无点动态功能组合 [英] Pointfree dynamic function composition

问题描述

我正试图将此功能重构为无点的.

I'm trying to refactor this function to be pointfree.

function siblings(me) {
  return R.pipe(family, R.reject(equalsMe(me)))(me);
}

我想将 me 和 family 返回的值一起传递给函数.

I'd like to pass me to a function down the pipe along with the value that family returns.

使用 R.useWith 或 R.converge 和 R.identity 或 R .__ (甚至不确定我是否应该使用该功能)，但没有任何效果.

Tried a few things with R.useWith or R.converge with R.identity or R.__ (not even sure if I should be using that) but found nothing to work.

推荐答案

如果我正确理解， family 是一个可以接受一个人并返回一个家庭成员(包括该人)列表的函数，例如

If I understand correctly family is a function that takes a person and returns a list of family members (including that person) e.g.

family(2);
//=> [1, 2, 3]

然后您要创建一个函数 siblings ，该函数接受一个人并仅返回其兄弟姐妹，例如

Then you want to create a function siblings which takes a person and returns only their siblings e.g.

siblings(2);
//=> [1, 3]

我个人认为，如果这样编写，您的函数会更好读:

Personally I think your function would read slightly better if it was written this way:

const siblings = me => reject(equals(me), family(me));

siblings(2);
//=> [1, 3]

如果您真的想要它的无点版本，则可以使用 converge ，但是我真的认为这没有什么更好的:

If you really wanted a pointfree version of it, you could use converge but I really don't think it is any better:

const siblings = converge(reject, [unary(equals), family]);

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