Question

我有此数据，如何返回唯一的数组-并非每个索引都重复的数组。

[ 
  [ 0, 1, 2 ], 
  [ 1, 0, 2 ], 
  [ 1, 1, 1 ], 
  [ 1, 2, 0 ], 
  [ 2, 0, 1 ], 
  [ 2, 1, 0 ] 
]

我想要的输出将是这样
0 1 2
1 2 0
2 0 1

Answer 1

这应该是您想要的。

console.clear()
arr = [ 
  [ 0, 1, 2 ], 
  [ 1, 0, 2 ], 
  [ 1, 1, 1 ], 
  [ 1, 2, 0 ], 
  [ 2, 0, 1 ], 
  [ 2, 1, 0 ] 
];

var res = arr.filter((duplicates = [], e => {
  // prefill duplicates with empty arrays
  e.forEach((k, i) => {
    duplicates[i] = duplicates[i] ||  [];
  })
  // check if there are duplicates and then set keep to false
  let keep = e.reduce((keep, val, i) => {
    return keep && (!duplicates[i].includes(val));
  }, true);
  // if keep, save this array to duplicates
  if (keep) {
    e.forEach((k, i) => {
      duplicates[i].push(k)
    })
  }
  return keep;
}))

res.forEach(k => {
  console.log(...k)
})

.as-console-wrapper { max-height: 100% !important; top: 0; }

Answer 2

尝试

let a= [ [0,1,2], [1,0,2], [1,1,1], [1,2,0 ], [2,0,1 ], [2,1,0 ] ];

let t=[{},{},{}];

let d= a.filter(e => 
  !e.reduce((g,h,i)=> g||(e[i] in t[i]),false) && e.map((x,i) => t[i][x]=x)
);

console.log(JSON.stringify(d));

更新

OP唯一性任务可以有多个解决方案（@NinaSholz在对此答案的评论中提到了什么）-例如对于OP示例，我们至少有两个独立的独特（按OP的描述）解决方案：

[0,1,2],[1,2,0],[2,0,1]
[1,0,2],[2,1,0]

OP示例中有趣的是，这两个解决方案都有不同数量的元素-我问有关寻找最佳解决方案（最短或最长）here的新问题。

Answer 3

尝试一下

let arr= [ [ 0, 1, 2 ], [ 1, 0, 2 ], [ 1, 1, 1 ], [ 1, 2, 0 ], [ 2, 0, 1 ], [ 2, 1, 0 ] ];

let index =[];
arr.forEach(function(element) {
  element.forEach(function (value, i) {
    if(index[i] == undefined)
    {
      var x = [value];
      index[i]= x;
    }
    else if (index[i].indexOf(value) ==-1)
    {
      index[i].push(value);
    }
  });
});
index.forEach(k => {
  console.log(...k)
})

Answer 4

此解决方案是在一个循环中检查每个数组的所有元素，如果每个数组中有三个元素（如您所示），那么此解决方案应该对您有用。

var arr = [ 
  [ 0, 1, 2 ], 
  [ 1, 0, 2 ], 
  [ 1, 1, 1 ], 
  [ 1, 2, 0 ], 
  [ 2, 0, 1 ], 
  [ 2, 1, 0 ] 
]

var indicesToRemove = arr.reduce((acc, eachElem, index) => {
  var isRemoveIndexPushed = false;
  if (acc.firstIndex.length !== 0 && acc.firstIndex.indexOf(eachElem[0]) != -1) {
    isRemoveIndexPushed = true
  }
  
  if (acc.secondIndex.length !==0 && acc.secondIndex.indexOf(eachElem[1]) != -1) {
    isRemoveIndexPushed = true;
  }
  
  
  if (acc.thirdIndex.length !== 0 && acc.thirdIndex.indexOf(eachElem[2]) != -1) {
    isRemoveIndexPushed = true
  } 
  
  if (!isRemoveIndexPushed) {
    acc.firstIndex.push(eachElem[0]);
    acc.secondIndex.push(eachElem[1]);
    acc.thirdIndex.push(eachElem[2]);
  } else {
    acc.removeIndex.push(index);
  }
  
  return acc;

  
}, {removeIndex: [], firstIndex: [], secondIndex: [], thirdIndex: []}).removeIndex;


var newArr = []
arr.forEach((elem, i) => {
  if (i === indicesToRemove[0]) {
    indicesToRemove.shift()
  } else {
    newArr.push(elem)
  }
})

arr = newArr
console.log(arr)

删除多维数组中的重复项

4 个答案: