如何获得 JavaScript 中具有重复值的两个数组之间的差异

Question

如何检查一个数组是否包含在另一个数组中以及 return 缺失值？我找到了执行此操作的方法 in this post 但 none 考虑了数组中的重复值。例如，我正在尝试做这样的事情：

getDiff([1, 2, 3], [1, 2, 3, 4]) --> []

getDiff([1, 2, 2, 3], [1, 2, 3, 4]) --> [2]

getDiff(["A", "B", "C"], ["B", "B", "A", "C"]) --> []

getDiff(["B", "B", "B"], [3, 2, 1]) --> ["B", "B", "B"]

Answer 1

一种可能的方法：

function getRepeatingDiff(source, diffed) {
  const diffedCounter = diffed.reduce((acc, el) => {
    acc[el] = (acc[el] || 0) + 1
    return acc;
  }, {});
  
  const diff = source.reduce((acc, el) => {
    return diffedCounter[el]-- > 0 ? acc : [ ...acc, el ];
  }, []);
  
  return diff;
}

console.log( getRepeatingDiff([1, 2, 3], [1, 2, 3, 4]) );
console.log( getRepeatingDiff([1, 3, 2, 2, 3], [1, 2, 3, 4]) );
console.log( getRepeatingDiff(["A", "B", "C"], ["B", "B", "A", "C"]) );
console.log( getRepeatingDiff(["B", "B", "B"], [3, 2, 1]) );

本质上，这是一个两步过程：计算 diffed 数组中的项目数，然后再次遍历 source 数组 - 并将新项目添加到 diff diffedCounter 中缺失的每个副本的结果数组。但是这个特定的实现有一个缺陷：因为它使用 Object 来收集计数，所以它在计算元素时不区分 3 作为数字和 '3' 作为字符串。

这可以通过两种不同的方式解决：切换到 Map（但这会使代码更复杂，因为没有诸如递减 map 值之类的东西）- 或者在创建计数器键时只使用类型前缀.这是基于前者的方法：

function getRepeatingDiff(source, diffed) {
  const diffedCounter = diffed.reduce((map, el) => {
    const prev = map.get(el);
    return map.set(el, (prev || 0 ) + 1);
  }, new Map());
  
  const diff = source.reduce((acc, el) => {
    const remainingCount = diffedCounter.get(el);
    if (remainingCount) diffedCounter.set(el, remainingCount - 1);
    return remainingCount ? acc : [ ...acc, el ];
  }, []);
  
  return diff;
}

console.log( getRepeatingDiff([1, 2, 3], [1, 2, 3, 4]) );
console.log( getRepeatingDiff([1, 3, 2, 2, 3], [1, 2, 3, 4]) );
console.log( getRepeatingDiff(["A", "B", "C"], ["B", "B", "A", "C"]) );
console.log( getRepeatingDiff(["B", "B", "B"], [3, 2, 1]) );

如何获得 JavaScript 中具有重复值的两个数组之间的差异

How to get the difference between two arrays in JavaScript with repeating values

javascript

arrays

array-comparison