數組'map'與'forEach'-函數式編程

Question

我有一個對象數組：

let reports = [{ inbound_calls: [...], outbound_calls: [...],  outbound_national_calls: [...] },...];

創建新數組並將其分配給變量的最佳方法是什么：

第一種方法-一個循環：

let inbound_calls = []; outbound_national_calls = [], outbound_calls = [];

reports.forEach((e) => {
 inbound_calls.push(e.inbound_calls);
 outbound_national_calls.push(e.outbound_national_calls);
 outbound_calls.push(e.outbound_calls);
})

第二種方法：

let inbound_calls = this.reports.map((report) => report.inbound_calls)
let outbound_national_calls = this.reports.map((report) => report.outbound_national_calls)
let outbound_calls = this.reports.map((report) => report.outbound_calls)

我開始學習函數式編程，並想將其應用到我的代碼中，我會采用第一種方法（一個循環），但是當我對函數式編程進行研究時，我認為第二種方法是正確的方法（更簡潔）但是，我不確定什么是更便宜的手術？

Answer 1

如果最終目標是在對象之外創建三個變量，則可以按以下方式使用對象分解。 無需循環。

 let reports = { inbound_calls: [1, 2, 3], outbound_calls: [4, 5, 6], outbound_national_calls: [7, 8, 9] }; let {inbound_calls, outbound_calls, outbound_national_calls} = reports; console.log(inbound_calls); console.log(outbound_calls); console.log(outbound_national_calls); 

Answer 2

如果要復制陣列，只需使用Array#slice （傳遞的0是可選的，因為它是默認的起始索引，因此您可以根據需要省略它），例如：

let inbound_calls = reports.inbound_calls.slice(0),
    outbound_national_calls = reports.outbound_national_calls.slice(0), 
    outbound_calls = reports.outbound_calls.slice(0);

或Array.from一樣：

let inbound_calls = Array.from(reports.inbound_calls),
    outbound_national_calls = Array.from(reports.outbound_national_calls), 
    outbound_calls = Array.from(reports.outbound_calls);

Answer 3

您實際上要做的是矩陣轉置：

 const report = (inbound_calls, outbound_calls, outbound_national_calls) => ({ inbound_calls, outbound_calls, outbound_national_calls }); const reports = [report(1,2,3), report(4,5,6), report(7,8,9)]; const transpose = reports => report( reports.map(report => report.inbound_calls) , reports.map(report => report.outbound_calls) , reports.map(report => report.outbound_national_calls) ); console.log(transpose(reports)); 

現在，根據您的應用程序，轉置矩陣的最快方法可能是根本不轉置矩陣。 例如，假設您有一個矩陣A及其轉置B 然后，對於所有索引i和j ， A[i][j] = B[j][i] 。 考慮：

 const report = (inbound_calls, outbound_calls, outbound_national_calls) => ({ inbound_calls, outbound_calls, outbound_national_calls }); const reports = [report(1,2,3), report(4,5,6), report(7,8,9)]; // This is equivalent to transpose(reports).outbound_calls[1] const result = reports[1].outbound_calls; console.log(result); 

話雖如此，您的第二種方法是恕我直言，最易讀。