通过功能链的功能来传递结果

本文介绍了通过功能链的功能来传递结果的处理方法，对大家解决问题具有一定的参考价值，需要的朋友们下面随着小编来一起学习吧！ 问题描述 29岁程序员，3月因学历无情被辞！ 如果我有一个数组 someArray ，我首先要做一些操作，然后将结果传递给函数 arrfun 将一个数组作为参数。像下面这样： let arr = someArray.filter（foo）.map（bar） let result = arrfun（在上面的场景中，我想避免必须分配一个中间变量来传递给中间变量 arrfun 。我想这样的东西。 Object.prototype.pipe = function（f）{return f（this）} result result = someArray.filter（foo）.map（bar）.pipe（arrfun） 代替 .pipe（）您如何解决这个问题？ 将该函数引入 Object ？ 是 pipe >这样的功能的最好的名字？ 链？ 传递？ 新示例 const pc = options =>选项 .join（''） .match（/ - [\w。] * [\w。] * / g） .map（s => s .slice（2）.split（''）） .map（（[key，value]）=>（{key，value}）） .map（nestObject） （（acc，val）=> Object.assign（acc，val），{}） const nestObject =（{key，value}）=> $ { .split（'。'） .reverse（） .reduce（（inner，key）=>（{[key]：inner}），value） 在上面的例子中，一个问题是 .match 返回 null 如果找不到匹配项。使用 .pipe 你可以解决它，只要将该行更改为 。 pipe（s => s.match（/ - [\ w。] * [\w。] * / g）|| []） 如果没有 pipe ？ 在上面的场景中，我想避免必须分配一个中间变量来传递给 arrfun 您是否忽略了这种简单明了的表达方式？ let result = arrfun（someArray.filter（foo）.map（bar）） 从右到左的功能组合 （从右到左）函数组合？ const compose =（f，... fs）=> x => f === undefined？ x：f（compose（... fs）（x））const filter = f => xs => xs.filter（x => f（x））const map = f => xs => xs.map（x => f（x））const foo = x => x> 3 const bar = x => x * x const arrfun = xs => xs.reverse（）const myfunc = compose（arrfun，map（bar），filter（foo））let someArray = [1,2,3,4,5,6] let result = myfunc（someArray）console.log（result ）// [36,25,16] 从左至右的函数组合 使用从左至右的函数组合 const compose =（f，... fs）=> x => f === undefined？ x：compose（... fs）（f（x））const filter = f => xs => xs.filter（x => f（x））const map = f => xs => xs.map（x => f（x））const foo = x => x> 3 const bar = x => x * x const arrfun = xs => xs.reverse（）//注意functionsconst的顺序myfunc = compose（filter（foo），map（bar），arrfun）let someArray = [1,2,3,4,5,6] let result = myfunc（someArray） console.log（result）// [36,25,16] 身份函子 你应该看到这个答案我写了关于身份仿函数 - 这给你一个可链接的接口，但没有触及原生原型 const Identity = x => （{runIdentity：x，map：f => Identity（f（x））}）const foo = x => x> 3 const bar = x => x * x const arrfun = xs => xs.reverse（）const myfunc = xs => Identity（xs）.map（xs => xs.filter（foo））.map（xs => xs.map（bar））.map（xs => arrfun（xs））.runIdentitylet someArray = [1 ，2,3,4,5,6] let result = myfunc（someArray）console.log（result）// [35,25,16] 当然，如果您保留过滤器和 map ，就像我们之前定义的那样，它清理了 myfunc const Identity = x => （{runIdentity：x，map：f => Identity（f（x））}）const filter = f => xs => xs.filter（x => f（x））const map = f => xs => xs.map（x => f（x））const foo = x => x> 3 const bar = x => x * x const arrfun = xs => xs.reverse（）const myfunc = x => Identity（x）.map（filter（foo））.map（map（bar））.map（arrfun）.runIdentitylet someArray = [1,2,3,4,5,6] let result = myfunc（someArray）console .log（result）// [35,25,16] 不要挂在先前定义的 foo 和 bar 上。我们可以直接在 myfunc 中使用lambda表达式，如果您想要的话 const myfunc = xs => .map（xs => xs.filter（x => x> 3）） .map（xs => xs.map（x = > x * x）） .map（arrfun） //或者跳过在其他地方定义arrfun并且只是... // .map（xs => xs.reverse（ ）） .runIdentity If I have an array someArray that I first want to do some operations on and then pass that result to a function arrfun that takes an array as an argument. Like the followinglet arr = someArray.filter(foo).map(bar)let result = arrfun(arr)In the above scenario I would like to avoid having to assign an intermediary variable to be passed to arrfun. I would like to have something like this.Object.prototype.pipe = function(f) {return f(this)}let result = someArray.filter(foo).map(bar).pipe(arrfun)In lieu of a .pipe() how would you solve this?Would it be sensible to introduce that function to Object?Is pipe the best name for such a function? chain? pass?New exampleconst pc = options => options .join(' ') .match(/--[\w.]