Question

我们花了将近一天半的时间来完成这项工作，但我仍然不确定它为什么会这样。如果有更好的方法可以实现，我很乐意听到它。在目前的状态下，我希望这有助于某人。

var newValuesArray = [];
var arrayIndex = [1, 4, 9];
var valuesArray = [["Zero", "One", "Two", "Three", "Four", "Five", "Six", "Seven", "Eight", "Nine"], 
              ["A", "B", "C", "D", "E", "F", "G", "H", "I", "J"]
              ];

var roots = valuesArray.map(function(num) {
    arrayIndex[num];
    return arrayIndex;
});

for (var i = 0, len = roots.length; i < len; i++) {
  newValuesArray.push(roots[i].map(function(num) {
     return valuesArray[i][num];
  }));
}

console.log(newValuesArray);

这是我正在寻找上面代码产生的结果：

[["One", "Four", "Nine"], ["B", "E", "J"]]

Answer 1

您可以使用 map() 和 filter()

＆＃13;

var arrayIndex = [1, 4, 9];
var valuesArray = [
  ["Zero", "One", "Two", "Three", "Four", "Five", "Six", "Seven", "Eight", "Nine"],
  ["A", "B", "C", "D", "E", "F", "G", "H", "I", "J"]
];

// iterate over main array using map()
var newValuesArray = valuesArray.map(function(v) {
  // iterate and filter values in inner array using filfer()
  return v.filter(function(v1, i) {
    // check index in arrayIndex
    return arrayIndex.indexOf(i) > -1;
  });
});

document.write('<pre>'+JSON.stringify(newValuesArray,null,3)+'</pre>');

＆＃13;

更新：如果您需要获取索引数组的相同顺序的值，请使用

＆＃13;

var arrayIndex = [4, 9, 1];
var valuesArray = [
  ["Zero", "One", "Two", "Three", "Four", "Five", "Six", "Seven", "Eight", "Nine"],
  ["A", "B", "C", "D", "E", "F", "G", "H", "I", "J"]
];

// iterate over main array using map()
var newValuesArray = valuesArray.map(function(v) {
  // iterate the index array
  return arrayIndex.map(function(v1) {
    // get value from inner array based on index
    return v[v1];
  });
});

document.write('<pre>'+JSON.stringify(newValuesArray,null,2)+'</pre>');

＆＃13;

Answer 2

您可以考虑使用数组方法Array.prototype.map来迭代valuesArray和arrayIndex。

var arrayIndex = [1, 4, 9],
    valuesArray = [
        ["Zero", "One", "Two", "Three", "Four", "Five", "Six", "Seven", "Eight", "Nine"],
        ["A", "B", "C", "D", "E", "F", "G", "H", "I", "J"]
    ],
    newValuesArray = valuesArray.map(function (a) {
        return arrayIndex.map(function (b) {
            return a[b];
        });
    });
document.write('<pre>' + JSON.stringify(newValuesArray, 0, 4) + '</pre>');

Answer 3

一个解决方案：

＆＃13;

var valuesArray = [
  ["Zero", "One", "Two", "Three", "Four", "Five", "Six", "Seven", "Eight", "Nine"], 
  ["A", "B", "C", "D", "E", "F", "G", "H", "I", "J"],
  ["blue" , "green" , "yellow" , "orange" , "black" , "white" , "violet" , "pink" , "purple" , "grey" ]
];

/* the function ---------------------------------------------*/
var fn = function(indexed){
  
  return indexed.reduce(function( trans , value){
  
     trans.arr.forEach(function(curArray , i){
        var curBranch = (trans.rep[i] = trans.rep[i] || []);
        curBranch.push( curArray[value]  )
     })
     return trans;
  } , {arr : valuesArray , rep : []}).rep;
 
};


/* the samples -----------------------------------------------*/

var arrayIndex = [1, 4, 9];   
document.write( JSON.stringify( fn(arrayIndex) ) );


arrayIndex = [9, 4, 1];
document.write('<br>' + JSON.stringify( fn(arrayIndex) ) );


arrayIndex = [9, 9, 3, 4, 1];
document.write('<br>' + JSON.stringify( fn(arrayIndex) ) );


arrayIndex = [99, 1, 2];
document.write('<br>' + JSON.stringify( fn(arrayIndex) ) );

＆＃13;

使用一个数组的元素作为索引创建新数组，以选择不同数组中的元素 - JavaScript

3 个答案: