用javascript实现数组交叉的最简单、无库代码是什么?我想写
intersection([1,2,3], [2,3,4,5])
并获得
[2, 3]
用javascript实现数组交叉的最简单、无库代码是什么?我想写
intersection([1,2,3], [2,3,4,5])
并获得
[2, 3]
当前回答
这是一个现代和简单的ES6方式来做,也非常灵活。 它允许您指定多个数组作为与主题数组进行比较的数组,并且可以在包含和独占模式下工作。
// =======================================
// The function
// =======================================
function assoc(subjectArray, otherArrays, { mustBeInAll = true } = {}) {
return subjectArray.filter((subjectItem) => {
if (mustBeInAll) {
return otherArrays.every((otherArray) =>
otherArray.includes(subjectItem)
);
} else {
return otherArrays.some((otherArray) => otherArray.includes(subjectItem));
}
});
}
// =======================================
// The usage
// =======================================
const cheeseList = ["stilton", "edam", "cheddar", "brie"];
const foodListCollection = [
["cakes", "ham", "stilton"],
["juice", "wine", "brie", "bread", "stilton"]
];
// Output will be: ['stilton', 'brie']
const inclusive = assoc(cheeseList, foodListCollection, { mustBeInAll: false }),
// Output will be: ['stilton']
const exclusive = assoc(cheeseList, foodListCollection, { mustBeInAll: true })
实例:https://codesandbox.io/s/zealous-butterfly-h7dgf?fontsize=14&hidenavigation=1&theme=dark
其他回答
使用jQuery:
var a = [1,2,3];
var b = [2,3,4,5];
var c = $(b).not($(b).not(a));
alert(c);
这是一个提议的标准:对于当前阶段2的提议https://github.com/tc39/proposal-set-methods,您可以使用
mySet.intersection(mySet2);
在此之前,你可以使用Immutable.js的Set,它激发了这个提议
Immutable.Set(mySet).intersect(mySet2)
这是一个现代和简单的ES6方式来做,也非常灵活。 它允许您指定多个数组作为与主题数组进行比较的数组,并且可以在包含和独占模式下工作。
// =======================================
// The function
// =======================================
function assoc(subjectArray, otherArrays, { mustBeInAll = true } = {}) {
return subjectArray.filter((subjectItem) => {
if (mustBeInAll) {
return otherArrays.every((otherArray) =>
otherArray.includes(subjectItem)
);
} else {
return otherArrays.some((otherArray) => otherArray.includes(subjectItem));
}
});
}
// =======================================
// The usage
// =======================================
const cheeseList = ["stilton", "edam", "cheddar", "brie"];
const foodListCollection = [
["cakes", "ham", "stilton"],
["juice", "wine", "brie", "bread", "stilton"]
];
// Output will be: ['stilton', 'brie']
const inclusive = assoc(cheeseList, foodListCollection, { mustBeInAll: false }),
// Output will be: ['stilton']
const exclusive = assoc(cheeseList, foodListCollection, { mustBeInAll: true })
实例:https://codesandbox.io/s/zealous-butterfly-h7dgf?fontsize=14&hidenavigation=1&theme=dark
“filter”和“indexOf”在IE中的Array上不支持。这个怎么样:
var array1 = [1, 2, 3];
var array2 = [2, 3, 4, 5];
var intersection = [];
for (i in array1) {
for (j in array2) {
if (array1[i] == array2[j]) intersection.push(array1[i]);
}
}
最简单、最快的O(n)和最短的方式:
函数交点(a, b) { const setA = new Set(a); 返回b.filter(value => setA.has(value)); } console.log(十字路口([1,2,3],[2、3、4、5)))
@nbarbosa几乎有相同的答案,但他将两个数组强制转换为Set,然后再转换回数组。不需要任何额外的选角。