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
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。
简单性:
// Usage
const intersection = allLists
.reduce(intersect, allValues)
.reduce(removeDuplicates, []);
// Implementation
const intersect = (intersection, list) =>
intersection.filter(item =>
list.some(x => x === item));
const removeDuplicates = (uniques, item) =>
uniques.includes(item) ? uniques : uniques.concat(item);
// Example Data
const somePeople = [bob, doug, jill];
const otherPeople = [sarah, bob, jill];
const morePeople = [jack, jill];
const allPeople = [...somePeople, ...otherPeople, ...morePeople];
const allGroups = [somePeople, otherPeople, morePeople];
// Example Usage
const intersection = allGroups
.reduce(intersect, allPeople)
.reduce(removeDuplicates, []);
intersection; // [jill]
好处:
泥土简单 以数据为中心的 适用于任意数量的列表 适用于任意长度的列表 适用于任意类型的值 适用于任意排序顺序 保留形状(在任何数组中首次出现的顺序) 尽可能早退出 内存安全,不影响函数/数组原型
缺点:
内存占用率较高 更高的CPU使用率 需要理解reduce 需要理解数据流
你不希望将其用于3D引擎或内核工作,但如果你在基于事件的应用程序中运行时遇到问题,那么你的设计就有更大的问题。
这是我使用的一个非常简单的实现。它是无损的,也确保不复制整个。
Array.prototype.contains = function(elem) {
return(this.indexOf(elem) > -1);
};
Array.prototype.intersect = function( array ) {
// this is naive--could use some optimization
var result = [];
for ( var i = 0; i < this.length; i++ ) {
if ( array.contains(this[i]) && !result.contains(this[i]) )
result.push( this[i] );
}
return result;
}
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
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)