给定一个数组，如 [[el1,el2],[el11,el22],[el111,el222]]

Question

当我有这样的数组时var test = [['11 may 2018',0],['11 may 2018',1],['12 may 2018',5],['13 may 2018',0],['14 may 2018',0],['15 may 2018',3],['15 may 2018',7],['16 may 2018',30]];我希望它最终看起来像这样var test2 = [['11 may 2018',1],['12 may 2018',5],['13 may 2018',0],['14 may 2018',0],['15 may 2018',10],['16 may 2018',30]]我不知道该怎么做。我做了类似的事情：    var test = [['11 may 2018',0],['11 may 2018',1],['12 may 2018',5],['13 may 2018',0],['14 may 2018',0],['15 may 2018',3],['15 may 2018',7],['16 may 2018',30]];var test2 = [['11 may 2018',1],['12 may 2018',5],['13 may 2018',0],['14 may 2018',0],['15 may 2018',10],['16 may 2018',30]]var testLength = test.length;for (let i = 0; i< testLength; i++){    if(i != testLength -1){        if(test[i][0] == test[i+1][0]){            test[i][1] += test[i+1][1];        }    }}console.log(test)哪个返回    [  [ '11 may 2018', 1 ],  [ '11 may 2018', 1 ],  [ '12 may 2018', 5 ],  [ '13 may 2018', 0 ],  [ '14 may 2018', 0 ],  [ '15 may 2018', 10 ],  [ '15 may 2018', 7 ],  [ '16 may 2018', 30 ]]我不知道如何在不弄乱索引的情况下删除重复数组的第二个实例

Answer 1

只需跟踪键入日期的对象中的日期，并在迭代列表时递增。最后Object.entries，对象将是您想要的：

var test = [['11 may 2018',0],['11 may 2018',1],['12 may 2018',5],['13 may 2018',0],['14 may 2018',0],['15 may 2018',3],['15 may 2018',7],['16 may 2018',30]];

let counts =  test.reduce((sums, [key, count]) => {

    sums[key] = (sums[key] || 0) + count

    return sums

}, {})

console.log(Object.entries(counts))

Answer 2

您可以使用一个对象来对相同的键进行分组，并从该对象中获取值作为结果。

var array = [['11 may 2018', 0], ['11 may 2018', 1], ['12 may 2018', 5], ['13 may 2018', 0], ['14 may 2018', 0], ['15 may 2018', 3], ['15 may 2018', 7], ['16 may 2018', 30]],

    result = Object.values(array.reduce((r, [key, value]) => {

        r[key] = r[key] || [key, 0];

        r[key][1] += value;

        return r;

    }, {}));

console.log(result);

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Answer 3

您实际上需要一个Map结构来解决此类问题。你可以这样做：

const buildMap = arr => {

    const map = {};

    arr.forEach(element => {

        const [key, value] = element;

        if (map[key]) map[key] += value;

        else {

            map[key] = value;

        }

    });

    return map;

};