Question

我有一个引用数组，其值为["a","b","c","d"]。我还有另一个数组，该数组是作为API的一部分而获得的，该数组的格式不是很一致。

case 1.`{
     names : ["a"],
     value : [ [0],[0],[2],[4],... ]
   }`
case 2. `{
     names : ["a","c"],
     value : [ [0,2],[0,0],[2,3],[4,4],... ]
    }`

结果可以是任意组合但我的要求是将传入结果的值分配给另一个数组索引与我的参考数组相同例如：在情况1

`
let finalArray = [["0",null,null,null],
                  ["0",null,null,null], 
                   ["2",null,null,null]....  ]
`

对于情况2：

`let finalArray = [["0",null,"2",null],
                  ["0",null,"0",null], 
                   ["2",null,"3",null]....   ]  
`

在下面的内容中附加一个小提琴

jsfiddle link to problem

有什么建议吗？我正在尝试使用最少的for循环进行性能优化

Answer 1

希望这会有所帮助。

var refArray = ["a","b","c","d"];

setTimeout(()=>{processResult({
     "names" : ["a"],
     "value" : [ [0],[0],[2],[4]]
   })},2000);
setTimeout(()=>{processResult(
{
     "names" : ["a","c"],
     "value" : [ [0,2],[0,0],[2,3],[4,4]]
})},4000);
setTimeout(()=>{processResult(
{
     "names" : ["d","c"],
     "value" : [ [0,2],[0,0],[2,3],[4,4]]
})},6000);


function processResult (result) {
    let res = result.value;
    let resArray = res.map((el)=>{
        let k=Array(refArray.length).fill(null);
        refArray.forEach((e,i)=>{
          let indx = result.names.indexOf(e);
          if(indx>=0){      	
                k[i] = el[indx]
            }      
        });
        return k;
    })
  console.log("result",resArray)	
}

Answer 2

下面是我认为需要最少迭代的情况。

    var refArray = ["a", "b", "c", "d"];
    setTimeout(()=>{processResult({
         "names" : ["a"],
         "value" : [ [0],[0],[2],[4]]
       })},2000);
    setTimeout(()=>{processResult(
    {
         "names" : ["a","c"],
         "value" : [ [0,2],[0,0],[2,3],[4,4]]
    })},4000);
    setTimeout(()=>{processResult(
    {
         "names" : ["d","c"],
         "value" : [ [0,2],[0,0],[2,3],[4,4]]
    })},6000);


    function processResult(result) {
      //This map will contain max names matched in the result
      var maxItemsFromResult = {};

      //Find the indexes in refArray and fill map
      //e.g. 1st- {0:0}, 2nd - {0:0, 1:2}, 3rd - {0:3, 1:2}
      result.names.forEach((item, index) => {
        let indexFound = refArray.indexOf(item);
        if (indexFound > -1) {
          maxItemsFromResult[index] = indexFound;
        }
      });

      //for performance if no key matched exit
      if (Object.keys(maxItemsFromResult).length < 1) {
        return;
      }
      //This will be final result
      let finalArray = [];

      //Because finalArray's length shuld be total items in value array loop through it
      result.value.forEach((item, itemIndex) => {
        //Create a row
        let valueArray = new Array(refArray.length).fill(null);
        //Below will only loop matched keys and fill respective position/column in row
        //i'm taking all the matched keys from current value[] before moving to next
        Object.keys(maxItemsFromResult).forEach((key, index) => {
          valueArray[maxItemsFromResult[key]] = item[index];//get item from matched key
        });
        finalArray.push(valueArray);
      });
      console.log(finalArray);
      return finalArray;
    }

根据另一个数组Javascript排列数组

2 个答案: