为什么我得到 NaN 以及如何获得预期的输出？

Question

我是 JS 新手。我想跟踪数组中重复对象的数量，在另一个对象中，所以我决定编写一个函数。例如：考虑下面的对象。

const fruits = [{name: 'orange'}, {name: 'orange'}, {name: 'apple'}, {name: 'banana'}];

我想跟踪上面数组中重复对象的数量，所以我写了下面的函数。

function calcDuplicates(fruitObj) {
  const duplicates = {};
  for (let i of fruitObj) {
    duplicates[`${i.name}`] !== "undefined"
      ? (duplicates[`${i.name}`] += 1)
      : (duplicates[`${i.name}`] = 1);
  }
  return duplicates;
}

我得到的输出是：

{orange: NaN, apple: NaN, banana: NaN}

(预期输出) 我想要的输出是：

{orange: 2, apple: 1, banana: 1}

如何获得预期的输出？我做错了什么？

Answer 1

您的错误是针对"undefined"（单词）而不是undefined（未定义值）进行测试：

const fruits = [{name: 'orange'}, {name: 'orange'}, {name: 'apple'}, {name: 'banana'}];

function calcDuplicates(fruitObj) {
  const duplicates = {};
  for (let i of fruitObj) {
    duplicates[i.name] !== undefined
      ? (duplicates[i.name] += 1)
      : (duplicates[i.name] = 1);
  }
  return duplicates;
}

console.log( calcDuplicates(fruits) );

Answer 2

在条件中使用原始值undefined代替字符串"undefined"：

duplicates[`${i.name}`] !== undefined

const fruits = [{name: 'orange'}, {name: 'orange'}, {name: 'apple'}, {name: 'banana'}];

function calcDuplicates(fruitObj) {
  const duplicates = {};
  for (let i of fruitObj) {
    duplicates[`${i.name}`] !== undefined
      ? (duplicates[`${i.name}`] += 1)
      : (duplicates[`${i.name}`] = 1);
  }
  return duplicates;
}
console.log(calcDuplicates(fruits))

Answer 3

1. 属性名称无需使用字符串文字。 i.name 就足够了。

2. 您检查的是“未定义”而不是 undefined。但是，真的，没有必要这样做。只需检查对象属性是否存在。

const fruits = [{name: 'orange'}, {name: 'orange'}, {name: 'apple'}, {name: 'banana'}];

function calcDuplicates(fruitObj) {
  const duplicates = {};
  for (let i of fruitObj) {
    duplicates[i.name]
      ? duplicates[i.name] += 1
      : duplicates[i.name] = 1;
  }
  return duplicates;
}

console.log(calcDuplicates(fruits));

Answer 4

只需从 undefined 中删除双引号 "

const fruits = [{name: 'orange'}, {name: 'orange'}, {name: 'apple'}, {name: 'banana'}];

function calcDuplicates(fruitObj) {
  const duplicates = {};
  for (let i of fruitObj) {
    duplicates[`${i.name}`] !== undefined
      ? (duplicates[`${i.name}`] += 1)
      : (duplicates[`${i.name}`] = 1);
  }
  return duplicates;
}

console.log( calcDuplicates(fruits) );

Answer 5

这是因为您的条件应该针对undefined 进行测试，而不是包含'undefined' 的字符串。

const fruits = [{name: 'orange'}, {name: 'orange'}, {name: 'apple'}, {name: 'banana'}];

function calcDuplicates(fruitObj) {
  const duplicates = {};
  for (let i of fruitObj) {
    duplicates[`${i.name}`] !== undefined
      ? (duplicates[`${i.name}`] += 1)
      : (duplicates[`${i.name}`] = 1);
  }
  return duplicates;
}

console.log( calcDuplicates(fruits) );

我就是这样写的：

const fruits = [{name: 'orange'}, {name: 'orange'}, {name: 'apple'}, {name: 'banana'}];

function calcDuplicates(fruitObj) {
  const duplicates = {};
  for (let i of fruitObj) {
    duplicates[i.name] = (duplicates[i.name] || 0) + 1
  }
  return duplicates;
}

console.log( calcDuplicates(fruits) );

几乎相同的东西，使用reduce 和解构map：

const fruits = [{name: 'orange'}, {name: 'orange'}, {name: 'apple'}, {name: 'banana'}];

const calcDuplicates = o => o
  .map(({ name }) => name)
  .reduce((a, b) => ({ ...a, [b]: (a[b] || 0) + 1 }), {})

console.log( calcDuplicates(fruits) );

Answer 6

这可以通过reduce 和对象解构来实现，如下所示：

const fruits = [{name: 'orange'}, {name: 'orange'}, {name: 'apple'}, {name: 'banana'}];


const duplicates = fruits.reduce( (acc,{name}) => {
  if (!acc[name]) acc[name] = 0 ; 
  acc[name]++;
  return acc;
},{});

console.log(duplicates);