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
如果你不关心原始数组,编辑它们没有问题,那么这是一个更快的算法:
let iterator = arrayA.values()
let result = []
for (entryA of iterator) {
if (!arrayB.includes(entryA)) {
result.push(entryA)
} else {
arrayB.splice(arrayB.indexOf(entryA), 1)
}
}
result.push(...arrayB)
return result
**这将根据'type'参数为任意2个数组返回一个唯一值数组,或一个重复值数组,或一个非重复值数组(difference)。**
let json1 = ['one', 'two']
let json2 = ['one', 'two', 'three', 'four']
function uniq_n_shit (arr1, arr2, type) {
let concat = arr1.concat(arr2)
let set = [...new Set(concat)]
if (!type || type === 'uniq' || type === 'unique') {
return set
} else if (type === 'duplicate') {
concat = arr1.concat(arr2)
return concat.filter(function (obj, index, self) {
return index !== self.indexOf(obj)
})
} else if (type === 'not_duplicate') {
let duplicates = concat.filter(function (obj, index, self) {
return index !== self.indexOf(obj)
})
for (let r = 0; r < duplicates.length; r++) {
let i = set.indexOf(duplicates[r]);
if(i !== -1) {
set.splice(i, 1);
}
}
return set
}
}
console.log(uniq_n_shit(json1, json2, null)) // => [ 'one', 'two', 'three', 'four' ]
console.log(uniq_n_shit(json1, json2, 'uniq')) // => [ 'one', 'two', 'three', 'four' ]
console.log(uniq_n_shit(json1, json2, 'duplicate')) // => [ 'one', 'two' ]
console.log(uniq_n_shit(json1, json2, 'not_duplicate')) // => [ 'three', 'four' ]
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 })
);
var a1 = ['a', 'b'];
var a2 = ['a', 'b', 'c', 'd'];
var diff = [];
for (var i in a2) {
var found = false;
for (var j in a1) {
if (a2[i] === a1[j]) found = true;
}
if (found === false) diff.push(a2[i]);
}
那么简单。也可以用于对象,检查对象的一个属性。 就像,
if (a2[i].id === a1[j].id) found = true;
您可以使用一个公共对象并计算第一个数组中每个值的频率。对于第二个数组,减少公共对象中的值。然后遍历所有键并添加所有值大于1的键。
常量差值= (a1, a2) => { Var obj = {}; a1。forEach(obj[v] = (obj[v] || 0) + 1); a2。forEach(v => obj[v] = (obj[v] || 0) - 1); 返回对象 . keys (obj) .reduce((r,k) => { If (obj[k] > 0) r = r.concat (Array.from({长度:obj [k]}) .fill (k)); 返回r; }, []); }; const =结果不同([' a ', ' ', ' b ', ' c ', ' d '], [a, b]); console.log(结果);