在坐标的相对位置显示矩阵的算法

Question

考虑x=0 and y=0作为原点，向下y-axis和向右x-axis作为正轴，是否有任何标准方法或算法用于空间中相对位置的矩阵变换。

[ [{x:36,y:14},{x:242,y:14}],
  [{x:36,y:133}],
  [{x:36,y:252}],
  [{x:36,y:371},{x:242,y:371},{x:446,y:371},{x:651,y:371}],
  [{x:242,y:490},{x:446,y:490},{x:651,y:490}] ]

现在因为这个数组数组的长度是 5 并且其中最长数组的长度是 4，所以我需要大小为 5 * 4 的变换矩阵 格式如下。

[ [{x:36,y:14},{x:242,y:14},null,null],
  [{x:36,y:133},null,null,null],
  [{x:36,y:252},null,null,null],
  [{x:36,y:371},{x:242,y:371},{x:446,y:371},{x:651,y:371}],
  [null,{x:242,y:490},{x:446,y:490},{x:651,y:490}] ]

在上述情况下，保留了相对位置。

提前致谢！！

Answer 1

解决方案首先将所有唯一的 x 值减少到排序的平面数组中。

然后遍历每一行数据，遍历每一行数组将null拼接到孔中

let data =[ [{x:36,y:14},{x:242,y:214}],
  [{x:36,y:133}],
  [{x:36,y:252}],
  [{x:36,y:371},{x:242,y:371},{x:446,y:371},{x:651,y:371}],
  [{x:242,y:490},{x:446,y:490},{x:651,y:490}] ]
  
  
let xVals  = [...new Set(data.reduce((a,c)=>a.concat(c.map(({x})=>x)),[]))].sort((a,b)=>a-b)

data.forEach(row=>{
   xVals.forEach((x,i)=>{
      if(row[i] === undefined  || row[i].x > x){
          row.splice(i,0, null)
      }
   });
});

 data.forEach(arr=>console.log(JSON.stringify(arr)))

Answer 2

检查这段代码。解释将在那里评论。

function normalize(array){

	// Get the largest sub-array. We will save this as a reference
	// to use it later
	var longest_value = array.reduce((a,b)=>a>b?a:b)


	// map each element in the main array
	return array.map(function(a){ 

		// for each item return a modified copy of the largest one.
		// To do this we map it
		return longest_value.map(function(b,i){

			// we the item with the same x position in the current main array item
			var v = a.filter(r=>r.x==b.x)

			//if there is, we return it, is not we return null
			return v.length? v[0] : null
		})
	})
}


console.log(normalize([ [{x:36,y:14},{x:242,y:214}],[{x:36,y:133}],[{x:36,y:252}],[{x:36,y:371},{x:242,y:371},{x:446,y:371},{x:651,y:371}],[{x:242,y:490},{x:446,y:490},{x:651,y:490}] ]))