Question

我在json中有以下对象数组，我想得到数组中对象的索引，该索引最接近给定数字值的天数属性。

在jQuery或JS中存在任何内置函数，或者我如何轻松解决它？

感谢您的帮助。

"warningAttributes": [
                        {
                            "id": "3",
                            "days": 15,                          
                            "color": "#FFFFFF"                 
                        },
                        {
                            "id": "4",
                            "days": 98,                          
                            "color": "#343434"                  
                        }
                    ]

Answer 1

var warningAttributes =  [{"id": "3", "days": 15, "color": "#FFFFFF"},
                          {"id": "4", "days": 98, "color": "#343434"}]

var target = 90; 

var nearest = warningAttributes.sort(function(a,b){ 
    return Math.abs(a.days-target) - Math.abs(b.days-target)
})[0];

如果您不在[0]

，这也会为您提供最近的列表

Answer 2

考虑到您正在尝试找到最接近的ID。

你可以这样做，http://jsfiddle.net/s4jq5kv7/。虽然你需要清理代码，但这是有效的。

var id=3;
var temp;
var prev;

var x={"warningAttributes": [
                        {
                            "id": "3",
                            "days": 15,                          
                            "color": "#FFFFFF"                 
                        },
                        {
                            "id": "4",
                            "days": 98,                          
                            "color": "#343434"                  
                        }
                    ]
      }
prev=x.warningAttributes[0].id;
val=prev;
$(x.warningAttributes).each(function(i,ele){

    if(id>parseInt(ele.id)){
        temp=id-parseInt(ele.id);
    }else{
                temp=parseInt(ele.id)-id;
    }
    if(prev>temp)
    {prev=temp;
     val=ele.id;}
})

console.log(val)

唯一的问题是，如果有一个数字与两个以上的数字有相同的差异，那么返回第一个数字。前3和4，如果您发现最接近3.5，则会获得第一个数字，即3。

Answer 3

如果我理解得很好：

    jQuery(document).ready(function () {
    var control=40;
    var json = {"warningAttributes": [
                        {
                            "id": "3",
                            "days": 15,                          
                            "color": "#FFFFFF"                 
                        },
                        {
                            "id": "4",
                            "days": 98,                          
                            "color": "#343434"                  
                        },
                         {
                            "id": "5",
                            "days": 40,                          
                            "color": "#343434"                  
                        }
                    ]};
   $.each(json.warningAttributes, function(k, v) {
       var diff;
       diff= control - v.days
       if(diff==0) alert("this is your element: " + v.id);
   })
});

Answer 4

这很简单：

function getNearest(arr,value)
{
  var diff = null;
  var index= 0;
  $.each(arr,function(i,item) {
    var current = Math.abs(item.days - value);
    if (diff == null) diff = current;
    if (current< diff) 
       index=i;
  });
  return index
}

Answer 5

克里斯几乎偷了这个（+1），但这里是我如何尝试它的细分：

function getDays(arr, value) {

    // get a list of the differences between the requested value
    // and the numbers in the array and make them all positive
    var abs = arr.map(function (el) { return Math.abs(el.days - value); });

    // find the smallest number
    var smallest = Math.min.apply(null, abs);

    // find out where that number is in the positive number array
    var index = abs.indexOf(smallest);

    // return an object that contains the index of the
    // object, and the number of days that were closest
    return { index: index, value: arr[index].days };
}

var result = getDays(arr, 35); // Object { index: 3, value: 17 }

使用result.index获取索引。

DEMO

Answer 6

我的去处是因为编写短代码很有趣。

var warningAttributes =  [{"id": "3", "days": 15, "color": "#FFFFFF"},
                          {"id": "4", "days": 98, "color": "#343434"}]

var n = 90; 
var distances = warningAttributes.map(function(c) {return Math.abs(n - c.days);})
var nearest_idx = distances.indexOf(Math.min.apply(Math, distances));

如何获取具有最接近值的数组项的索引？

6 个答案: