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
所选的答案只对了一半。您必须比较数组的两种方式才能得到完整的答案。
const ids_exist = [
'1234',
'5678',
'abcd',
]
const ids_new = [
'1234',
'5678',
'efjk',
'9999',
]
function __uniq_Filter (__array_1, __array_2) {
const one_not_in_two = __array_1.filter(function (obj) {
return __array_2.indexOf(obj) == -1
})
const two_not_in_one = __array_2.filter(function (obj) {
return __array_1.indexOf(obj) == -1
})
return one_not_in_two.concat(two_not_in_one)
}
let uniq_filter = __uniq_Filter(ids_exist, ids_new)
console.log('uniq_filter', uniq_filter) // => [ 'abcd', 'efjk', '9999' ]
其他回答
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)) ]
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 })
);
基于思考者的答案,但允许重复。
映射在映射值出现时增加它们,如果它们在另一个数组中,则减少它们。
任何剩余的部分都将包括在差额中。
function diff(curr, prev) { let a = curr.split('').sort(), b = prev.split('').sort(), c = arrDiff(a, b); console.log(JSON.stringify(a), '-', JSON.stringify(b), '=', JSON.stringify(c)); return c; } function arrDiff(larger, smaller) { var entries = {}; for (var i = 0; i < larger.length; i++) { entries[larger[i]] = (entries[larger[i]] || 0) + 1; } for (var i = 0; i < smaller.length; i++) { if (entries[smaller[i]]) { entries[smaller[i]] -= 1; } else { entries[smaller[i]] = (entries[smaller[i]] || 0) + 1; } } return Object.keys(entries).sort().reduce((diff, key) => { if (entries[key] > 0) { for (var i = 0; i < entries[key]; i++) { diff.push(key); } } return diff; }, []); } // Smaller is a subset of Larger console.log('Result:', JSON.stringify(diff('ENLIGHTEN', 'LENGTHEN'))); // [ I ] console.log('Result:', JSON.stringify(diff('STRENGTH', 'TENTHS'))); // [ G, R ] // Both have a unique value console.log('Result:', JSON.stringify(diff('BUBBLE', 'RUBBLE'))); // [ B, R ] .as-console-wrapper { top: 0; max-height: 100% !important; }
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}
我一直在寻找一个不涉及使用不同库的简单答案,我想出了我自己的答案,我想这里没有提到过。 我不知道它的效率如何,但它确实有效;
function find_diff(arr1, arr2) {
diff = [];
joined = arr1.concat(arr2);
for( i = 0; i <= joined.length; i++ ) {
current = joined[i];
if( joined.indexOf(current) == joined.lastIndexOf(current) ) {
diff.push(current);
}
}
return diff;
}
对于我的代码,我也需要删除副本,但我想这并不总是可取的。
我想主要的缺点是它可能会比较许多已经被拒绝的选择。