1

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人：有朋友和没有朋友；此外，我希望提升的谓词，Array.filter如下所示：

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这样的by函数来实现此行为：

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，我需要显式地写一个全新的函数peopleWithoutFriends

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

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

我不想把我的by函数写两次。我宁愿将较小的功能组合在一起。

问题：

如何在. _ _flow Booleanget curry notwithFriendswithOutFriendspeople

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);
4

1 回答 1

2

由于有/无朋友函数的结果是布尔值，因此您可以否定（或补充）一个结果以获得另一个。此外，函数的数量为 1（它们操作的对象）。

罗达什/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>

笔记：

  1. 两者都_.flow()执行R.pipe从左到右（从上到下）的序列。功能_.compose()和R.compose顺序是相反的。
  2. 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）/。
  3. Ramda 和 Lodash 都有一个拒绝方法，它与过滤器相反，如果谓词返回true，则删除该项目。例如，R.reject(foo, xs)等价于R.filter(R.complement(foo), xs)。（@ScottSauyet 在此评论中指出）
于 2019-05-18T19:18:29.487 回答