是否有一种方法可以在JavaScript中返回两个数组之间的差异?

例如:

var a1 = ['a', 'b'];
var a2 = ['a', 'b', 'c', 'd'];

// need ["c", "d"]

当前回答

function diffArray(arr1, arr2) {
  var newArr = arr1.concat(arr2);
  return newArr.filter(function(i){
    return newArr.indexOf(i) == newArr.lastIndexOf(i);
  });
}

这对我很有用

其他回答

类似于Ian Grainger的解决方案(但在打字稿中):

function findDiffs(arrayOne: string[], arrayTwo: string[]) {

    let onlyInArrayOne = []
    let onlyInArrayTwo = []
    let share = []
    let [arrayOneCopy, arrayTwoCopy] = [[...arrayOne], [...arrayTwo]]

    arrayOneCopy.sort(); arrayTwoCopy.sort()

    while (arrayOneCopy.length !== 0 && arrayTwoCopy.length !== 0) {
        if (arrayOneCopy[0] == arrayTwoCopy[0]) {
            share.push(arrayOneCopy[0])
            arrayOneCopy.splice(0, 1)
            arrayTwoCopy.splice(0, 1)
        }
        if (arrayOneCopy[0] < arrayTwoCopy[0]) {
            onlyInArrayOne.push(arrayOneCopy[0])
            arrayOneCopy.splice(0, 1)
        }
        if (arrayOneCopy[0] > arrayTwoCopy[0]) {
            onlyInArrayTwo.push(arrayTwoCopy[0])
            arrayTwoCopy.splice(0, 1)
        }
    }
    onlyInArrayTwo = onlyInArrayTwo.concat(arrayTwoCopy)
    onlyInArrayOne = onlyInArrayOne.concat(arrayOneCopy)

    return {
        onlyInArrayOne,
        onlyInArrayTwo,
        share,
        diff: onlyInArrayOne.concat(onlyInArrayTwo)
    }
}

// arrayOne: [ 'a', 'b', 'c', 'm', 'y' ] 
// arrayTwo: [ 'c', 'b', 'f', 'h' ]
//
// Results: 
// { 
//    onlyInArrayOne: [ 'a', 'm', 'y' ],
//    onlyInArrayTwo: [ 'f', 'h' ],
//    share: [ 'b', 'c' ],
//    diff: [ 'a', 'm', 'y', 'f', 'h' ] 
// }

您可以使用一个公共对象并计算第一个数组中每个值的频率。对于第二个数组,减少公共对象中的值。然后遍历所有键并添加所有值大于1的键。

常量差值= (a1, a2) => { Var obj = {}; a1。forEach(obj[v] = (obj[v] || 0) + 1); a2。forEach(v => obj[v] = (obj[v] || 0) - 1); 返回对象 . keys (obj) .reduce((r,k) => { If (obj[k] > 0) r = r.concat (Array.from({长度:obj [k]}) .fill (k)); 返回r; }, []); }; const =结果不同([' a ', ' ', ' b ', ' c ', ' d '], [a, b]); console.log(结果);

对我来说,把它作为部分函数处理比较容易。很惊讶没有看到函数式编程的解决方案,这是我在ES6中的:

const arrayDiff = (a, b) => {
  return diff(b)(a);
}

const contains = (needle) => (array) => {
  for (let i=0; i < array.length; i++) {
    if (array[i] == needle) return true;
  }

  return false;
}

const diff = (compare) => {
    return (array) => array.filter((elem) => !contains(elem)(compare))
}
var arrayDifference = function(arr1, arr2){
  if(arr1 && arr1.length){
    if(arr2 && arr2.length > 0){
      for (var i=0, itemIndex; i<arr2.length; i++){
        itemIndex = arr1.indexOf(arr2[i]);
        if(itemIndex !== -1){
          arr1.splice(itemIndex, 1);
        }
      }
    }
    return arr1;
  }
  return [];
};

arrayDifference([1,2,3,4,5], [1,5,6]);

ES2015的函数方法

计算两个数组之间的差值是Set操作之一。这个术语已经表明应该使用本机Set类型,以便提高查找速度。不管怎样,当你计算两个集合之间的差值时,有三种排列:

[+left difference] [-intersection] [-right difference]
[-left difference] [-intersection] [+right difference]
[+left difference] [-intersection] [+right difference]

下面是反映这些排列的功能性解决方案。

离开的区别:

// small, reusable auxiliary functions const apply = f => x => f(x); const flip = f => y => x => f(x) (y); const createSet = xs => new Set(xs); const filter = f => xs => xs.filter(apply(f)); // left difference const differencel = xs => ys => { const zs = createSet(ys); return filter(x => zs.has(x) ? false : true ) (xs); }; // mock data const xs = [1,2,2,3,4,5]; const ys = [0,1,2,3,3,3,6,7,8,9]; // run the computation console.log( differencel(xs) (ys) );

正确的区别:

差异是微不足道的。这与翻转的参数不同。为了方便,你可以写一个函数:const differencer = flip(difference)。这是所有!

对称的区别:

现在我们有了左边和右边,实现对称的差异也变得微不足道:

// small, reusable auxiliary functions const apply = f => x => f(x); const flip = f => y => x => f(x) (y); const concat = y => xs => xs.concat(y); const createSet = xs => new Set(xs); const filter = f => xs => xs.filter(apply(f)); // left difference const differencel = xs => ys => { const zs = createSet(ys); return filter(x => zs.has(x) ? false : true ) (xs); }; // symmetric difference const difference = ys => xs => concat(differencel(xs) (ys)) (flip(differencel) (xs) (ys)); // mock data const xs = [1,2,2,3,4,5]; const ys = [0,1,2,3,3,3,6,7,8,9]; // run the computation console.log( difference(xs) (ys) );

我想这个例子是一个很好的起点,可以让你了解函数式编程的含义:

使用可以以许多不同方式组合在一起的构建块进行编程。