数组的Map方法详解

var numbers = [1, 4, 9]
var roots = numbers.map(item => {
    return Math.sqrt(item)
})
// roots的值为[1, 2, 3], numbers的值仍为[1, 4, 9]

因为Math.sqrt只有一个参数，所以可简写为
var numbers = [1, 4, 9]
var roots = numbers.map(Math.sqrt)
// roots的值为[1, 2, 3], numbers的值仍为[1, 4, 9]

需要注意 parseInt 方法是有两个参数的

["1", "2", "3"].map(parseInt); // 期望输出 [1, 2, 3], 而实际结果是 [1, NaN, NaN].

解决方法

function returnInt(element) {
  return parseInt(element, 10);
}

['1', '2', '3'].map(returnInt); // [1, 2, 3]

var kvArray = [{key: 1, value: 10},
               {key: 2, value: 20},
               {key: 3, value: 30}];

var reformattedArray = kvArray.map(function(obj) {
   var rObj = {};
   rObj[obj.key] = obj.value;
   return rObj;
});

// reformattedArray 数组为：[{1: 10}, {2: 20}, {3: 30}],

// kvArray 数组未被修改:
// [{key: 1, value: 10},
//  {key: 2, value: 20},
//  {key: 3, value: 30}

var numbers = [1, 2, 3, 4];
var filteredNumbers = numbers.map(function(num, index) {
  if(index < 3) {
     return num;
  }
});
//index goes from 0,so the filterNumbers are 1,2,3 and undefined.
// filteredNumbers is [1, 2, 3, undefined]
// numbers is still [1, 2, 3, 4]