数组'map'与'forEach'-函数式编程

Question

我有一个对象数组：

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

创建新数组并将其分配给变量的最佳方法是什么：

第一种方法-一个循环：

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);
})

第二种方法：

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)

我开始学习函数式编程，并想将其应用到我的代码中，我会采用第一种方法（一个循环），但是当我对函数式编程进行研究时，我认为第二种方法是正确的方法（更简洁）但是，我不确定什么是更便宜的手术？

Answer 1

如果最终目标是在对象之外创建三个变量，则可以按以下方式使用对象分解。 无需循环。

 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

如果要复制阵列，只需使用Array#slice （传递的0是可选的，因为它是默认的起始索引，因此您可以根据需要省略它），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一样：

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

您实际上要做的是矩阵转置：

 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)); 

现在，根据您的应用程序，转置矩阵的最快方法可能是根本不转置矩阵。 例如，假设您有一个矩阵A及其转置B 然后，对于所有索引i和j ， A[i][j] = B[j][i] 。 考虑：

 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); 

话虽如此，您的第二种方法是恕我直言，最易读。