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
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]
其他回答
这是我用来获得两个数组之间的差异的函数-它适用于数值,字符串,混合num/字符串数组。在数组/多维数组中不是对象文字
function diff(arr1, arr2) {
var x,
t;
function uniq(a, b) {
t = b;
if( (b === 0 && x[b+1]!==a) ||
(t > 0 && a !== x[b+1] && a !== x[b-1]) ) {
return a;
}
}
x = arr1.concat(arr2).sort();
return x.filter(uniq);
}
var a1 = ['a', 'b', 'e', 'c'],
a2 = ['b', 'a', 'c', 'f' ];
diff(a1, a2);
一个衬垫
const unique = (a) => [...new Set(a)]; const uniqueBy = (x,f)=>Object.values(x.reduce((a,b)=>((a[f(b)]=b),a),{})); const intersection = (a, b) => a.filter((v) => b.includes(v)); const diff = (a, b) => a.filter((v) => !b.includes(v)); const symDiff = (a, b) => diff(a, b).concat(diff(b, a)); const union = (a, b) => diff(a, b).concat(b); const a = unique([1, 2, 3, 4, 5, 5]); console.log(a); const b = [4, 5, 6, 7, 8]; console.log(intersection(a, b), diff(a, b), symDiff(a, b), union(a, b)); console.log(uniqueBy( [ { id: 1, name: "abc" }, { id: 2, name: "xyz" }, { id: 1, name: "abc" }, ], (v) => v.id )); const intersectionBy = (a, b, f) => a.filter((v) => b.some((u) => f(v, u))); console.log(intersectionBy( [ { id: 1, name: "abc" }, { id: 2, name: "xyz" }, ], [ { id: 1, name: "abc" }, { id: 3, name: "pqr" }, ], (v, u) => v.id === u.id )); const diffBy = (a, b, f) => a.filter((v) => !b.some((u) => f(v, u))); console.log(diffBy( [ { id: 1, name: "abc" }, { id: 2, name: "xyz" }, ], [ { id: 1, name: "abc" }, { id: 3, name: "pqr" }, ], (v, u) => v.id === u.id ));
打印稿
操场上的链接
const unique = <T>(array: T[]) => [...new Set(array)];
const intersection = <T>(array1: T[], array2: T[]) =>
array1.filter((v) => array2.includes(v));
const diff = <T>(array1: T[], array2: T[]) =>
array1.filter((v) => !array2.includes(v));
const symDiff = <T>(array1: T[], array2: T[]) =>
diff(array1, array2).concat(diff(array2, array1));
const union = <T>(array1: T[], array2: T[]) =>
diff(array1, array2).concat(array2);
const intersectionBy = <T>(
array1: T[],
array2: T[],
predicate: (array1Value: T, array2Value: T) => boolean
) => array1.filter((v) => array2.some((u) => predicate(v, u)));
const diffBy = <T>(
array1: T[],
array2: T[],
predicate: (array1Value: T, array2Value: T) => boolean
) => array1.filter((v) => !array2.some((u) => predicate(v, u)));
const uniqueBy = <T>(
array: T[],
predicate: (v: T, i: number, a: T[]) => string
) =>
Object.values(
array.reduce((acc, value, index) => {
acc[predicate(value, index, array)] = value;
return acc;
}, {} as { [key: string]: T })
);
转换为字符串对象类型:
[1, 1].toString() === [1, 1].toString(); // true
对称和线性复杂度。需要ES6。
function arrDiff(arr1, arr2) {
var arrays = [arr1, arr2].sort((a, b) => a.length - b.length);
var smallSet = new Set(arrays[0]);
return arrays[1].filter(x => !smallSet.has(x));
}
CoffeeScript版本:
diff = (val for val in array1 when val not in array2)