如果重复，如何比较2D数组中的相邻元素?

Question

我有一个2D数组:

var arr1 = [["Egypt","Grid",50],["Egypt","Grid",10],["Nigeria","Grid",20],["Ghana","Grid",60],["Egypt","Grid",30]]

我想检查每个数组的第一个元素，例如埃及、尼日利亚和加纳，看看它们是否重复。但我只想让它检查下一个元素而不是数组中的所有元素。如果一个或多个元素相邻重复，我希望返回一个数组，将位置2(50,10,20,60,30)的元素相加

结果应该是:

var arr2 = [["Egypt","Grid",60],["Nigeria","Grid",20],["Ghana","Grid",60]],["Egypt","Grid",30]]

请注意，尽管有3个带有"Egypt"的数组，它只在它们相邻的地方组合。

我已经尝试了map和for loop，但没有任何接近工作的东西。这是我开始的内容:

arr1.map(function (unit, index, array) {
var length = array.length
if (index < length - 2) {
var next = array[index + 1][0]
if (unit[0] === next) {
}
}
})

用于检查元素是否重复。我想我接下来需要把重复到一个单独的阵列，我可以然后reduce，但我不知道如何做到这一点。任何能给我指出正确方向的帮助都会很好。谢谢!

Answer 1

您可以使用一个简单的for循环遍历除最后一个数组外的每个数组，并检查当前迭代元素的第一个元素是否与当前+1的第一个元素匹配。

如果匹配，则对每个数组的第三个元素求和并赋值给当前迭代数组，删除current+1(此处使用splice())并减少i以再次检查相同的索引。

var arr1 = [
['Egypt', 'Grid', 50],
['Egypt', 'Grid', 10],
['Nigeria', 'Grid', 20],
['Ghana', 'Grid', 60],
['Egypt', 'Grid', 30],
];
for (let i = 0; i < arr1.length - 1; i++) {
if (arr1[i][0] === arr1[i + 1][0]) {
arr1[i][2] = arr1[i][2] + arr1[i + 1][2];
arr1.splice(i + 1, 1);
i--;
}
}
console.log(arr1);

Answer 2

你可以创建一个grouping数组，只有当country和category的最后一个元素与grouping数组的最后一个元素不匹配或者grouping中没有元素

可以使用for..ofloop

var arr1 = [
["Egypt", "Grid", 50],
["Egypt", "Grid", 10],
["Nigeria", "Grid", 20],
["Ghana", "Grid", 60],
["Egypt", "Grid", 30],
];
const grouping = [];
for (let [country, category, num] of arr1) {
const last = grouping[grouping.length - 1];
const [lastCountry, lastCategory, lastNum] = last ?? [];
lastCountry && lastCountry === country && lastCategory === category
? (last[2] += num)
: grouping.push([country, category, num]);
}
console.log(grouping);

/* This is not a part of answer. It is just to give the output fill height. So IGNORE IT */
.as-console-wrapper { max-height: 100% !important; top: 0; }

Answer 3

以下是reduce的解决方案。为了简化解构，我选择首先将结果前置到累加器，然后最后使用reverse:

反转该结果。

let arr = [["Egypt","Grid",60],["Nigeria","Grid",20],["Ghana","Grid",60],["Egypt","Grid",30]];
let result = arr.slice(1).reduce(([[x, y, z], ...acc], [a, b, c]) =>
a === x && b == y ? [[a, b, z + c], ...acc]
: [[a, b, c], [x, y, z], ...acc]
, [arr[0]]).reverse();
console.log(result);

Answer 4

var arr1 = [
["Egypt","Grid",50],
["Egypt","Grid",10],
["Egypt","Grid",10],
["Nigeria","Grid",20],
["Nigeria","Grid",20],
["Nigeria","Grid",20],
["Nigeria","Grid",20],
["Nigeria","Grid",20],
["Ghana","Grid",60],
["Egypt","Grid",30]
];
var i=0;
var j=1;
while (j<arr1.length){
if (arr1[i][0] === arr1[j][0]){
arr1[i][2] += arr1[j][2];
arr1.splice(j,1);
continue
}
i++;
j++;
}
console.log(arr1)