JavaScript的Reduce方法：创建累加器对象

Question

var numbers = [5, 3, 8, 6, 9, 1, 0, 2, 2];
var oddEvenCounts = numbers.reduce(function(counts, number) {
  if (isOdd(number)) {
    counts[odd]++;
  } else {
    counts[even]++;
  }
}, {});

我正在寻找这段代�中的错误（还在学习reduce方法 😉）——我哪里出了问题？

- marie_antoinette

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变量even和odd中存储了什么值？ - Thomas

Answer 1

带有注释的工作代码：

var numbers = [5, 3, 8, 6, 9, 1, 0, 2, 2];

function isOdd(n) {
  return !!(n % 2);
}

var oddEvenCounts = numbers.reduce(function(counts, number) {
  if (isOdd(number)) {
    counts.odd++; // use dot notation or ['odd']
  } else {
    counts.even++;  // use dot notation or ['even']
  }
  
  return counts; // return the accumulator
}, { odd: 0, even: 0 }); // set the initial values of odd and even

console.log(oddEvenCounts);

你可以使用方括号符号和三元运算符来缩短代码：

var numbers = [5, 3, 8, 6, 9, 1, 0, 2, 2];

function isOdd(n) {
  return !!(n % 2);
}

var oddEvenCounts = numbers.reduce(function(counts, number) {  
  counts[isOdd(number) ? 'odd' : 'even']++;
  
  return counts;
}, { odd: 0, even: 0 });

console.log(oddEvenCounts);

- Ori Drori

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谢谢，非常有帮助：我知道我的问题所在——没有返回语句，我也没有正确设置计数器对象，也没有正确调用它。 - marie_antoinette

Answer 2

返回您的累加器：

var numbers = [5, 3, 8, 6, 9, 1, 0, 2, 2];
var oddEvenCounts = numbers.reduce(function(counts, number) {
  if (isOdd(number)) {
    counts[odd]++;
  } else {
    counts[even]++;
  }
  return counts;
}, {});

- Joe

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谢谢，这是我最常犯的错误。 - Amesys