Ramda.js 函数式编程详解：万物皆可柯里化与 API 实战

var square = n => n * n;
_.map([4, 8], square)

var R = require('ramda');
R.map(square, [4, 8])
// 等同于
var R = require('ramda');
R.map(square)([4, 8])

var R = require('ramda');
const odd = x => x%2 === 1
const data = [3, 5, 6];
R.filter(odd, data); // [3, 5]
// 等同于
R.filter(odd)(data); // [3, 5]
// 也可以延迟调用
const filter1 = R.filter(odd);
// filter1 等待参数的传入
// 后续再传入 data
const filter2 = filter1(data)

const _curry = f => a => b => f(a, b)
const odd = x => x%2 === 1
const _filter = _curry( (fn, arr) => arr.filter(fn) );
_filter(odd)([3,5,6]) // [3, 5]

const double = x => x * 2;
R.map(double, [1, 2, 3]); //=> [2, 4, 6]
R.map(double, {x: 1, y: 2, z: 3}); //=> {x: 2, y: 4, z: 6}

const isEven = n => n % 2 === 0;
R.filter(isEven, [1, 2, 3, 4]); //=> [2, 4]
R.filter(isEven, {a: 1, b: 2, c: 3, d: 4}); //=> {b: 2, d: 4}

R.add(2, 3);       //=>  5
R.add(7)(10);      //=> 17

R.multiply(2)(5)  // 10

R.compose(Math.abs, R.add(1), R.multiply(2))(-4)
// |-4*2 + 1|，等于 7

var negative = x => -1 * x;
var increaseOne = x => x + 1;
var f = R.pipe(Math.pow, negative, increaseOne)(3,4);
// -(3^4) + 1 ，等于 -80

const addFourNumbers = (a, b, c, d) => a + b + c + d;
const curriedAddFourNumbers = R.curry(addFourNumbers);
curriedAddFourNumbers(1, 2)(3)(4)