首頁猿問 JavaScript - 基于 3...

JavaScript - 基于 3 個數組形成數組的函數（合并）

JavaScript

慕碼人8056858 2021-06-30 15:54:09

給定以下 js 數組：var prices_A =[['company','100g','200g','300g'],['business_a','40€','50€',''],['business_b','50€','61€','42€'],['business_c','66€','','31€']]var prices_B =[['company','100g','200g','300g'],['business_b','40€','50€',''],['business_d','50€','61€','42€'],['business_e','66€','','31€']]var prices_C =[['company','100g','200g','300g'],['business_e','40€','50€',''],['business_b','66€','','31€']]如何使函數傳遞兩個變量：grams 和 price_type，返回一個基于輸入（傳遞的變量）合并數組的數組。price_type 可以是：prices_A、prices_B、prices_C 或prices_all。當prices_all 被引入時，它應該查詢合并所有的價格（prices_A、prices_B 和prices_C）示例 1：var prices = myfunction(200,prices_C){}結果：prices =[['company','200g'],['business_e','50€']]示例 2：var prices = myfunction(100,prices_all){}結果：prices =[['company','100g_A','100g_B','100g_C'],['business_a','40€',''],['business_b','50€','40€','66€'],['business_c','66€',''],['business_d','','50€',''],['business_e','','66€','40€'],]

查看完整描述

1 回答

翻閱古今

TA貢獻1780條經驗獲得超5個贊

這是一個解決方案，它獲取一個組的價格，如果給定或所有組，然后將這些組分配給單個數組。

function getPrices(weight, category) {

function getByGroup(group) {

var index = group[0].indexOf(weight);

return group

.map(a => [0, index].map(i => a[i]))

.filter(a => a[1]);

}

return category in allPrices

? getByGroup(allPrices[category])

: Object.entries(allPrices).reduce((r, [k, group], i) => {

var temp = getByGroup(group),

index;

if (!temp.length) return r;

temp[0][1] += '_' + k.slice(-1);

if (!r) return temp;

index = r[0].push(temp[0][1]) - 1;

temp.forEach(([company, price]) => {

var row = r.find(([c]) => c === company);

if (!row) {

r.push(row = [company]);

}

row[index] = price;

});

r.forEach(a => a[index] = a[index] || '');

return r;

}, undefined);

}

var prices_A = [['company', '100g', '200g', '300g'], ['business_a', '40€', '50€', ''], ['business_b', '50€', '61€', '42€'], ['business_c', '66€', '', '31€']],

prices_B = [['company', '100g', '200g', '300g'], ['business_b', '40€', '50€', ''], ['business_d', '50€', '61€', '42€'], ['business_e', '66€', '', '31€']],

prices_C = [['company', '100g', '200g', '300g'], ['business_e', '40€', '50€', ''], ['business_b', '66€', '', '31€']],

allPrices = { prices_A, prices_B, prices_C };

console.log(getPrices('200g', 'prices_C'));

console.log(getPrices('100g', 'prices_all'));

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

反對回復 2021-07-08

1 回答
0 關注
215 瀏覽

關注

添加回答

舉報

0/150

提交

取消

亚洲在线久爱草,狠狠天天香蕉网,天天搞日日干久草,伊人亚洲日本欧美

熱搜

最近搜索清空

JavaScript - 基于 3 個數組形成數組的函數（合并）

JavaScript - 基于 3 個數組形成數組的函數（合并）

1 回答

添加回答