如何一起提升和组合功能？

Question

我有这个结构中的人员列表：

const people = [
  {name: 'jenny', friends: ['jeff']},
  {name: 'frank', friends: ['jeff', 'ross']},
  {name: 'sarah', friends: []},
  {name: 'jeff', friends: ['jenny', 'frank']},
  {name: 'russ', friends: []},
  {name: 'calvin', friends: []},
  {name: 'ross', friends: ['frank']},
];

我想通过两种方式过滤掉人：有朋友和没有朋友；而且我想要 Predicate of the Array.filter to be lifted，像这样：

const peopleWithoutFriends = people.filter(withoutFriends);
console.log(peopleWithoutFriends);

const peopleWithFriends = people.filter(withFriends);
console.log(peopleWithFriends);

我可以通过显式编写 by 函数来实现此行为，如下所示：

const by = x => i => {
  return Boolean(get(i, x));
};
const withFriends = by('friends.length');
const peopleWithFriends = people.filter(withFriends);
console.log(peopleWithFriends);

问题： 如果我想要逆函数，我需要明确地为 peopleWithoutFriends

编写一个全新的函数

const notBy = x => i => {
  return !Boolean(get(i, x));
};

const withOutFriends = notBy('friends.length');
const peopleWithoutFriends = people.filter(withOutFriends);

我不想将 by 函数写两次。我宁愿将较小的函数组合在一起。

问题：

如何编写和使用小函数，例如：flow Booleanget curry not 并为我的 Array.filter 列表编写 withFriends 和 withOutFriends 谓词people.

回复：https://repl.it/@matthewharwood/ChiefWelloffPaintprogram

const {flow, get, curry} = require('lodash');

const people = [
  {name: 'jenny', friends: ['jeff']},
  {name: 'frank', friends: ['jeff', 'ross']},
  {name: 'sarah', friends: []},
  {name: 'jeff', friends: ['jenny', 'frank']},
  {name: 'russ', friends: []},
  {name: 'calvin', friends: []},
  {name: 'ross', friends: ['frank']},
];
const not = i => !i;

const withFriends = i => flow(
  Boolean,
  get(i, 'friends.length'), // arity of this is 2 so might be harder to lift, is it possible tho with curry?
); // No idea what i'm doing here.


const peopleWithFriends = people.filter(withFriends);
console.log(peopleWithFriends);

const withoutFriends = flow(not, withFriends);
const peopleWithoutFriends = people.filter(withoutFriends);
console.log(peopleWithoutFriends);

Answer 1

由于 with/without 朋友函数的结果是一个布尔值，您可以取反（或补码）其中一个的结果以获得另一个。此外，函数的元数为 1（它们所操作的对象）。

Lodash/fp:

const { flow, get, isEmpty, negate } = _;

const people = [
  {name: 'jenny', friends: ['jeff']},
  {name: 'frank', friends: ['jeff', 'ross']},
  {name: 'sarah', friends: []},
  {name: 'jeff', friends: ['jenny', 'frank']},
  {name: 'russ', friends: []},
  {name: 'calvin', friends: []},
  {name: 'ross', friends: ['frank']},
];

const withoutFriends = flow(get('friends'), isEmpty); // create a function that gets the friends array, and check if it is empty
const withFriends = negate(withoutFriends); // negate the result of withoutFriends

const peopleWithFriends = people.filter(withFriends);
console.log(peopleWithFriends);

const peopleWithoutFriends = people.filter(withoutFriends);
console.log(peopleWithoutFriends);

<script src='https://cdn.jsdelivr.net/g/lodash@4(lodash.min.js+lodash.fp.min.js)'></script>

拉姆达：

const { pipe, prop, isEmpty, complement } = R;

const people = [
  {name: 'jenny', friends: ['jeff']},
  {name: 'frank', friends: ['jeff', 'ross']},
  {name: 'sarah', friends: []},
  {name: 'jeff', friends: ['jenny', 'frank']},
  {name: 'russ', friends: []},
  {name: 'calvin', friends: []},
  {name: 'ross', friends: ['frank']},
];

const withoutFriends = pipe(prop('friends'), isEmpty); // create a function that gets the friends array, and check if it is empty
const withFriends = complement(withoutFriends); // negate the result of withoutFriends

const peopleWithFriends = people.filter(withFriends);
console.log(peopleWithFriends);

const peopleWithoutFriends = people.filter(withoutFriends);
console.log(peopleWithoutFriends);

<script src="https://cdnjs.cloudflare.com/ajax/libs/ramda/0.26.1/ramda.js"></script>

备注：

_.flow()和R.pipe都从左到右（从上到下）执行序列。函数 _.compose() 和 R.compose 顺序颠倒。
flow/pipe/compose 中的第一个函数获取传递给组合函数的所有内容。序列中的其他函数总是得到一个参数（前一个函数的结果）/.
Ramda和Lodash都有一个reject方法，与filter相反，如果predicatereturnstrue，item被移除。例如，R.reject(foo, xs) 等同于 R.filter(R.complement(foo), xs)。（@ScottSauyet 在此 comment 中指出）

如何一起提升和组合功能？

How to lift and compose functions together?

javascript

functional-programming

lifting

lodash

ramda.js

问题：