映射數組並根據對象鍵返回一個新數組

Question

這是原始數組：

let list = [
    {
        city: "new york", 
        current_time: "123", 
        time1: "456", 
        time2: "789",
    },
    {
        city: "london",
        current_time: "123",
        time1: "456",
        time2: "789",
    },
    {
        city: "tokyo",
        current_time: "123",
        time1: "456",
        time2: "789",
    }
]

我正在嘗試創建一個數組數組，其中每個內部數組包含帶有城市和時間的對象。 但是按current_time ， time1或time2進行組織 。

預期結果：

result = [
    [{
        city: "new york",
        time: "123"
    }, {
        city: "london",
        time: "123"
    }, {
        city: "tokyo",
        time: "123"
    }],
    [{
        city: "new york",
        time: "456"
    }, {
        city: "london",
        time: "456"
    }, {
        city: "tokyo",
        time: "456"
    }], [{
        city: "new york",
        time: "789"
    }, {
        city: "london",
        time: "789"
    }, {
        city: "tokyo",
        time: "789"
    }]
]

我嘗試使用map函數，但是我只能使用current_time創建數組，我可能需要迭代鍵，但是如果我需要使用forEach或更好的方法進行迭代，我會感到困惑。

result = list.map((element, index) => {
    if(element.current_time) {
        return { city: element.city, time: element.current_time };
});

Answer 1

您可以使用一個數組作為鍵，並映射給定數組的映射值的結果。

 var list = [{ city: "new york", current_time: "123", time1: "456", time2: "789" }, { city: "london", current_time: "123", time1: "456", time2: "789" }, { city: "tokyo", current_time: "123", time1: "456", time2: "789" }], keys = ["current_time", "time1", "time2"], result = keys.map(k => list.map(o => ({ city: o.city, time: o[k] }))); console.log(result); 

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另一種方法可能是迭代數據並通過迭代密鑰來簡化數據，以使用密鑰索引對對象進行賦值。

 var list = [{ city: "new york", current_time: "123", time1: "456", time2: "789" }, { city: "london", current_time: "123", time1: "456", time2: "789" }, { city: "tokyo", current_time: "123", time1: "456", time2: "789" }], keys = ["current_time", "time1", "time2"], result = list.reduce((r, o) => { keys.forEach((k, i) => (r[i] = r[i] || []).push({ city: o.city, time: o[k] })); return r; }, []); console.log(result); 

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略有更改的數據版本，其中包含一些缺少的鍵。

 var list = [{ city: "new york", current_time: "123", time1: "456", time2: "789" }, { city: "london", time1: "456", time2: "789" }, { city: "tokyo", current_time: "123", time1: "456" }], keys = ["current_time", "time1", "time2"], result = list.reduce((r, o) => { keys.forEach((k, i) => k in o && (r[i] = r[i] || []).push({ city: o.city, time: o[k] })); return r; }, []); console.log(result); 

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Answer 2

我希望這可以幫助你

 let list = [{ city: "new york", current_time: "123", time1: "456", time2: "789" }, { city: "london", current_time: "123", time1: "456", time2: "789" }, { city: "tokyo", current_time: "123", time1: "456", time2: "789" }, { city: "tokyo", current_time: "1223", time1: "456", time2: "789" }, { city: "tokyo", current_time: "1223", time1: "456", time2: "789" }], obj = {}; list.forEach(function(itm) { if (!obj.hasOwnProperty(itm.current_time)) { obj[itm.current_time] = []; } obj[itm.current_time].push(itm); }); let result = []; var keys = Object.keys(obj); for (var i = 0; i < keys.length; i++) { result.push(obj[keys[i]]); } console.log(result); 

Answer 3

函數map會創建一個與原始數組具有相同長度的新數組，因此，您需要做的是對該數組進行循環並將這些值分組。

一種替代方法是使用函數reduce

 let list = [{ city: "new york", current_time: "123", time1: "456", time2: "789"}, { city: "london", current_time: "123", time1: "456", time2: "789"},{ city: "tokyo", current_time: "123", time1: "456", time2: "789"}], keys = ['current_time', 'time1', 'time2'], result = Object.values(list.reduce((a, c) => { keys.forEach(k => (a[c[k]] || (a[c[k]] = [])).push({ city: c.city, time: c[k] })); return a; }, {})); console.log(result); 

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Answer 4

 var list = [{ city: "new york", current_time: "123", time1: "456", time2: "789", }, { city: "london", current_time: "123", time1: "456", time2: "789", }, { city: "tokyo", current_time: "123", time1: "456", time2: "789", } ]; let timeKeys = Object.keys(list[0]); timeKeys.shift(); let finalArray = timeKeys.map(val => list.map(nestedVal => ({ city: nestedVal.city, time: nestedVal[val] }))); console.log(finalArray); 

Answer 5

我采用了這種方法來定義一個新的空數組，並將每個元素重置為空。 通過遍歷不是城市的每個元素屬性的循環並推送具有屬性city和time的新對象來填充此數組。

希望對您有所幫助。

 var list = [ { city: "new york", current_time: "123", time1: "456", time2: "789", }, { city: "london", current_time: "123", time1: "456", time2: "789", }, { city: "tokyo", current_time: "123", time1: "456", time2: "789", }]; var arr = [] var result = list.map((ele) => { arr = [] for(var prop in ele){ if(prop != 'city') arr.push({city: ele['city'], time: ele[prop]}) } return arr }) console.log('result =', result)