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
function diffArray(newArr, oldArr) {
var newSet = new Set(newArr)
var diff = []
oldArr.forEach((a) => {
if(!newSet.delete(a))diff.push(a)
})
return diff.concat(Array.from(newSet))
}
其他回答
function array_diff(array1, array2) {
let returnArray = [];
$.each(array1, function(index, value) {
let findStatus = false;
if (Array.isArray(array2)) {
$.each(array2, function(index2, value2) {
if (value == value2) findStatus = true;
});
} else {
if (value == array2) {
findStatus = true;
}
}
if (findStatus == false) {
returnArray.push(value);
}
});
return returnArray;
}
我同意@luis-sieira的解决方案
我创建了位自解释函数,便于初学者一步一步理解:
function difference(oneArr, twoArr){
var newArr = [];
newArr = oneArr.filter((item)=>{
return !twoArr.includes(item)
});
console.log(newArr)
let arr = twoArr.filter((item)=>{
return !oneArr.includes(item)
});
newArr = newArr.concat(arr);
console.log(newArr)
}
difference([1, 2, 3, 5], [1, 2, 3, 4, 5])
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"]。
ES6中一个更简洁的方法是以下解决方案。
var a1 = ['a', 'b'];
var a2 = ['a', 'b', 'c', 'd'];
区别
a2.filter(d => !a1.includes(d)) // gives ["c", "d"]
十字路口
a2.filter(d => a1.includes(d)) // gives ["a", "b"]
析取联合(对称差分)
[ ...a2.filter(d => !a1.includes(d)),
...a1.filter(d => !a2.includes(d)) ]
只是修整字符串以确保....空格不会影响差异
function arr_diff(a1, a2) {
var a=[], diff=[];
for(var i=0;i<a1.length;i++)
a[a1[i]]=true;
for(var i=0;i<a2.length;i++)
if(a[a2[i].trim()]) delete a[a2[i].trim()];
else a[a2[i].trim()]=true;
for(var k in a)
diff.push(k);
return diff;
}