I need to check whether one array contains all of the elements of another array, including the same duplicates. The second array can have extra elements. I'm using every...includes, but it's not catching that the second array doesn't have the right duplicates.
For example:
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
}
Thanks!
Edit: arr1 is one of many arrays that I am looping through which are coming out of a graph traversal algorithm, so I'd like to avoid restructuring them into an object to create a dictionary data structure if possible.
CodePudding user response:
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);
CodePudding user response:
Does this solve your problem?
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);
CodePudding user response:
Try creating this function:
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;
}
Then invoke it like this:
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
CodePudding user response:
You indicate order does not matter in your comments. That makes this very simple.
- Sort both arrays
- Check if corresponding elements are equal
- consider errors associated with sparse or short arrays
- Use
.reduce()
to boil it down to a single result
So this really comes down to a single statement once the arrays are sorted:
matcher.reduce((acc, value , idx)=>matcher[idx] === test[idx], false);
You also mentioned testing this against many arrays. So the full example below does that for demo purposes.
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}`);
});