功能组合与休息操作员，减速器和映射器 [英] function composition with rest operator, reducer and mapper

问题描述

我正在关于 JavaScript中的传感器以及特别是我定义了以下函数

I'm following an article about Transducers in JavaScript, and in particular I have defined the following functions

const reducer = (acc, val) => acc.concat([val]);
const reduceWith = (reducer, seed, iterable) => {
  let accumulation = seed;

  for (const value of iterable) {
    accumulation = reducer(accumulation, value);
  }

  return accumulation;
}
const map =
  fn =>
    reducer =>
      (acc, val) => reducer(acc, fn(val));
const sumOf = (acc, val) => acc + val;
const power =
  (base, exponent) => Math.pow(base, exponent);
const squares = map(x => power(x, 2));
const one2ten = [1, 2, 3, 4, 5, 6, 7, 8, 9, 10];
res1 = reduceWith(squares(sumOf), 0, one2ten);
const divtwo = map(x => x / 2);

现在我想定义一个合成运算符

Now I want to define a composition operator

const more = (f, g) => (...args) => f(g(...args));

，我看到它在以下情况下工作

and I see that it is working in the following cases

res2 = reduceWith(more(squares,divtwo)(sumOf), 0, one2ten);
res3 = reduceWith(more(divtwo,squares)(sumOf), 0, one2ten);

相当于

res2 = reduceWith(squares(divtwo(sumOf)), 0, one2ten);
res3 = reduceWith(divtwo(squares(sumOf)), 0, one2ten);

整个脚本是在线。

我不明白为什么我不能连接最后一个函数（ sumOf ）与组合运算符（更多）。理想情况下，我想写

I don't understand why I can't concatenate also the last function (sumOf) with the composition operator (more). Ideally I'd like to write

res2 = reduceWith(more(squares,divtwo,sumOf), 0, one2ten);
res3 = reduceWith(more(divtwo,squares,sumOf), 0, one2ten);

但不行。

修改

很明显，我的初始尝试是错误的，但即使我将组合定义为

It is clear that my initial attempt was wrong, but even if I define the composition as

const compose = (...fns) => x => fns.reduceRight((v, fn) => fn(v), x);

我仍然无法替换 compose（divtwo，squares）（sumOf） 与 compose（divtwo，squares，sumOf）

推荐答案

最后，我找到了一种方法来实现似乎工作正常的组合

Finally I've found a way to implement the composition that seems to work fine

const more = (f, ...g) => {
  if (g.length === 0) return f;
  if (g.length === 1) return f(g[0]);
  return f(more(...g));
}

更好的解决方案

这里是另一个解决方案，使用reducer并且没有递归

Better solution

Here it is another solution with a reducer and no recursion

const compose = (...fns) => (...x) => fns.reduceRight((v, fn) => fn(v), ...x);
const more = (...args) => compose(...args)();

使用：

res2 = reduceWith(more(squares,divtwo,sumOf), 0, one2ten);
res3 = reduceWith(more(divtwo,squares,sumOf), 0, one2ten);

完整脚本在线

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