英文:
How can I efficiently retrieve all possible values of the 'subjects' property from an object in JavaScript?
问题
我有一个具有以下结构的对象：
[{"event_id": 1,"subjects": [{"id": 12,"name": "化学"},{"id": 13,"name": "物理"},{"id": 14,"name": "心理学"},{"id": 16,"name": "历史"}]},{"event_id": 2,"subjects": [{"id": 11,"name": "数学"},{"id": 12,"name": "化学"},{"id": 14,"name": "生物学"},{"id": 15,"name": "地理"}]},{"event_id": 3,"subjects": [{"id": 14,"name": "生物学"},{"id": 15,"name": "地理"},{"id": 16,"name": "历史"}]}]
在这个选项中，获取subjects属性的所有可能值的最快方式是循环遍历每个子对象，并将其值推送到“计数数组”中，如果它尚未存在。
英文:
I had an object with this structure:
[{"event_id": 1,"subjects": [{"id": 12,"name": "Chemistry"},{"id": 13,"name": "Physics"},{"id": 14,"name": "Psychology"},{"id": 16,"name": "History"}]},{"event_id": 2,"subjects": [{"id": 11,"name": "Maths"},{"id": 12,"name": "Chemistry"},{"id": 14,"name": "Biology"},{"id": 15,"name": "Geography"}]},{"event_id": 3,"subjects": [{"id": 14,"name": "Biology"},{"id": 15,"name": "Geography"},{"id": 16,"name": "History"}]}]
What will be the fastest way to get all possible values of subjects prop in this option? Is this only possible to achieve by looping through each single sub-object and push value to "counting array" if it not already there?
答案1
得分: 1
The simplest and most readable way would indeed be to loop through the subjects and add them to the array if they don't exist.
const data = [{"event_id": 1,"subjects": [{"id": 12,"name": "Chemistry"},{"id": 13,"name": "Physics"},{"id": 14,"name": "Psychology"},{"id": 16,"name": "History"}]},{"event_id": 2,"subjects": [{"id": 11,"name": "Maths"},{"id": 12,"name": "Chemistry"},{"id": 14,"name": "Biology"},{"id": 15,"name": "Geography"}]},{"event_id": 3,"subjects": [{"id": 14,"name": "Biology"},{"id": 15,"name": "Geography"},{"id": 16,"name": "History"}]}]const all = []for (const { subjects } of data) {subjects.forEach((s) => {if (all.indexOf(s.name) === -1) {all.push(s.name)}})}console.log(all)
Another useful way is to use a Set with spread syntax to deduplicate the values in the array.
const data = [{"event_id": 1,"subjects": [{"id": 12,"name": "Chemistry"},{"id": 13,"name": "Physics"},{"id": 14,"name": "Psychology"},{"id": 16,"name": "History"}]},{"event_id": 2,"subjects": [{"id": 11,"name": "Maths"},{"id": 12,"name": "Chemistry"},{"id": 14,"name": "Biology"},{"id": 15,"name": "Geography"}]},{"event_id": 3,"subjects": [{"id": 14,"name": "Biology"},{"id": 15,"name": "Geography"},{"id": 16,"name": "History"}]}]const all = []for (const { subjects } of data) {subjects.forEach((s) => {all.push(s.name)})}const s = [...new Set(all)]console.log(s)
英文:
The simplest and most readable way would indeed be to loop through the subjects and add them to the array if they don't exist.
<!-- begin snippet: js hide: false console: true babel: false -->
<!-- language: lang-js -->
const data = [{"event_id": 1,"subjects": [{"id": 12,"name": "Chemistry"},{"id": 13,"name": "Physics"},{"id": 14,"name": "Psychology"},{"id": 16,"name": "History"}]},{"event_id": 2,"subjects": [{"id": 11,"name": "Maths"},{"id": 12,"name": "Chemistry"},{"id": 14,"name": "Biology"},{"id": 15,"name": "Geography"}]},{"event_id": 3,"subjects": [{"id": 14,"name": "Biology"},{"id": 15,"name": "Geography"},{"id": 16,"name": "History"}]}]const all = []for (const { subjects } of data) {subjects.forEach((s) => {if (all.indexOf(s.name) === -1) {all.push(s.name)}})}console.log(all)
<!-- end snippet -->
Another useful way is to use a Set with spread syntax to deduplicate the values in the array.
