I have an array which contains nested arrays that are dynamically created, it looks like this:
[['1', '2'],['1','3', '4'],['1', '3']]
I am trying to implement AND logic by getting only duplicate values from these arrays. My expected output here would be ['1'] since all nested arrays must contain the same value.
// [['1', '2'],['1','3', '4'],['1', '3']]
const arrays = [...new Set(response)].filter(newSet => newSet.length > 0);
const builder = []; // array of all the id's no longer needed
// [[],[],[]]
arrays.forEach(arr => {
// []
arr.forEach(value => {
// [[], [], []]
const found = arrays.filter(a => a.find(v => v === value));
if (found.length === 0)
builder.push(value);
});
});
console.log(builder); // empty []
This gives me an empty array because of the filter(). How could I return an array of values that all (3 in this case but could be more) arrays contain?
Expected output with the above input would be ["1"]. Any help appreciated.
from what I understand you need the common elements from all array
let response1 = [['1', '2'],['1','3', '4'],['1', '3']]
let response2 = [['1', '2', '3'],['1','3', '4'],['1', '3']]
const getCommon = res => [...new Set(res.flat())].filter(a => res.every(c => c.includes(a)));
console.log(getCommon(response1))
console.log(getCommon(response2))UPDATE
Apparently the set transformation is unnecessary since anyway it has to give the elements common to every array so filtering from res[0] should do
let response1 = [['1', '2'],['1','3', '4'],['1', '3']]
let response2 = [['1', '2', '3'],['1','3', '4'],['1', '3']]
const getCommon = res => res[0].filter(a => res.every(c => c.includes(a)));
console.log(getCommon(response1))
console.log(getCommon(response2))You can make a count object that has the frequency of each number in it, and just check if the frequency of a number is equal to the length of the original array.
const getIntersectVals = (arrayOfVals)=>{
const freqs = {};
for(let arr of arrayOfVals){
for(let val of arr){
if(freqs[val]) freqs[val]++;
else freqs[val] = 1;
}
}
const uniqueVals = Object.keys(freqs);
const correctVals = uniqueVals.filter(elem=>{
return freqs[elem] === arrayOfVals.length;
})
return correctVals;
}
const arrayOfVals = [['1', '2'],['1','3', '4'],['1', '3']];
console.log(getIntersectVals(arrayOfVals))Lodash intesection if you don't mind
const arrayOfVals = [['1', '2'],['1','3', '4'],['1', '3']];
const result = _.intersection(...arrayOfVals);
console.log(result);.as-console-wrapper{min-height: 100%!important; top: 0}<script src="https://cdn.jsdelivr.net/npm/lodash@4.17.21/lodash.min.js"></script>