607294

有没有办法在我的 JSON 的每个对象中将所有值（甚至单独的 value1、value2...）四舍五入到小数点后 2 位？

喜欢

0:Object
        $id:"1"
        Value1:18.44
        GeoName:"EAST"
        Value2:87.89
        Value3:16.73
        Value4:90.72
1:Object
        $id:"2"
        Value1:18.44
        GeoName:"WEST"
        Value2:87.89
        Value3:16.73
        Value4:90.71

我尝试过使用

Obj = JSON.parse(JSON.stringify(MyObj, function (GeoName, Value1) {
                return Value1.toFixed ? Number(Value1.toFixed(2)) : Value1;
            }));

这对我不起作用。

有没有办法通过调整上面的代码行来实现这一点，或者有更好的方法来解决这个问题？

Edit1（根据 wostex 建议的方法）

var numKeys = ['ShareValue','ReachValue','DepthValue','WidthValue'];
MyObj=Obj.map(e => Object.assign(e,Object.keys(e).filter(k => numKeys.includes(k)).map(v => Object.assign({[v]: parseFloat(e[v]).toFixed(2)}))))

这就是我的对象的样子

0:Object
       0:Object
        ReachValue:87.88
       1:Object
        DepthValue:16.73
       2:Object
        ShareValue:18.44
       3:Object
        WidthValue:90.71
        $id:"1"
        ShareValue:18.44377295716579
        GeoName:"EAST"
        ReachValue:87.88221970554928
        DepthValue:16.732108234801206
        WidthValue:90.71955219607294

编辑2

Obj = JSON.parse(JSON.stringify(BubbleObj));
var numKeys = ['ShareValue','ReachValue','DepthValue','WidthValue'];
            var MyObj=Obj.map(e => Object.assign(e,Object.keys(e).filter(k => numKeys.includes(k)).map(v => Object.assign({[v]: parseFloat(e[v]).toFixed(2)}))))

这里 Obj 有来自问题第一行的 JSON 字符串，而 BubbleObj 有我在问题中提到为 MyObj 的 Object。

Answer 1

简单的方法就是使用map - 只要你的 Value1-4 总是被称为

var input = [{"$id":"1","GeoName":"EAST","Value2":87.88221970554928,"Value4":90.71955219607294,"Value1":18.44377295716579,"Value3":16.732108234801206},{"$id":"2","GeoName":"WEST","Value2":87.88221970554928,"Value4":90.71955219607294,"Value1":18.44377295716579,"Value3":16.732108234801206}];

var output = input.map(function(e) {
    return {
      $id: e.$id,
      GeoName: e.GeoName,
      Value1: Math.round(e.Value1*100) /100,
      Value2: Math.round(e.Value2*100) /100,
      Value3: Math.round(e.Value3*100) /100,
      Value4: Math.round(e.Value4*100) /100
    }
});

console.log(output);

如果您不想对其进行硬编码，则可以查找以 Value 开头的任何密钥 - 它有点复杂。

var input = [{"$id":"1","GeoName":"EAST","Value2":87.88221970554928,"Value4":90.71955219607294,"Value1":18.44377295716579,"Value3":16.732108234801206},{"$id":"2","GeoName":"WEST","Value2":87.88221970554928,"Value4":90.71955219607294,"Value1":18.44377295716579,"Value3":16.732108234801206}];

var output = input.map(function(e) {
    return Object.keys(e).reduce(function(p,n){
       if(n.startsWith("Value"))
           p[n] = Math.round(e[n]*100)/100;
       else
           p[n] = e[n];
       return p;
    },{})
});

console.log(output);

Answer 2

顺便说一句，您的第一个字符串中有重复的键“Value3”，除此之外，您可以执行以下操作：

var j = [{"$id":"1","GeoName":"EAST","ShareValue":87.88221970554928,"ReachValue":90.71955219607294,"DepthValue":18.44377295716579,"WidthValue":16.732108234801206},{"$id":"2","GeoName":"WEST","ShareValue":87.88221970554928,"ReachValue":90.71955219607294,"DepthValue":18.44377295716579,"WidthValue":16.732108234801206}];

// your numeric keys
var numKeys = ['ShareValue','ReachValue','DepthValue','WidthValue'];

var res = j.map(e => Object.assign(e, // rewrite each object
  ...Object.keys(e).filter(k => numKeys.indexOf(k) > -1) // pick keys which include 'Value'
    .map(v => Object.assign({[v]: parseFloat(e[v]).toFixed(2)})) // chop numbers
  ))
  
console.log(res)