检查一个数组是否包含另一个数组的所有元素，包括重复项是否出现两次

Question

我需要检查一个数组是否包含另一个数组的所有元素，包括相同的重复元素。第二个数组可以有额外的元素。我正在使用每。。。包括，但第二个数组没有正确的重复项，这一点并不明显。

例如：

const arr1 = [1, 2, 2, 3, 5, 5, 6, 6]
const arr2 = [1, 2, 3, 5, 6, 7]
if(arr1.every(elem => arr2.includes(elem))){
return true     // should return false because arr2 does not have the same duplicates
}

谢谢！

编辑：arr1是我循环使用的许多数组之一，这些数组来自图遍历算法，所以如果可能的话，我希望避免将它们重组为对象来创建字典数据结构。

Answer 1

尝试创建此函数：

function containsAll (target, toTest) {
const dictionary = {}
target.forEach(element => {
if (dictionary[element] === undefined) {
dictionary[element] = 1;
return;
}
dictionary[element]++;
});

toTest.forEach(element => {
if (dictionary[element] !== undefined)
dictionary[element]--;
})
for (let key in dictionary) {
if (dictionary[key] > 0) return false;
}
return true;
}

然后像这样调用它：

const arr1 = [1, 2, 2, 3, 5, 5, 6, 6]
const arr2 = [1, 2, 3, 5, 6, 7]

console.log(containsAll(arr1, arr2)) // returns false

Answer 2

const arr1 = [1, 2, 2, 3, 5, 5, 6, 6];
//const arr2 = [1, 2, 3, 5, 6, 7];
const arr2 = [1, 2, 2, 3, 5, 5];
let includesAll1 = true;
let includesAll2 = true;
const checkObj1 = {
};
const checkObj2 = {
};
arr1.forEach((el)=> {
if(checkObj1[el] === undefined) {
checkObj1[el] = 1;
} else {
checkObj1[el]++;
}
});
arr2.forEach((el)=> {
if(checkObj2[el] === undefined) {
checkObj2[el] = 1;
} else {
checkObj2[el]++;
}
});
const check1Keys = Object.keys(checkObj1);
const check2Keys = Object.keys(checkObj2);
if(check1Keys.length > check2Keys.length) {
includesAll2 = false;
check2Keys.forEach((key)=> {
const value1 = checkObj1[key];
const value2 = checkObj2[key];
if(!arr1.includes(parseInt(key)) || value1 != value2) {
includesAll1 = false;
}
});
} else {
includesAll1 = false;
check1Keys.forEach((key)=> {
const value1 = checkObj1[key];
const value2 = checkObj2[key];
console.log(value1, value2, key);
if(!arr2.includes(parseInt(key)) || value1 != value2) {
includesAll2 = false;
}
});
}
console.log(includesAll1);
console.log(includesAll2);

Answer 3

这能解决您的问题吗？

const arr = [1, 2, 3, 5, 6, 7, 2, 10, 2, 3, 2];
const subArr = [1, 2, 2, 3, 2] 
const contains = subArr.every(num => subArr.filter(n => n == num).length <= arr.filter(n => n== num).length);
console.log(contains);

Answer 4

您在评论中指出顺序无关紧要。这让事情变得非常简单。

1. 对两个数组进行排序
2. 检查相应元素是否相等
   • 考虑与稀疏或短数组相关的错误
3. 使用.reduce()将其归结为单个结果

因此，一旦对数组进行排序，这实际上就归结为一条语句：

matcher.reduce((acc, value , idx)=>matcher[idx] === test[idx], false);

您还提到针对许多阵列进行测试。下面的完整示例就是为了演示目的。

let toMatch = [1, 2, 2, 3, 5, 5, 6, 6]
let arrayOfArrays = [[1,2],[1, 2, 3, 5, 6, 7, 3, 9, 8, 2, 7],[1, 2, 3, 3, 6, 7],[1, 3, 3, 5, 6, 7],[1, 2, 3, 5, 6, 6], [3,5,2,1,6,2,5,6]];
let toMatchSorted = toMatch.slice().sort();
arrayOfArrays.forEach(arr=>{
let toTestSorted = arr.slice().sort();
let out = toMatchSorted.reduce((acc, value , idx)=>toMatchSorted[idx] === toTestSorted[idx], false);
console.log(`Input: ${arr}, Result: ${out}`);
});