* [\w.]*/g) .map(s => s.slice(2).split(' ')) .map(([key, value]) => ({key, value})) .map(nestObject) .reduce((acc, val) => Object.assign(acc, val), {})const nestObject = ({key, value}) => key .split('.') .reverse() .reduce((inner, key) => ({[key]: inner}), value)In the above example a problem is that .match returns null if no match is found. Using .pipe you could solve it ny changing that line to.pipe(s => s.match(/--[\w.]* [\w.]*/g) || [])How would you solve this one without pipe? 解决方案 Are you overlooking this simple, straightforward expression ?let result = arrfun(someArray.filter(foo).map(bar))right-to-left function compositionOr maybe you wish for classic (right-to-left) function composition?const compose = (f,...fs) => x => f === undefined ? x : f(compose(...fs)(x))const filter = f => xs => xs.filter(x => f(x))const map = f => xs => xs.map(x => f(x))const foo = x => x > 3const bar = x => x * xconst arrfun = xs => xs.reverse()const myfunc = compose(arrfun, map(bar), filter(foo))let someArray = [1,2,3,4,5,6]let result = myfunc(someArray)console.log(result)// [ 36, 25, 16 ]left-to-right function compositionThe same answer as above using left-to-right function compositionconst compose = (f,...fs) => x => f === undefined ? x : compose(...fs)(f(x))const filter = f => xs => xs.filter(x => f(x))const map = f => xs => xs.map(x => f(x))const foo = x => x > 3const bar = x => x * xconst arrfun = xs => xs.reverse()// notice order of functionsconst myfunc = compose(filter(foo), map(bar), arrfun)let someArray = [1,2,3,4,5,6]let result = myfunc(someArray)console.log(result)// [ 36, 25, 16 ]Identity functorYou should see this answer I wrote about the Identity functor - This gives you a chainable interface but doesn't touch native prototypesconst Identity = x => ({ runIdentity: x, map: f => Identity(f(x))})const foo = x => x > 3const bar = x => x * xconst arrfun = xs => xs.reverse()const myfunc = xs => Identity(xs) .map(xs => xs.filter(foo)) .map(xs => xs.map(bar)) .map(xs => arrfun(xs)) .runIdentitylet someArray = [1,2,3,4,5,6]let result = myfunc(someArray)console.log(result)// [ 35, 25, 16 ]Of course if you keep filter and map as we defined before, it cleans up the definition of myfuncconst Identity = x => ({ runIdentity: x, map: f => Identity(f(x))})const filter = f => xs => xs.filter(x => f(x))const map = f => xs => xs.map(x => f(x))const foo = x => x > 3const bar = x => x * xconst arrfun = xs => xs.reverse()const myfunc = x => Identity(x) .map(filter(foo)) .map(map(bar)) .map(arrfun) .runIdentitylet someArray = [1,2,3,4,5,6]let result = myfunc(someArray)console.log(result)// [ 35, 25, 16 ]And don't get hung up on foo and bar being defined up front. We can use lambda expressions directly within myfunc if you wanted toconst myfunc = xs => Identity(xs) .map(xs => xs.filter(x => x > 3)) .map(xs => xs.map(x => x * x)) .map(arrfun) // or skip defining arrfun somewhere else and just ... // .map(xs => xs.reverse()) .runIdentity 这篇关于通过功能链的功能来传递结果的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持！ 上岸，阿里云！