Let, I have two arrays with strings.
let a = ["ABC", "DEF"] let b = ["ABC", "DEF", "GHI"] I want to compare the above two arrays and make a new array called c which contains the following.
c = ["GHI"] So, I want to compare a and b arrays and push the non-identical elements into a new array.
How to do this in React Native?
You can do that in following steps:
a which are not in bb which are not in aconcat() both the arrays.let a = ["ABC", "DEF"] let b = ["ABC", "DEF", "GHI"] const res = a.filter(x => !b.includes(x)).concat(b.filter(x=> !a.includes(x))) console.log(res)Consider the line
a.filter(x => !b.includes(x)) filter() is array method which takes a callback. If callback returns true then the element will be added to result array otherwise not.
So now in above code x is element of array b through which we are iterating. b.includes(x) will return true if x is present in array b otherwise false.
! operator converts true to false and vice verse
So if x will be inside b it will return true. So by ! it will be become false and false will be returned from callback. So the element x will not be included in result/filtered array.
The sentence for the line above line is that "It get only those items of array a which are not present in b"
The second line
b.filter(x=> !a.includes(x)) Gets those elements of b which are not in a.
Atlast concat() is used to join both array.
You can also use Array.reduce and Array.filter like so:
const findUniques = (a,b) => [...a,...b].reduce((r,c,i,a) => { a.filter(x => x===c).length === 1 ? r.push(c) : null return r }, []) console.log(findUniques(["ABC", "DEF"], ["ABC", "DEF", "GHI"])) console.log(findUniques(["ABC", "DEF"], ["ABC"])) console.log(findUniques(["ABC", "DEF", "AAA"], ["AAA", "DEF", "GHI"]))If you are using lodash this becomes really trivial with xor:
console.log(_.xor(["ABC", "DEF"], ["ABC", "DEF", "GHI"])) console.log(_.xor(["ABC", "DEF"], ["DEF"])) console.log(_.xor(["ABC", "DEF", "AAA"], ["AAA", "DEF", "GHI"]))<script src="https://cdnjs.cloudflare.com/ajax/libs/lodash.js/4.17.11/lodash.min.js"></script>
x===c. Then those elements we push to the accumulator r. For those filtered values which have more than one hit we do nothing. Hence null. Hope this helps.
If you want to prevent going through the arrays O(n²) times, due to having to search each array for duplicate values, you could take advantage of the performance of JavaScript Objects.
const a = ["ABC", "DEF", "JKL"] const b = ["ABC", "DEF", "GHI"] const getUniqueValues = (a, b) => { const uniqueValuesObject = [...a, ...b].reduce((uniqueValuesObject, value) => { // if the value already exists, then it is not unique if (uniqueValuesObject[value]) { uniqueValuesObject[value] = false } else { uniqueValuesObject[value] = true } return uniqueValuesObject }, {}) return Object.entries(uniqueValuesObject).reduce((uniqueValues, [value, isUnique]) => { if (isUnique) { uniqueValues.push(value) } return uniqueValues }, []) } const c = getUniqueValues(a, b) console.log(c)This can be shortened by using delete, however I am unsure if that will cause a performance hit compared to the solution above.
const a = ["ABC", "DEF", "JKL"] const b = ["ABC", "DEF", "GHI"] const getUniqueValues = (a, b) => { const uniqueValuesObject = [...a, ...b].reduce((uniqueValuesObject, value) => { // delete if the value already exists if (uniqueValuesObject[value]) { delete uniqueValuesObject[value] } else { uniqueValuesObject[value] = true } return uniqueValuesObject }, {}) return Object.keys(uniqueValuesObject) } const c = getUniqueValues(a, b) console.log(c)
Objects are implemented in whatever JavaScript engine you are using. For simplicity, if we assume that Objects are simply hash tables, then with the first solution it is possible to have an Object lookup of O(1), making the solution O(n) since all it does it traverse both arrays twice. Hash tables can have a worst time complexity of O(n) if many elements are hashed to the same key, which could give this O(n²) complexity, but it's unlikely. At the very least, it is much faster than searching the array using filter.
balways contain every element ofa?a=['ABC','ABC']andb = ['DEF']what will be result?