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
对我来说,把它作为部分函数处理比较容易。很惊讶没有看到函数式编程的解决方案,这是我在ES6中的:
const arrayDiff = (a, b) => {
return diff(b)(a);
}
const contains = (needle) => (array) => {
for (let i=0; i < array.length; i++) {
if (array[i] == needle) return true;
}
return false;
}
const diff = (compare) => {
return (array) => array.filter((elem) => !contains(elem)(compare))
}
其他回答
使用http://phrogz.net/JS/ArraySetMath.js你可以:
var array1 = ["test1", "test2","test3", "test4"];
var array2 = ["test1", "test2","test3","test4", "test5", "test6"];
var array3 = array2.subtract( array1 );
// ["test5", "test6"]
var array4 = array1.exclusion( array2 );
// ["test5", "test6"]
根据之前的答案…取决于你是想要一个高效的还是“漂亮的联机”解决方案。
一般有三种方法……
"manual iterative" (using indexOf) - naive with O(n2) complexity (slow) var array_diff_naive = function(a,b){ var i, la = a.length, lb = b.length, res = []; if (!la) return b; else if (!lb) return a; for (i = 0; i < la; i++) { if (b.indexOf(a[i]) === -1) res.push(a[i]); } for (i = 0; i < lb; i++) { if (a.indexOf(b[i]) === -1) res.push(b[i]); } return res; } "abstract iterative" (using filter and concat library methods) - syntactic sugar for manual iterative (looks nicer, still sucks) var array_diff_modern = function(a1,a2){ return a1.filter(function(v) { return !a2.includes(v); } ) .concat(a2.filter(function(v) { return !a1.includes(v);})); } "using hashtable" (using object keys) - much more efficient - only O(n), but has slightly limited range of input array values var array_diff_hash = function(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]]) { delete a[a2[i]]; } else { a[a2[i]] = true; } } for (var k in a) { diff.push(k); } return diff; }
在jsperf上可以看到 https://jsperf.com/array-diff-algo
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 })
);
对于想要从一个数组中减去另一个数组的人的回应…
如果不超过1000个元素试试这个…
设置一个新变量来复制Array01,并将其命名为Array03。
现在,使用冒泡排序算法比较Array01和Array02的元素,并在找到匹配时进行比较 对Array03执行以下操作…
if (Array01[x]==Array02[y]) {Array03.splice(x,1);}
注意:我们正在修改Array03而不是Array01,这样就不会搞砸冒泡排序的嵌套循环!
最后,用一个简单的赋值操作将Array03的内容复制到Array01,就完成了。
function arrayDiff(a, b) {
return a.concat(b).filter(val => !(b.includes(val)));
//(or) return a.concat(b).filter(val => !(a.includes(val) && b.includes(val)));
}