Question

我有以下对象：

var data = {
        2: [[1,2],[3,4,5],[6,7]],
        3: [[1,2],[3,4],[6,7]],
        4: [[1,2],[3,4,5],[6,7]],
        5: [[10,2],[3,4,5],[60,7]]
    }

我想从对象中获取密钥2，因为它的值在数组中是重复的。可能有多个这样的重复键。

我使用lodash并尝试以下但它不会工作。我不知道如何开始：

var dups = [];
_.map(formData,function(key, value){
    dups.push(value);
})
var duplicateArray = _.uniqWith(dups, _.isEqual);

Answer 1

这段代码将匹配整个数组，而不管值序列和值计数。请考虑以下示例

//代码在这里

Plunker：https://plnkr.co/edit/6jEByZHQhcwUopqqG4a2?p=preview （单击上面的链接并打开控制台以查看结果）

这将输出[“2”，“3”]因为“2”matches“4”和“3”matches“6”。

//代码在这里

var data = {
  2: [ [1, 2], [3, 4, 5], [6, 7]],
  3: [[1, 2],[3, 4],[6, 7]],
  4: [[2, 1, 2, 1],[5, 3, 4],[6, 7]],
  5: [[10, 2],[3, 4, 5],[60, 7]],
  6: [[2, 1],[3, 4],[7, 6, 6]]
}

var dups = [];

//sorted and uniq buffer.
var buff = _.clone(data);
_.forEach(buff, function(lvl1, key) {
  buff[key] = _.map(lvl1, function(val) {
    return _.uniq(_.sortBy(val));
  });
});

_.forEach(buff, function(lvl1, key) {

  //Match each row with all others.
  //i.e. 2 with 3, 2 with 4, 2 with 5 ..... 6 with 5.
  _.forEach(buff, function(_lvl1, _key) {
    //Skip entries like 2 with 2, 3 with 3 and so on
    if (key !== _key) {
      console.log('compare ' + key + ' with ' + _key);
      var x = _.filter(_lvl1, function(val, index) {
        //Below condition are only true because ary are sorted and uniq.
              //If both must be of same len 
        return val.length === lvl1[index].length
               //If lenght of intersection of both values must be exactly equal 
               && _.intersection(val, lvl1[index]).length === val.length;
      });
      //If count matches and key and _key is not in dups
      // key and _key not in dups because we only want following:
      // 2 with 4 -> 2
      // 4 with 2 -> ignore! (we dont wanna include 4)
      //Hope it makes sense
      if (x.length === _lvl1.length && _.indexOf(dups, key) === -1 && _.indexOf(dups, _key) === -1) {
         //Mark the key as duplicate
         dups.push(key);
      }
    }

  });
});


console.log(dups)

希望这有帮助！

Answer 2

我不确定正确理解你想要什么，但我认为答案看起来就像那样（sample）：

<强>使用Javascript：

  var data = {
        2: [[1,2],[3,4,5],[6,7]],
        3: [[1,2],[3,4],[6,7]],
        4: [[1,2],[3,4,5],[6,7]],
        5: [[10,2],[3,4,5],[60,7]]
    };

  var find = function(input){
    var res = [];
    for(var prop in input){

      var start = false;
      for(var prop2 in input){
        if(start){
          var flag = JSON.stringify(input[prop]) == JSON.stringify(input[prop2]);
          if(flag){
            var flag3 = true;
            for(var j = 0; j < res.length; j++)
              if(JSON.stringify(input[prop]) == JSON.stringify(input[res[j]])){
                  flag3 = false
                  break;
              }
            if(flag3)
              res.push(prop);
            break;
          }
        }
        else{
         if(prop == prop2)
          start = true; 
        }
      }

    }
    return res;
  }

  var answer = find(data);

使用lodash查找深重复值的对象键

2 个答案: