JavaScript函数参数翻转——连接多个数组——zip、zipwith

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1.

let f = {};
f.flip = (fn)=>(...args)=>fn(...args.reverse());
var divide = (a,b)=>a / b;
var flip = f.flip(divide);
flip(10, 5)
// 0.5
flip(1, 10)
// 10
var three = (a,b,c)=>[a, b, c];
var flip = f.flip(three);
flip(1, 2, 3);
// => [2, 1, 3]

 2.

let f = {};


f.concat =
(...xs) => xs.reduce((a, b) => a.concat(b));
f.concatMap = (g, ...xs) => f.concat(xs.map(g));

f.concat([5], [27], [3]) // [5, 27, 3]
f.concatMap(x => 'hi ' + x, 1, [[2]], 3) // ['hi 1', 'hi 2', 'hi 3']

3.

let f = {};

f.zip = (...xs)=>{
    let r = [];
    let nple = [];
    let length = Math.min.apply(null, xs.map(x=>x.length));
    for (var i = 0; i < length; i++) {
        xs.forEach(x=>nple.push(x[i]));
        r.push(nple);
        nple = [];
    }
    return r;
}
;
f.zipWith = (op,...xs)=>f.zip.apply(null, xs).map((x)=>x.reduce(op));
let a = [0, 1, ];
let b = [3, 4, 6];
let c = [6, 7, 8];
console.log('--')
console.log(f.zip(a, b))
// [[0, 3], [1, 4], [2, 5]]
console.log(f.zipWith((a,b)=>a + b, a, b, c))

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