Javascript map array then compare keys if statement

Question

I have an array of the alphabet and 2 sets of values, Im trying to check if a value is between the 2 given values then print out the letter. My code:

var alphabet_num = {
"alphabet": ["A", "B", "C", "D", "E", "F", "G", "H", "I", "J", "K", "L", "M", "N", "O", "P"],
"start": ["0", "150", "300", "450", "600", "750", "900", "1050", "1200", "1350", "1500", "1650", "1800", "1950", "2100", "2250"],
"end": ["149", "299", "449", "599", "749", "899", "1049", "1199", "1349", "1499", "1649", "1799", "1949", "2099", "2249", "2399"]
};

var xValue = 76;
var result = alphabet_num.alphabet.map(
(k, i) => [
    k, alphabet_num.start[i], alphabet_num.end[i]

    if (xValue >= alphabet_num.start[i] && xValue <= alphabet_num.end[i]) {
        //a
    }
]
);

Hi yeah I need the if statement part, so if xValue = 76 then it should check start number and end number then print out the letter A, xValue = 150 then its B etc etc — Chris
– Chris, Commented Mar 26, 2022 at 6:10
Mao would return you an array of element. Are you sure you want to use map. — Praveen Kumar Yadav
– Praveen Kumar Yadav, Commented Mar 26, 2022 at 6:20
You should use destructuring to obtain conditional array elements. Something like this: ...(cond ? ['a'] : []) . More about that here : 2ality.com/2017/04/conditional-literal-entries.html — Goran.it
– Goran.it, Commented Mar 26, 2022 at 6:21
I run this code and getting error .map( (k, i) => [.... so this should be .map( (k, i) => { .... and console result show as you want. — Raeesh Alam
– Raeesh Alam, Commented Mar 26, 2022 at 6:23

davidies · Accepted Answer · 2022-03-26 06:22:45Z

2

If you are looking for a specific value, you don't need the map function, you could solve it with find.

const result = alphabet_num.alphabet.find(
  (_letter, i) =>
    parseInt(alphabet_num.start[i]) <= xValue &&
    xValue <= parseInt(alphabet_num.end[i])
);

answered Mar 26, 2022 at 6:22

davidies

564 bronze badges

Sign up to request clarification or add additional context in comments.

Comments

quangdang · Accepted Answer · 2022-03-26 06:27:28Z

1

<script>
var alphabet_num = {
"alphabet": ["A", "B", "C", "D", "E", "F", "G", "H", "I", "J", "K", "L", "M", "N", "O", "P"],
"start": ["0", "150", "300", "450", "600", "750", "900", "1050", "1200", "1350", "1500", "1650", "1800", "1950", "2100", "2250"],
"end": ["149", "299", "449", "599", "749", "899", "1049", "1199", "1349", "1499", "1649", "1799", "1949", "2099", "2249", "2399"]
};


var xValue = 350;

var result = alphabet_num.alphabet.map(
    function(k, i){
        if (xValue >= alphabet_num.start[i] && xValue <= alphabet_num.end[i]) {
            console.log(alphabet_num.alphabet[i]);
            return alphabet_num.alphabet[i];
        }else{
            return '0';
        }
    }

);
// with 350 the array result
// result = ['0', '0', 'C', '0', '0', '0', '0', '0', '0', '0', '0', '0', '0', '0', '0', '0']
console.log(result);

// user findIndex function to find indx
let index = result.findIndex(function (value){
    return value!='0';
});

// the index of 'C' letter in array result index =2
console.log(index);

// get result[2] so I get 'C' letter
console.log(result[index]);
</script>

answered Mar 26, 2022 at 6:27

quangdang

4724 silver badges8 bronze badges

1 Comment

Ziur Olpa Over a year ago

Thanks for your answer, don't forget to include a short description, welcome to SO!

Praveen Kumar Yadav · Accepted Answer · 2022-03-26 06:29:13Z

0

You can use this

    alphabet_numBetweenStartAndEnd = function (alphabet_num, xValue) {
    for (var i = 0; i < alphabet_num.start.length; i++) {
        if (xValue >= alphabet_num.start[i] && xValue <= alphabet_num.end[i]) {
            return alphabet_num.alphabet[i];
        }
    }
}

answered Mar 26, 2022 at 6:29

Praveen Kumar Yadav

1051 silver badge6 bronze badges

Collectives™ on Stack Overflow

Javascript map array then compare keys if statement

3 Answers 3

Comments

1 Comment

Comments

Your Answer

Hot Network Questions

Collectives™ on Stack Overflow

3 Answers 3

Comments

1 Comment

Comments

Your Answer

Sign up or log in

Post as a guest

Related