javascript - 如何比较一个数组中的元素到另一个数组中

Question

如果 choices 数组中的任何 choice 对象在任何 results 数组中都找不到，则隐藏 choice 本例中的元素 {"choice": "f"} 和 {"choice": "g"} 在这 3 个结果中都不可用，所以我需要隐藏它。我该怎么做？

{"choices":[
    {"choice": "a"},
    {"choice": "b"},
    {"choice": "c"},
    {"choice": "d"},
    {"choice": "e"},
    {"choice": "f"},
    {"choice": "g"}
],
"results":[
    {"result":["a", "b", "c"]},             
    {"result":["a", "b", "d"]},
    {"result":["a", "b", "e"]},

Answer 1

您可以从结果和过滤器选项中收集所有值。

var object = { choices: [{ choice: "a" }, { choice: "b" }, { choice: "c" }, { choice: "d" }, { choice: "e" }, { choice: "f" }, { choice: "g" }], results: [{ result: ["a", "b", "c"] }, { result: ["a", "b", "d"] }, { result: ["a", "b", "e"] }] },
    hash = Object.create(null),
    notIn = [];

object.results.forEach(function (a) {
    a.result.forEach(function (b) {
        hash[b] = true;
    });
});

notIn = object.choices.filter(function (a) {
    return !hash[a.choice];
});

console.log(notIn);

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ES6 与 Set

var object = { choices: [{ choice: "a" }, { choice: "b" }, { choice: "c" }, { choice: "d" }, { choice: "e" }, { choice: "f" }, { choice: "g" }], results: [{ result: ["a", "b", "c"] }, { result: ["a", "b", "d"] }, { result: ["a", "b", "e"] }] },
    results = object.results.reduce((s, a) => new Set([...s, ...a.result]), new Set),
    notIn = object.choices.filter(a => !results.has(a.choice));

console.log(notIn);

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