如何使用功能编程

Question

过滤到多条路径，我有以下伪代码：

let array = getData(); 
array.filter(x => condition1(x)).doSomething1... 
array.filter(x => condition2(x)).doSomething2... 
array.filter(x => condition3(x)).doSomething3... 

显然，这不是有效的，因为它可以迭代阵列3次。

我在想，如果我有办法做这样的事情：这样的数组被遍历一次

array.filterMany([ 
    x => condition1(x).doSomething1..., 
    x => condition2(x).doSomething2..., 
    x => condition3(x).doSomething3... 
]) 

？

Answer 1

您可以使用数组缩减功能。

例如：

const arr = [1, 2, 3, 4, 5, 6, 7, 8, 9, 10]; 
 
    
 
    const split = arr.reduce(([odd, even], current) => { 
 
     if (current % 2 === 0) { 
 
      even.push(current); 
 
     } else { 
 
      odd.push(current); 
 
     } 
 
    
 
     return [odd, even]; 
 
    }, [[], []]); 
 
    
 
    console.log('Odd/Even', split);

Answer 2

怎么这样呢？

const condition1 = x => x === 1; 
 
const condition2 = x => x === 2; 
 
const condition3 = x => x === 3; 
 

 
[1, 2, 3].map(n => { 
 
    condition1(n) && console.log('foo'); 
 
    condition2(n) && console.log('bar'); 
 
    condition3(n) && console.log('baz'); 
 
})

Answer 3

你可以采取的条件到一个数组中，并核对与Array#every。

var conditions = [condition1, condition2, condition3], 
    filtered = array.filter(a => conditions.every(c => c(a)));