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
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' ]
// }
其他回答
下面是另一个可以返回差异的解决方案,就像git diff一样:(它已经用typescript编写,如果你不使用typescript版本,只需删除类型)
/**
* util function to calculate the difference between two arrays (pay attention to 'from' and 'to'),
* it would return the mutations from 'from' to 'to'
* @param { T[] } from
* @param { T[] } to
* @returns { { [x in string]: boolean } } it would return the stringified version of array element, true means added,
* false means removed
*/
export function arrDiff<T>(from: T[], to: T[]): { [x in string]: boolean } {
var diff: { [x in string]: boolean } = {};
var newItems: T[] = []
diff = from.reduce((a, e) => ({ ...a, [JSON.stringify(e)]: true }), {})
for (var i = 0; i < to.length; i++) {
if (diff[JSON.stringify(to[i])]) {
delete diff[JSON.stringify(to[i])]
} else {
newItems.push(to[i])
}
}
return {
...Object.keys(diff).reduce((a, e) => ({ ...a, [e]: false }), {}),
...newItems.reduce((a, e) => ({ ...a, [JSON.stringify(e)]: true }), {})
}
}
下面是一个用法示例:
arrDiff(['a', 'b', 'c'], ['a', 'd', 'c', 'f']) //{"b": false, "d": true, "f": true}
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]);
困难的方法(如果你想做一些比.indexOf更奇特的东西)
var difference = function (source, target) {
return source.reduce(function (diff, current) {
if (target.indexOf(current) === -1) {
diff.push(current);
}
return diff;
}, []);
}
简单的方法
var difference = function (source, target) {
return source.filter(function (current) {
return target.indexOf(current) === -1;
});
}
使用额外的内存来做到这一点。这样你可以用更少的时间复杂度来求解,O(n)而不是O(n *n)
function getDiff(arr1,arr2){
let k = {};
let diff = []
arr1.map(i=>{
if (!k.hasOwnProperty(i)) {
k[i] = 1
}
}
)
arr2.map(j=>{
if (!k.hasOwnProperty(j)) {
k[j] = 1;
} else {
k[j] = 2;
}
}
)
for (var i in k) {
if (k[i] === 1)
diff.push(+i)
}
return diff
}
getDiff([4, 3, 52, 3, 5, 67, 9, 3],[4, 5, 6, 75, 3, 334, 5, 5, 6])
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' ]