Question

我有两个这样的字符串列表：

let previous_id = ["5b7b5498ab3f510e307e7d04", "5b7ae97cc6d75e1331e28d9c", "5b7a0c207fab2722a2081caf"];
let current_id = ["5b7b5498ab3f510e307e7d04", "5b83e3b4412f370bd7b9a05d"];

我想比较两个列表中的所有元素，以便知道列表中仍存在哪个，哪个是新的，以及哪个已被删除。

结果将是：

delete_id = ["5b7ae97cc6d75e1331e28d9c", "5b7a0c207fab2722a2081caf"];
new_id = ["5b83e3b4412f370bd7b9a05d"];
existing_id = ["5b7b5498ab3f510e307e7d04"];

有什么功能吗？

我想避免像这样的多个if条件进行两个循环：

let previous_id = ["5b7b5498ab3f510e307e7d04", "5b7ae97cc6d75e1331e28d9c", "5b7a0c207fab2722a2081caf"];
let current_id = ["5b7b5498ab3f510e307e7d04", "5b83e3b4412f370bd7b9a05d"];

delete_id = [];
new_id = [];
existing_id = [];

for(let item of previous_id){
   let index = current_id.indexOf(item);
   if(index == -1){
       delete_id.push(item);
   } else {
       existing_id.push(item);
   }
}

for(let item of current_id){
   let index = previous_id.indexOf(item);
   if(index == -1){
      new_id.push(item);
   }
}


console.log(delete_id)
console.log(new_id)
console.log(existing_id)

Answer 1

基本设置操作应该起作用：

// Copied directly from MDN
function intersection(setA, setB) {
    var _intersection = new Set();
    for (var elem of setB) {
        if (setA.has(elem)) {
            _intersection.add(elem);
        }
    }
    return _intersection;
}

function difference(setA, setB) {
    var _difference = new Set(setA);
    for (var elem of setB) {
        _difference.delete(elem);
    }
    return _difference;
}


let previous_id = ["5b7b5498ab3f510e307e7d04", "5b7ae97cc6d75e1331e28d9c", "5b7a0c207fab2722a2081caf"],
    current_id = ["5b7b5498ab3f510e307e7d04", "5b83e3b4412f370bd7b9a05d"],
    previous_set = new Set(previous_id),
    current_set = new Set(current_id),

    // perform set operations on the two sets
    existing_ids = intersection(previous_set, current_set),
    deleted_ids = difference(previous_set, current_set),
    new_ids = difference(current_set, previous_set);
    
console.log("Existing ids: ", Array.from(existing_ids));
console.log("Deleted ids: ", Array.from(deleted_ids));
console.log("New ids: ", Array.from(new_ids));

集合操作比列表操作更有效，因此应在有意义的任何时候使用集合。

Answer 2

我会在设置的数组之上构建数组操作。由于不应有重复项，因此这是内部计算的正确结构。我不知道为什么Set.prototype至少没有union，intersection和difference，但是自己编写它们是很简单的。

// Set operations

const intersection = (set1, set2) => new Set([...set1].filter(x => set2.has(x)))
const difference = (set1, set2) => new Set([...set1].filter(x => !set2.has(x)))


// Array operations, builtt using the set ones
// NB: Arrays are NOT Sets, and there is some information lost in the conversion.
// But Sets are the proper data structure for unordered collections of unique values.

const intersectionA = (arr1, arr2) =>  Array.from(intersection(new Set(arr1), new Set(arr2)))
const differenceA = (arr1, arr2) => Array.from(difference(new Set(arr1), new Set(arr2)))


// Main code

const breakdown = (prev, curr) => ({
  delete_id: differenceA(prev, curr),
  new_id: differenceA(curr, prev),
  existing_id: intersectionA(prev, curr)
})

let previous_id = ["5b7b5498ab3f510e307e7d04", "5b7ae97cc6d75e1331e28d9c", "5b7a0c207fab2722a2081caf"];
let current_id = ["5b7b5498ab3f510e307e7d04", "5b83e3b4412f370bd7b9a05d"];

console.log(breakdown(previous_id, current_id))

如果您对多个循环的反对与代码膨胀有关，那么这可能会有所帮助。如果您认为多个循环是应用程序中的性能瓶颈（您已经测试过了吧？），那么这根本无济于事，因为它会针对所需的每个输出单独循环，并且在数组之间的转换上还有其他工作和集。但是如果这些引起应用程序的严重瓶颈，我会感到非常惊讶。

比较两个列表以查找新元素，已删除元素和现有元素

2 个答案: