ES2015的函数式方法

函数式方法必须考虑只使用没有副作用的纯函数，每个函数只与单个作业有关。

这些限制增强了所涉及函数的可组合性和可重用性。

//小的，可重用的辅助函数 const createSet = xs => new Set(xs); Const filter = f => xs => xs.filter(apply(f)); Const apply = f => x => f(x); / /十字路口 Const相交= xs => ys => { const zs =创建集(ys); 返回过滤器(x => zs.has(x)) ？ 真正的 :假 ) (x); }; //模拟数据 Const xs = [1,2,2,3,4,5]; Const ys = [0,1,2,3,3,3,6,7,8,9]; //运行 Console.log (intersect(xs) (ys));

请注意，使用本机Set类型，这有一个优点 查找性能。

避免重复

显然，第一个数组中重复出现的项将被保留，而第二个数组将被去重。这可能是也可能不是理想的行为。如果你需要一个唯一的结果，只需对第一个参数应用重复数据删除:

// auxiliary functions const apply = f => x => f(x); const comp = f => g => x => f(g(x)); const afrom = apply(Array.from); const createSet = xs => new Set(xs); const filter = f => xs => xs.filter(apply(f)); // intersection const intersect = xs => ys => { const zs = createSet(ys); return filter(x => zs.has(x) ? true : false ) (xs); }; // de-duplication const dedupe = comp(afrom) (createSet); // mock data const xs = [1,2,2,3,4,5]; const ys = [0,1,2,3,3,3,6,7,8,9]; // unique result console.log( intersect(dedupe(xs)) (ys) );

计算任意数量数组的交集

如果你想计算任意数量的数组的交点，只需用compose intersect和foldl。这是一个方便函数:

// auxiliary functions const apply = f => x => f(x); const uncurry = f => (x, y) => f(x) (y); const createSet = xs => new Set(xs); const filter = f => xs => xs.filter(apply(f)); const foldl = f => acc => xs => xs.reduce(uncurry(f), acc); // intersection const intersect = xs => ys => { const zs = createSet(ys); return filter(x => zs.has(x) ? true : false ) (xs); }; // intersection of an arbitrarily number of Arrays const intersectn = (head, ...tail) => foldl(intersect) (head) (tail); // mock data const xs = [1,2,2,3,4,5]; const ys = [0,1,2,3,3,3,6,7,8,9]; const zs = [0,1,2,3,4,5,6]; // run console.log( intersectn(xs, ys, zs) );

下面是underscore.js的实现:

_.intersection = function(array) {
  if (array == null) return [];
  var result = [];
  var argsLength = arguments.length;
  for (var i = 0, length = array.length; i < length; i++) {
    var item = array[i];
    if (_.contains(result, item)) continue;
    for (var j = 1; j < argsLength; j++) {
      if (!_.contains(arguments[j], item)) break;
    }
    if (j === argsLength) result.push(item);
  }
  return result;
};

来源:http://underscorejs.org/docs/underscore.html部分- 62

简单性:

// Usage
const intersection = allLists
  .reduce(intersect, allValues)
  .reduce(removeDuplicates, []);


// Implementation
const intersect = (intersection, list) =>
  intersection.filter(item =>
    list.some(x => x === item));

const removeDuplicates = (uniques, item) =>
  uniques.includes(item) ? uniques : uniques.concat(item);


// Example Data
const somePeople = [bob, doug, jill];
const otherPeople = [sarah, bob, jill];
const morePeople = [jack, jill];

const allPeople = [...somePeople, ...otherPeople, ...morePeople];
const allGroups = [somePeople, otherPeople, morePeople];

// Example Usage
const intersection = allGroups
  .reduce(intersect, allPeople)
  .reduce(removeDuplicates, []);

intersection; // [jill]

好处:

泥土简单 以数据为中心的 适用于任意数量的列表 适用于任意长度的列表 适用于任意类型的值 适用于任意排序顺序 保留形状(在任何数组中首次出现的顺序) 尽可能早退出 内存安全，不影响函数/数组原型

缺点:

内存占用率较高 更高的CPU使用率 需要理解reduce 需要理解数据流

你不希望将其用于3D引擎或内核工作，但如果你在基于事件的应用程序中运行时遇到问题，那么你的设计就有更大的问题。

我写了一个相交函数，它甚至可以根据对象的特定属性来检测对象数组的交集。

例如,

if arr1 = [{id: 10}, {id: 20}]
and arr2 =  [{id: 20}, {id: 25}]

我们想要基于id属性的交集，那么输出应该是:

[{id: 20}]

因此，相同(注:ES6代码)的函数为:

const intersect = (arr1, arr2, accessors = [v => v, v => v]) => {
    const [fn1, fn2] = accessors;
    const set = new Set(arr2.map(v => fn2(v)));
    return arr1.filter(value => set.has(fn1(value)));
};

你可以这样调用这个函数:

intersect(arr1, arr2, [elem => elem.id, elem => elem.id])

还要注意:该函数查找交集时考虑到第一个数组是主数组，因此交集结果将是主数组的结果。

破坏性似乎是最简单的，特别是如果我们可以假设输入是排序的:

/* destructively finds the intersection of 
 * two arrays in a simple fashion.  
 *
 * PARAMS
 *  a - first array, must already be sorted
 *  b - second array, must already be sorted
 *
 * NOTES
 *  State of input arrays is undefined when
 *  the function returns.  They should be 
 *  (prolly) be dumped.
 *
 *  Should have O(n) operations, where n is 
 *    n = MIN(a.length, b.length)
 */
function intersection_destructive(a, b)
{
  var result = [];
  while( a.length > 0 && b.length > 0 )
  {  
     if      (a[0] < b[0] ){ a.shift(); }
     else if (a[0] > b[0] ){ b.shift(); }
     else /* they're equal */
     {
       result.push(a.shift());
       b.shift();
     }
  }

  return result;
}

非破坏性的要稍微复杂一点，因为我们要跟踪指标:

/* finds the intersection of 
 * two arrays in a simple fashion.  
 *
 * PARAMS
 *  a - first array, must already be sorted
 *  b - second array, must already be sorted
 *
 * NOTES
 *
 *  Should have O(n) operations, where n is 
 *    n = MIN(a.length(), b.length())
 */
function intersect_safe(a, b)
{
  var ai=0, bi=0;
  var result = [];

  while( ai < a.length && bi < b.length )
  {
     if      (a[ai] < b[bi] ){ ai++; }
     else if (a[ai] > b[bi] ){ bi++; }
     else /* they're equal */
     {
       result.push(a[ai]);
       ai++;
       bi++;
     }
  }

  return result;
}

这是我使用的一个非常简单的实现。它是无损的，也确保不复制整个。

Array.prototype.contains = function(elem) {
    return(this.indexOf(elem) > -1);
};

Array.prototype.intersect = function( array ) {
    // this is naive--could use some optimization
    var result = [];
    for ( var i = 0; i < this.length; i++ ) {
        if ( array.contains(this[i]) && !result.contains(this[i]) )
            result.push( this[i] );
    }
    return result;
}