英文:
Addition of values inside an Array according to certain value
问题
我有一个类似这样的数组：
let array = [[0, 1, 4.75], [0, 1, 2.12], [0, 3, 8.1]];
期望的输出是：
let expectedOutput = [[0, 1, 6.87], [0, 3, 8.1]];
在这种情况下，4.75 + 2.12 已经被相加，因为前两个值匹配 [0, 1]。
我想查找子数组中的第一个和第二个值，并将具有相同第一个和第二个值的所有子数组的第三个值相加。你能帮助我吗？
英文:
I have an array like this:
let array = [[0, 1, 4.75], [0, 1, 2.12], [0, 3, 8.1]];
Expected output:
let expectedOutput = [[0, 1, 6.87], [0, 3, 8.1]];
In this case 4.75 + 2.12 has been summed up because first two values were matching [0, 1].
I want to lookup the first and second value in the sub-array und sum the third value of all the sub-arrays that has the same first and second value. Can you please help me out?
答案1
得分: 1
const array = [[0, 1, 4.75],[0, 1, 2.12],[0, 3, 8.1]];// 根据第一和第二项对数组进行分组const groups = {}for (const [a1, a2, a3] of array) {if (!(a1 in groups)) {groups[a1] = {}}if (!(a2 in groups[a1])) {groups[a1][a2] = []}groups[a1][a2].push(a3)}// 计算第三项的总和const result = []for (const a1 in groups) {for (const a2 in groups[a1]) {const sum = groups[a1][a2].reduce((s, c) => s + c, 0)result.push([+a1, +a2, sum])}}console.log(result)
英文:
<!-- begin snippet: js hide: false console: true babel: false -->
<!-- language: lang-js -->
const array = [[0, 1, 4.75],[0, 1, 2.12],[0, 3, 8.1]];// groups arrays by first and second itemconst groups = {}for (const [a1, a2, a3] of array) {if (!(a1 in groups)) {groups[a1] = {}}if (!(a2 in groups[a1])) {groups[a1][a2] = []}groups[a1][a2].push(a3)}// sum the third itemconst result = []for (const a1 in groups) {for (const a2 in groups[a1]) {const sum = groups[a1][a2].reduce((s, c) => s + c, 0)result.push([+a1, +a2, sum])}}console.log(result)
<!-- end snippet -->
答案2
得分: 0
构建结果时，请检查当前元素是否与先前的元素匹配，然后要么仅添加该值，要么添加整个元素：
let array = [[0, 1, 4.75], [0, 1, 2.12], [0, 3, 8.1]];const result = array.reduce((r, [x, y, v], i) => {i = r.find(e => e[0] === x && e[1] === y);if (i) {i[2] += v;} else {r.push([x, y, v]);}return r;}, []);console.log(result);
英文:
When building the result, check if the current element matches a previous one, and then either add only the value or the whole element:
<!-- begin snippet: js hide: false console: true babel: false -->
<!-- language: lang-js -->
let array = [[0, 1, 4.75], [0, 1, 2.12], [0, 3, 8.1]];const result = array.reduce( (r, [x,y,v], i) => (i = r.find(e => e[0] === x && e[1] === y), i ? i[2] += v : r.push([x,y,v]), r), [])console.log(result)
<!-- end snippet -->
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