匹配两个javascript数组以形成交集结果

Question

找到交集的速记和最佳方式是什么？

f = ["A","B","C","D","E","F"]; //might be less than 8 
b = [1,0,0,1,0,0,0,0]; //always 8 elements 

期望的结果数组[ “A”， “d”]

Answer 1

你可以使用Array#filter

var f = ["A", "B", "C", "D", "E", "F"], 
 
    b = [1, 0, 0, 1, 0, 0, 0, 0], 
 
    r = f.filter((_, i) => b[i]); 
 

 
console.log(r);

Answer 2

假设你f阵列比你b阵列

f.filter((item, index) => b[index] === 1); 

如果你想这永远不会再完全速记你可以重命名item和index并删除=== 1：

f.filter((a, i) => b[i]); 

---

var f = ["A","B","C","D","E","F"]; //might be less than 8 
 
var b = [1,0,0,1,0,0,0,0]; //always 8 elements 
 
console.log(f.filter((a, i) => b[i]));

Answer 3

var f = ["A","B","C","D","E","F"]; 
 
var b = [1,0,0,1,0,0,0,0]; 
 

 
var res = f.filter(function(e, i) { 
 
    return b[i]; // short for return b[i] === 1; 
 
}); 
 

 
console.log(res);

甚至使用箭头功能这样短：

var f = ["A","B","C","D","E","F"]; 
 
var b = [1,0,0,1,0,0,0,0]; 
 

 
var res = f.filter((e, i) => b[i]); 
 

 
console.log(res);

Answer 4

另一种方式：

$(function(){ 
 
    f = ["A","B","C","D","E","F"]; 
 
    b = [1,0,0,1,0,0,0,0]; 
 
    x = []; 
 
    $.each(b,function(key, value){ 
 
    value?x.push(f[key]):''; 
 
    }); 
 
    console.log(x) 
 
});

Answer 5

由于for循环比过滤方法更快我的建议是：

var results = []; 
for(var i=0;i<b.length;i++){ 
    if (b[i]) results.push(f[i]); 
}