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"]
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
还有另一个答案,但似乎没有人提到jsperf,他们比较了几种算法和技术支持:https://jsperf.com/array-difference-javascript似乎使用过滤器得到了最好的结果。谢谢
其他回答
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。
为了获得对称差异,您需要以两种方式比较数组(或在多个数组的情况下以所有方式比较)
ES7 (ECMAScript 2016)
// diff between just two arrays:
function arrayDiff(a, b) {
return [
...a.filter(x => !b.includes(x)),
...b.filter(x => !a.includes(x))
];
}
// diff between multiple arrays:
function arrayDiff(...arrays) {
return [].concat(...arrays.map( (arr, i) => {
const others = arrays.slice(0);
others.splice(i, 1);
const unique = [...new Set([].concat(...others))];
return arr.filter(x => !unique.includes(x));
}));
}
ES6(2015年ECMAScript)
// diff between just two arrays:
function arrayDiff(a, b) {
return [
...a.filter(x => b.indexOf(x) === -1),
...b.filter(x => a.indexOf(x) === -1)
];
}
// diff between multiple arrays:
function arrayDiff(...arrays) {
return [].concat(...arrays.map( (arr, i) => {
const others = arrays.slice(0);
others.splice(i, 1);
const unique = [...new Set([].concat(...others))];
return arr.filter(x => unique.indexOf(x) === -1);
}));
}
ES5 (ECMAScript 5.1)
// diff between just two arrays:
function arrayDiff(a, b) {
var arrays = Array.prototype.slice.call(arguments);
var diff = [];
arrays.forEach(function(arr, i) {
var other = i === 1 ? a : b;
arr.forEach(function(x) {
if (other.indexOf(x) === -1) {
diff.push(x);
}
});
})
return diff;
}
// diff between multiple arrays:
function arrayDiff() {
var arrays = Array.prototype.slice.call(arguments);
var diff = [];
arrays.forEach(function(arr, i) {
var others = arrays.slice(0);
others.splice(i, 1);
var otherValues = Array.prototype.concat.apply([], others);
var unique = otherValues.filter(function (x, j) {
return otherValues.indexOf(x) === j;
});
diff = diff.concat(arr.filter(x => unique.indexOf(x) === -1));
});
return diff;
}
例子:
// diff between two arrays:
const a = ['a', 'd', 'e'];
const b = ['a', 'b', 'c', 'd'];
arrayDiff(a, b); // (3) ["e", "b", "c"]
// diff between multiple arrays
const a = ['b', 'c', 'd', 'e', 'g'];
const b = ['a', 'b'];
const c = ['a', 'e', 'f'];
arrayDiff(a, b, c); // (4) ["c", "d", "g", "f"]
对象数组之间的差异
function arrayDiffByKey(key, ...arrays) {
return [].concat(...arrays.map( (arr, i) => {
const others = arrays.slice(0);
others.splice(i, 1);
const unique = [...new Set([].concat(...others))];
return arr.filter( x =>
!unique.some(y => x[key] === y[key])
);
}));
}
例子:
const a = [{k:1}, {k:2}, {k:3}];
const b = [{k:1}, {k:4}, {k:5}, {k:6}];
const c = [{k:3}, {k:5}, {k:7}];
arrayDiffByKey('k', a, b, c); // (4) [{k:2}, {k:4}, {k:6}, {k:7}]
随着ES6的到来,带有sets和splat操作符(当时只在Firefox中工作,请检查兼容性表),你可以编写以下一行代码:
var a = ['a', 'b', 'c', 'd'];
var b = ['a', 'b'];
var b1 = new Set(b);
var difference = [...new Set(a.filter(x => !b1.has(x)))];
结果是["c", "d"]。
以上Joshaven Potter的回答非常棒。但是它返回数组B中不在数组C中的元素,而不是反过来。例如,如果var a=[1,2,3,4,5,6]。diff([3、4、5、7]);然后它将输出:==>[1,2,6],而不是[1,2,6,7],这是两者之间的实际差异。你仍然可以使用上面的Potter代码,但也可以简单地向后做一次比较:
Array.prototype.diff = function(a) {
return this.filter(function(i) {return !(a.indexOf(i) > -1);});
};
////////////////////
// Examples
////////////////////
var a=[1,2,3,4,5,6].diff( [3,4,5,7]);
var b=[3,4,5,7].diff([1,2,3,4,5,6]);
var c=a.concat(b);
console.log(c);
这应该输出:[1,2,6,7]
const a1 = ['a', 'b', 'c', 'd'];
const a2 = ['a', 'b'];
const diffArr = a1.filter(o => !a2.includes(o));
console.log(diffArr);
输出:
[ 'a', 'b' ]