Question

我有一个矩阵算法：

输入：

const input = [
    ['Camry', 'Toyota', 'Jan', 'Nowhere Town', '50'],
    ['Camry', 'Toyota', 'Feb', 'Nowhere Town', '70'],
    ['Camry', 'Toyota', 'Jan', 'Random City', '3000'],
    ['Prius', 'Toyota', 'Jan', 'Nowhere Town', '60'],
    ['Prius', 'Toyota', 'Jan', 'Random Town', '60'],
    ['Prius', 'Toyota', 'Mar', 'Nowhere Town', '50'],
    ['Civic', 'Honda', 'Jan', 'Nowhere Town', '10'],
    ['Civic', 'Honda', 'Feb', 'Nowhere Town', '10'],
    ['Civic', 'Honda', 'Mar', 'Random Town', '10'],
    ['Civic', 'Honda', 'Mar', 'Random Town', '20'],
]

预期产出：

const output = [
    ['S', 'Camry', 'Toyota', 'Jan', '3050'],
    ['D', 1, 'Camry', 'Nowhere Town', '50'],
    ['D', 2, 'Camry', 'Random City', '3000'],
    ['S', 'Camry', 'Toyota', 'Feb', '70'],
    ['D', 1, 'Camry', 'Nowhere Town', '70'],
    ['S', 'Prius', 'Toyota', 'Jan', '120'],
    ['D', 1, 'Prius', 'Nowhere Town', '60'],
    ['D', 2, 'Prius', 'Random Town', '60'],
    ['S', 'Prius', 'Toyota', 'Mar', '50'],
    ['D', 1, 'Prius', 'Nowhere Town', '50'],
    ['S', 'Civic', 'Honda', 'Jan', '10'],
    ['D', 1, 'Civic', 'Nowhere Town', '10'],
    ['S', 'Civic', 'Honda', 'Feb', '10'],
    ['D', 1, 'Civic', 'Nowhere Town', '10'],
    ['S', 'Civic', 'Honda', 'Mar', '20'],
    ['D', 1, 'Civic', 'Random Town', '10'],
    ['D', 2, 'Civic', 'Random Town', '10'],
]

用语言说：如果行包含相同的品牌，则相同的制作和相同的月会在总销售额的顶部添加摘要行，并为每个行添加列出的订单细节行。

我有一个旧代码：

const groupReport = arr => {
    let grouped = [].concat(...arr.reduce((acc, cur) => {
        var data = acc.get(cur[1]) || [['P', cur[1], '0']]
        data.push(['D', data.length, cur[0], cur[3], cur[4]])
        data[0][2] = (+data[0][2] + +cur[4]).toString()
        return acc.set(cur[0], data);
    }, new Map)
    .values()
    )
    return grouped
}

它不起作用，因为它只比较一个col（品牌），而不是Make和Month。

Answer 1

您可以使用组合键和破坏数组。

var array = [['Camry', 'Toyota', 'Jan', 'Nowhere Town', '50'], ['Camry', 'Toyota', 'Feb', 'Nowhere Town', '70'], ['Camry', 'Toyota', 'Jan', 'Random City', '3000'], ['Prius', 'Toyota', 'Jan', 'Nowhere Town', '60'], ['Prius', 'Toyota', 'Jan', 'Random Town', '60'], ['Prius', 'Toyota', 'Mar', 'Nowhere Town', '50'], ['Civic', 'Honda', 'Jan', 'Nowhere Town', '10'], ['Civic', 'Honda', 'Feb', 'Nowhere Town', '10'], ['Civic', 'Honda', 'Mar', 'Random Town', '10'], ['Civic', 'Honda', 'Mar', 'Random Town', '20']],
    result = [].concat(...array
        .reduce((m, [brand, make, month, town, amount]) => {
            var key = [brand, make, month].join('|'),
                data = m.get(key) || [['S', brand, make, month, '0']];

            data.push(['D', data.length, brand, town, amount]);
            data[0][4] = (+data[0][4] + +amount).toString();
            return m.set(key, data);
        }, new Map)
        .values()
    );

console.log(result.map(a => a.join(' ')));

.as-console-wrapper { max-height: 100% !important; top: 0; }

Answer 2

您可以使用reduce按品牌，制作和月份创建对象，然后使用扩展语法和Object.values来获取数组。

const input = [['Camry', 'Toyota', 'Jan', 'Nowhere Town', '50'],['Camry', 'Toyota', 'Feb', 'Nowhere Town', '70'],['Camry', 'Toyota', 'Jan', 'Random City', '3000'],['Prius', 'Toyota', 'Jan', 'Nowhere Town', '60'],['Prius', 'Toyota', 'Jan', 'Random Town', '60'],['Prius', 'Toyota', 'Mar', 'Nowhere Town', '50'],['Civic', 'Honda', 'Jan', 'Nowhere Town', '10'],['Civic', 'Honda', 'Feb', 'Nowhere Town', '10'],['Civic', 'Honda', 'Mar', 'Random Town', '10'],['Civic', 'Honda', 'Mar', 'Random Town', '20'],]

const obj = input.reduce((r, [brand, make, month, city, value]) => {
  let key = `${brand}|${make}|${month}`;
  if(!r[key]) r[key] = [
    ['S', brand,  make, month, +value],
    ["D", 1, make, city, value]
  ]
  else {
    r[key][0][4] += +value;
    let prev = r[key].slice(-1)[0]
    r[key].push(["D",  prev[1] + 1, make, city, value]);
  }
  return r;
}, {})

const result = [].concat(...Object.values(obj))

console.log(result)

Answer 3

以下中断分两步完成输入。它大量使用es6功能Map和扩展运算符。

第一步是通过生成键值对，然后将每个项目添加到由生成的键键入的项目数组的Map中，将项目转换为组。

第二步采用每个组的值来创建标题，然后创建一个新标头，标题作为第一个项目。然后将这些组合并为一个数组。

＆＃13;

const input = [['Camry', 'Toyota', 'Jan', 'Nowhere Town', '50'], ['Camry', 'Toyota', 'Feb', 'Nowhere Town', '70'], ['Camry', 'Toyota', 'Jan', 'Random City', '3000'], ['Prius', 'Toyota', 'Jan', 'Nowhere Town', '60'], ['Prius', 'Toyota', 'Jan', 'Random Town', '60'],    ['Prius', 'Toyota', 'Mar', 'Nowhere Town', '50'],    ['Civic', 'Honda', 'Jan', 'Nowhere Town', '10'], ['Civic', 'Honda', 'Feb', 'Nowhere Town', '10'], ['Civic', 'Honda', 'Mar', 'Random Town', '10'], ['Civic', 'Honda', 'Mar', 'Random Town', '20']]

const group = new Map();
input.map(x => [ x.slice(0,3).join('::'), ['D', ...x.slice(0,4), +x[4]] ])
  .forEach(([key, val]) => (group.get(key) || group.set(key, []).get(key)).push(val));

const result = [].concat(... Array.from(group.values())
  .map(x => [['S', ...x[0].slice(1,4), x.reduce((a, c) => a + c[5], 0)], 
  	...x.map(y => [...y.slice(0, 3), ... y.slice(4,6)])]));

console.log(result.map(x => x.toString()));

＆＃13;

为cols添加新的摘要行共享2个不同cols

3 个答案: