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
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 })
);
其他回答
function arrayDiff(a, b) {
return a.concat(b).filter(val => !(b.includes(val)));
//(or) return a.concat(b).filter(val => !(a.includes(val) && b.includes(val)));
}
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
这个问题很老了,但仍然是javascript数组减法的热门问题,所以我想添加我正在使用的解决方案。适用于以下情况:
var a1 = [1,2,2,3]
var a2 = [1,2]
//result = [2,3]
下面的方法将产生预期的结果:
function arrayDifference(minuend, subtrahend) {
for (var i = 0; i < minuend.length; i++) {
var j = subtrahend.indexOf(minuend[i])
if (j != -1) {
minuend.splice(i, 1);
subtrahend.splice(j, 1);
}
}
return minuend;
}
需要注意的是,该函数不包括减数中没有被减数的值:
var a1 = [1,2,3]
var a2 = [2,3,4]
//result = [1]
Array.prototype.diff =函数(a) { return this.filter(function(i) {return a.f index_of (i) < 0;}); }; ////////////// //示例// ////////////// Const dif1 =[1,2,3,4,5,6]。Diff ([3,4,5]); console.log (dif1);// => [1,2,6] const dif2 =[“test1”、“test2”,“test3”,“test4”,“test5”,“test6”].diff([“test1”、“test2”,“test3”、“test4”)); console.log (dif2);// => ["test5", "test6"]
注意:. indexof()和.filter()在IE9之前是不可用的。
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]);