我想从两个-dimensional数组中删除相同的部分

Question

我想比较以下两种安排的脚本删除重复安排。

我使用了过滤器并包含函数或语句，但失败了。

我该怎么办？

let ground = [
  [1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1],
  [1,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,1],
  [1,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,1],
  [1,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,1],
  [1,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,1],
  [1,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,1],
  [1,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,1],
  [1,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,1],
  [1,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,1],
  [1,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,1],
  [1,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,1],
  [1,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,1],
  [1,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,1],
  [1,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,1],
  [1,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,1],
  [1,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,1],
  [1,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,1],
  [1,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,1],
  [1,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,1],
  [1,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,1],
  [1,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,1],
  [1,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,1],
  [1,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,1],
  [1,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,1],
  [1,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,1],
  [1,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,1],
  [1,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,1],
  [1,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,1],
  [1,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,1],
  [1,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,1],
  [1,3,3,7,7,7,7,8,8,4,4,4,2,2,3,7,1],
  [1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1],
]

let deleteBlock = [[1,3,3,7,7,7,7,8,8,4,4,4,2,2,3,7,1]]

我试过这样做

let ans = ground.filter((r,idx) => {
    for(let i =0; i < deleteBlock.length;i++) {
     if(r === deleteBlock[i]) {
        ground.splice(idx, 1)
     }
    }
})

ground.filter((r) => deleteBlock.forEach(ele=> r.includes(ele)))

Answer 1

您可以创建一个比较两个数组的arrayEqual函数

function arrayEqual(arr1, arr2) {
  if (arr1.length !== arr2.length) return false;

  for (let i = 0; i < arr1.length; i++)
    if (arr1[i] !== arr2[i]) return false;  //adapt equality test if you have complex objects

  return true;
}

然后像下面这样使用它

let ground = [...];
let filtered = ground.filter(x => deleteBlock.every(y => !arrayEqual(x,y)));

Answer 2

如果将两个数组与相同的属性进行比较，则始终会得到false。例如，1 === 1总是返回false，因为它们都是一个全新的数组对象，内容完全相同。

因此，您应该循环两个数组，并逐一比较它们的每个属性，或者在您的情况下，因为我假设类型不是必需的，您可以解构它们，然后像下面这样比较字符串：

const filtered = ground.filter((row) => { return row.join() !== deleteBlock.join(); });