javascript - 尝试使用 reduce 查找偶数和奇数计数

Question

我正在尝试使用 reduce 解决以下问题，但我无法获得对象中偶数和奇数的正确计数。

有人可以指导我了解我的代码有什么问题吗？

Create a function countBy that accepts an array and a callback, andreturns an object. countBy will iterate through the array and performthe callback on each element. Each return value from the callback willbe saved as a key on the object. The value associated with each keywill be the number of times that particular return value was returned

function countBy(arr, fn) {
  return arr.reduce(function(acc, nums) {
    // console.log(nums);
    let oddCount = 0
    let evenCount = 0
    console.log(nums, fn(nums))
    if(fn(nums) === "even"){
      evenCount++;
      acc['even'] = evenCount;
    } else {
      oddCount++;
      acc['odd'] = oddCount;
    }
    return acc
  }, {}, 0)
}

function evenOdd(n) {
 if (n % 2 === 0) return "even";
 else return "odd";
}

var nums = [1, 2, 3, 4, 5];
console.log(countBy(nums, evenOdd)); // should log: { odd: 3, even: 2 }

Answer 1

您正在将 oddCount 和 evenCount 初始化为 0 inside reduce 回调，所以在每次迭代中，你的

evenCount++;
acc['even'] = evenCount;

只会将 evenCount 或 oddCount 递增到 1。相反，在回调之外初始化计数，以便对它们的更改在 reduce 回调的多次调用中保持不变:

function countBy(arr, fn) {
  let oddCount = 0
  let evenCount = 0
  return arr.reduce(function(acc, nums) {
    // console.log(nums);
    console.log(nums, fn(nums))
    if (fn(nums) === "even") {
      evenCount++;
      acc['even'] = evenCount;
    } else {
      oddCount++;
      acc['odd'] = oddCount;
    }
    return acc
  }, {}, 0)

}

function evenOdd(n) {
  if (n % 2 === 0) return "even";
  else return "odd";
}
var nums = [1, 2, 3, 4, 5];
console.log(countBy(nums, evenOdd)); // should log: { odd: 3, even: 2 }

或者，您可以通过检查累加器上已有的属性值来完全避免外部变量:

const countBy = (arr, fn) => arr.reduce((acc, num) => {
  const prop = fn(num);
  acc[prop] = (acc[prop] || 0) + 1;
  return acc;
}, {});

function evenOdd(n) {
  if (n % 2 === 0) return "even";
  else return "odd";
}
var nums = [1, 2, 3, 4, 5];
console.log(countBy(nums, evenOdd)); // should log: { odd: 3, even: 2 }