<!-- begin snippet: js hide: false console: true babel: false -->
<!-- language: lang-js -->
const data = [{"event_id": 1,"subjects": [{"id": 12,"name": "Chemistry"},{"id": 13,"name": "Physics"},{"id": 14,"name": "Psychology"},{"id": 16,"name": "History"}]},{"event_id": 2,"subjects": [{"id": 11,"name": "Maths"},{"id": 12,"name": "Chemistry"},{"id": 14,"name": "Biology"},{"id": 15,"name": "Geography"}]},{"event_id": 3,"subjects": [{"id": 14,"name": "Biology"},{"id": 15,"name": "Geography"},{"id": 16,"name": "History"}]}]const all = []for (const { subjects } of data) {subjects.forEach((s) => {all.push(s.name)})}const s = [...new Set(all)]console.log(s)
<!-- end snippet -->
答案2
得分: 0
以下是代码的翻译部分：
data.reduce((accumulator, currentValue) => {accumulator.push(currentValue.subjects.map(x => x.name));return [...new Set(accumulator.flatMap(x => x))];}, [])// 输出结果// ['化学', '物理', '心理学', '历史', '数学', '生物学', '地理学']
const data = [{"event_id": 1,"subjects": [{"id": 12,"name": "化学"},{"id": 13,"name": "物理"},{"id": 14,"name": "心理学"},{"id": 16,"name": "历史"}]},{"event_id": 2,"subjects": [{"id": 11,"name": "数学"},{"id": 12,"name": "化学"},{"id": 14,"name": "生物学"},{"id": 15,"name": "地理学"}]},{"event_id": 3,"subjects": [{"id": 14,"name": "生物学"},{"id": 15,"name": "地理学"},{"id": 16,"name": "历史"}]}];console.log(data.reduce((accumulator, currentValue) => {accumulator.push(currentValue.subjects.map(x => x.name));return [...new Set(accumulator.flatMap(x => x))];}, []));
希望这能帮助您理解代码和结果。
英文:
just because there's several ways to do the same things, here's one with reduce
data.reduce((accumulator, currentValue) => {accumulator.push(currentValue.subjects.map(x => x.name));return [...new Set(accumulator.flatMap(x => x))];}, [])// will output// ['Chemistry', 'Physics', 'Psychology', 'History', 'Maths', 'Biology', 'Geography']
<!-- begin snippet: js hide: false console: true babel: false -->
<!-- language: lang-js -->
const data = [{"event_id": 1,"subjects": [{"id": 12,"name": "Chemistry"},{"id": 13,"name": "Physics"},{"id": 14,"name": "Psychology"},{"id": 16,"name": "History"}]},{"event_id": 2,"subjects": [{"id": 11,"name": "Maths"},{"id": 12,"name": "Chemistry"},{"id": 14,"name": "Biology"},{"id": 15,"name": "Geography"}]},{"event_id": 3,"subjects": [{"id": 14,"name": "Biology"},{"id": 15,"name": "Geography"},{"id": 16,"name": "History"}]}];console.log(data.reduce((accumulator, currentValue) => {accumulator.push(currentValue.subjects.map(x=>x.name));return [...new Set(accumulator.flatMap(x=>x))];}, []));
<!-- end snippet -->
答案3
得分: 0
const data = [{"event_id": 1,"subjects": [{"id": 12,"name": "化学"},{"id": 13,"name": "物理"},{"id": 14,"name": "心理学"},{"id": 16,"name": "历史"}]},{"event_id": 2,"subjects": [{"id": 11,"name": "数学"},{"id": 12,"name": "化学"},{"id": 14,"name": "生物学"},{"id": 15,"name": "地理"}]},{"event_id": 3,"subjects": [{"id": 14,"name": "生物学"},{"id": 15,"name": "地理"},{"id": 16,"name": "历史"}]}];const allSubjects = data.map(x => x.subjects).reduce((a, x) => [...a, ...x], []);const uniqueSubjects = allSubjects.filter((e, i) => allSubjects.findIndex(a => a.id === e.id) === i);console.log(allSubjects, uniqueSubjects);
英文:
const data = [{"event_id": 1,"subjects": [{"id": 12,"name": "Chemistry"},{"id": 13,"name": "Physics"},{"id": 14,"name": "Psychology"},{"id": 16,"name": "History"}]},{"event_id": 2,"subjects": [{"id": 11,"name": "Maths"},{"id": 12,"name": "Chemistry"},{"id": 14,"name": "Biology"},{"id": 15,"name": "Geography"}]},{"event_id": 3,"subjects": [{"id": 14,"name": "Biology"},{"id": 15,"name": "Geography"},{"id": 16,"name": "History"}]}];const allSubjects = data.map(x=> x.subjects).reduce((a,x) => [...a, ...x] ,[]);const uniqueSubjects = allSubjects.filter((e,i) => allSubjects.findIndex(a => a.id == e.id) == i);console.log(allSubjects, uniqueSubjects);
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