How to check for a non existing value in a map typescript?

Question

I have the below map in typescript:

myMap = {"a" => 1, "b" => 2, "c" => 4, "d" => 5}

The values can be 1 , 2 , 3, 4, or 5 only.

I need to check for the values in the myMap from 1 to 5 and if any of the value is missing, i want to log it and exit the loop. Eg: In the above map, value 3 is missing. So, i would want to log that value.

Can anyone let me know, how this could be achieved in typescript ?

Answer 1

You could create a Set of the values in Map . Then filter the numbers array and check if the set has each number

 const numbers = [1, 2, 3, 4, 5], map = new Map([ ["a", 1], ["b", 2], ["c", 4] ]), set = new Set(map.values()), missing = numbers.filter(n => !set.has(n)) console.log(missing)

Answer 2

That something more javascript specific then typescript.

var myMap = new Map();
myMap.set("a", 1);
myMap.set("b", 2);

let hasThree = [...myMap.values()].includes(3);

console.log(hasThree);

You can use .values() and spread it into an array and use then .includes

Here is an image that shows all methods of the Map prototype.

Answer 3

Taking into consideration that values are 1 to 5 , you could change array as well and use some other values with specific range and use it with map with this -

 var map = new Map(); map.set('a', 1) map.set('b', 2) map.set('c',4) map.set('d', 5) var arr = []; for(var singleVal of map.values()) arr.push(singleVal) var originalArr = [ 1, 2, 3, 4, 5]; // taking into consideration that you have some original array like this, where '3' is present for(var i=0; i<originalArr.length; i++) if(!arr.includes(originalArr[i])) { console.log(originalArr[i]); break; }