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JavaScript - 基于 3 個數(shù)組形成數(shù)組的函數(shù)（合并）

JavaScript - 基于 3 個數(shù)組形成數(shù)組的函數(shù)（合并）

慕碼人8056858 2021-06-30 15:54:09
給定以下 js 數(shù)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(shù)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(shù)組的數(shù)組。price_type 可以是：prices_A、prices_B、prices_C 或prices_all。當(dāng)prices_all 被引入時，它應(yīng)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(jié)果：prices =[['company','200g'],['business_e','50€']]示例 2：var prices = myfunction(100,prices_all){}結(jié)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€'],]
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翻閱古今

TA貢獻(xiàn)1780條經(jīng)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(shù)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; }


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反對 回復(fù) 2021-07-08
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