我有两个数组apple = [1,5,10,15,20]
,bottle = [1,5,10,15,20,25]
使用Lodash或任何JavaScript函数,我想要一个具有唯一元素c= [25]
的数组C。更确切地说,我想要将"苹果"数组与"瓶子"数组进行比较时的所有元素的列表,以显示唯一的元素/
您可以将Array#filter
与相对数组的Set
一起使用。
此提案使用一个补体功能,如果元素A不在集合b。
中,则返回true
对于对称差异,必须将带回调的过滤用于双方。
function getComplement(collection) {
// initialize and close over a set created from the collection passed in
var set = new Set(collection);
// return iterator callback for .filter()
return function (item) {
return !set.has(item);
};
}
var apple = [1,5,10,15,20],
bottle = [1,5,10,15,20,25],
unique = [
...apple.filter(getComplement(bottle)),
...bottle.filter(getComplement(apple))
];
console.log(unique);
您可以使用reduce()
和filter()
创建自己的功能。
var apple = [1,5,10,15,20], bottle = [1,5,10,15,20,25]
function diff(a1, a2) {
//Concat array2 to array1 to create one array, and then use reduce on that array to return
//one object as result where key is element and value is number of occurrences of that element
var obj = a1.concat(a2).reduce(function(result, element) {
result[element] = (result[element] || 0) + 1
return result
}, {})
//Then as function result return keys from previous object where value is == 1 which means that
// that element is unique in both arrays.
return Object.keys(obj).filter(function(element) {
return obj[element] == 1
})
}
console.log(diff(apple, bottle))
带有ES6箭头功能的同一代码的较短版本。
var apple = [1,5,10,15,20], bottle = [1,5,10,15,20,25]
function diff(a1, a2) {
var obj = a1.concat(a2).reduce((r, e) => (r[e] = (r[e] || 0) + 1, r), {})
return Object.keys(obj).filter(e => obj[e] == 1)
}
console.log(diff(apple, bottle))