JS如果有两个以上重复项，则在数组中查找重复值的索引 [英] JS Find indices of duplicate values in array if there are more than two duplicates

问题描述

我要在连续游戏中创建三个坐标平面，所以我必须找出数组BUT WITHOUT sorting array中是否存在三个具有相同值的数字，因为该数组表示添加到坐标点上的点的x坐标在比赛中协调飞机... 例如，假设我已经向x坐标存储在下一个数组中的坐标平面添加了6个点:

I'm creating coordinate plane Three in a row game so I have to find out if there are three numbers of the same value in the array BUT WITHOUT sorting array because the array represents the x-coordinates of the points added to the coordinate plane during the game... For example, let's say that I've added 6 points to the coordinate plane with x-coordinates stored in next array:

var arr = [2,2,3,2,7,3];

我需要一个循环，该循环将仅对值2的出现进行计数，因为数字2在数组中出现3次，因此输出应为一个新数组(例如名为indexs的数组)，其精确索引为nmb 2出现的次数在……...

I need the loop that will count only the occurrences of the value 2 because the number 2 occurs 3 times in array, so the output should be a new array (for example array named indices) with the exact indices of nmb 2 occurrences in arr...

indices = [0,1,3]

如果某个值的出现次数少于3次，则循环应在到达arr末尾时重置".

The loop should therefor "reset" when comes to the end of the arr if the number of occurrences of some value is less than 3...

我已经尝试了下一个代码，但是它不能像我上面描述的那样工作，因为它也计算3的出现次数...因此，如果value，它将不会重置"数量少于2 ...

I've tried the next code, but it doesn't work like I've described above because it counts the number of occurrences of the number 3 as well... so, it won't "reset" if value count is less than 2...

var arr = [2,2,3,2,7,3];
var index = [];
var count = 0;
var current;
for(var i = 0;i<arr.length;i++){
    current = arr[i];
    //-------------------------------
    for(var j=i+1;j<arr.length;j++){
        if(arr[j]===current){
            count++;
            index.push(j);
            //+++++++++++
            if(j===arr.length-1){
                if(count<2){
                    count = 0;
                    index = [];
                    i++;
                }
                else{
                    index.unshift(i);
                    break;
                }
            }
            //+++++++++++
        }
    }
    //-------------------------------
}
alert(index);

感谢您的帮助或建议...

Thanks for any help or advice...

亚历山德拉(Aleksandra)

Aleksandra

推荐答案

是的，使用一些逻辑...

Yes, using some logic...

var arr = [2, 2, 3, 2, 7, 3];

function showDupPos(arr, mindups) {
  mindups = mindups || 2;
  var result = [];
  var positions = {};
  // collect all positions
  arr.forEach(function(value, pos) {
    positions[value] = positions[value] || [];
    positions[value].push(pos);
  });
  //check how much of same value in string
  Object.keys(positions).forEach(function(value) {
    var posArray = positions[value];
    if (posArray.length > mindups) {
      result = result.concat(posArray);
    }
  });
  return result.sort();
}
console.log(showDupPos(arr));

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