Question

给定此对象数组：

[ { source_q_id: 'Q2B', target_q_id: null },
  { source_q_id: 'Q2B', target_q_id: null },
  { source_q_id: 'Q2B', target_q_id: null },
  { source_q_id: 'Q5C', target_q_id: 'Q7' },
  { source_q_id: 'Q5C', target_q_id: 'Q7' },
  { source_q_id: 'Q5C', target_q_id: 'Q7' },
  { source_q_id: 'Q5D', target_q_id: 'Q7' },
  { source_q_id: 'Q5D', target_q_id: 'Q7' },
  { source_q_id: 'Q5D', target_q_id: 'Q7' },
  { source_q_id: 'Q6A1', target_q_id: 'Q7' },
  { source_q_id: 'Q6A1', target_q_id: 'Q7' },
  { source_q_id: 'Q6A1', target_q_id: 'Q7' },
  { source_q_id: 'Q6A2', target_q_id: null },
  { source_q_id: 'Q6A2', target_q_id: null },
  { source_q_id: 'Q6A3', target_q_id: 'Q7' },
  { source_q_id: 'Q6A3', target_q_id: 'Q7' },
  { source_q_id: 'Q6A3', target_q_id: 'Q7' },
  { source_q_id: 'Q6B', target_q_id: 'Q6A2' },
  { source_q_id: 'Q6B', target_q_id: 'Q6A2' },
  { source_q_id: 'Q7', target_q_id: null },
  { source_q_id: 'Q7', target_q_id: null }]

对于键/值，我需要一个新的独特dupe数组：

  [ { source_q_id: 'Q2B', target_q_id: null },
    { source_q_id: 'Q5C', target_q_id: 'Q7' },
    { source_q_id: 'Q5D', target_q_id: 'Q7' },
    { source_q_id: 'Q6A1', target_q_id: 'Q7' },
    { source_q_id: 'Q6A2', target_q_id: null },
    { source_q_id: 'Q6A3', target_q_id: 'Q7' },
    { source_q_id: 'Q6B', target_q_id: 'Q6A2' },
    { source_q_id: 'Q7', target_q_id: null }]

我使用了此SO answer中的代码，但它删除了太多target_q_id个独特的欺骗，因为它不计算source_q_id欺骗：<\ n / p>

[ { source_q_id: 'Q2B', target_q_id: null },
  { source_q_id: 'Q5C', target_q_id: 'Q7' },
  { source_q_id: 'Q6B', target_q_id: 'Q6A2' } ]

代码：

function dupesOnly(arr, 'target_q_id') {
    var seen = {},
        ret = [];

    arr.forEach(function(item) {
        var key = item[field],
            val = seen[key];

        if (!val) {
            seen[key] = val = {
                initial: item,
                count: 0
            }
        }

        if (val.count === 1) {
            ret.push(val.initial);
        }
        ++val.count;
    });

    return ret;
}

如何修改代码以查找source_q_id和target_q_id的唯一欺骗？

Answer 1

您可以简单地组合这些属性，并将其用作您的集合中的键另一个注意事项：如果您.push中有.forEach项，那么您最有可能做错了。

尝试使用Array.prototype.filter：

＆＃13;

Array.prototype.uniqueBy = function(keyBuilder) {
    var seen = {};
    return this.filter(function(o) {
      // build a filter key using a provided function
      var key = keyBuilder(o); 

      // if item already exists - do not add to the result
      if (seen[key]) 
        return false;
      
      // add item to the set and add item to the result
      // shortand for: 
      // seen[key] = true; return true;
      return (seen[key] = true);
    });
}

var obj = [{source_q_id:'Q2B',target_q_id:null},{source_q_id:'Q2B',target_q_id:null},{source_q_id:'Q2B',target_q_id:null},{source_q_id:'Q5C',target_q_id:'Q7'},{source_q_id:'Q5C',target_q_id:'Q7'},{source_q_id:'Q5C',target_q_id:'Q7'},{source_q_id:'Q5D',target_q_id:'Q7'},{source_q_id:'Q5D',target_q_id:'Q7'},{source_q_id:'Q5D',target_q_id:'Q7'},{source_q_id:'Q6A1',target_q_id:'Q7'},{source_q_id:'Q6A1',target_q_id:'Q7'},{source_q_id:'Q6A1',target_q_id:'Q7'},{source_q_id:'Q6A2',target_q_id:null},{source_q_id:'Q6A2',target_q_id:null},{source_q_id:'Q6A3',target_q_id:'Q7'},{source_q_id:'Q6A3',target_q_id:'Q7'},{source_q_id:'Q6A3',target_q_id:'Q7'},{source_q_id:'Q6B',target_q_id:'Q6A2'},{source_q_id:'Q6B',target_q_id:'Q6A2'},{source_q_id:'Q7',target_q_id:null},{source_q_id:'Q7',target_q_id:null}];
var objFiltered = obj.uniqueBy(function(o) { 
    return o.source_q_id + "~~~" + o.target_q_id;
});
console.log(objFiltered);

＆＃13;

Javascript：查找匹配多个字段的对象数组值dupes

1 个答案: