CODEWITHSUNDEEP
javascriptstringfor-looptypeof
AziCode
AziCode

Reputation: 2692

Looping through an array of numbers and letters and extract only the numbers?

I would like to loop through an array of numbers and letters and extract only the numbers . I tried using the typeof operator but my code doesn't seems to work

str = "55abcdef6ghij78"

function NumberAddition(str) {
    var numArr = [];
    str = str.split("");

    for (var i = 0; i < str.length; i++) {
        if (typeof str[i] === 'number') {
            numArr.push(str[i]);
        }
        return numArr;
    }
}

Upvotes: 0

Views: 1338

Answers (3)

Kedar
Kedar

Reputation: 1698

Try parsing the character as an Integer first, and then check the type.

if (parseInt(str[i])) {
    numArr.push(str[i]);
}

As pointed out. typeof NaN is also number. So typeof parseInt('a') won't work.

Upvotes: 1

RobG
RobG

Reputation: 147373

Some good answers, but the simplest way to me is to remove the non–digits first, then create the array, so:

var str = '55abcdef6ghij78';
str.replace(/\D/g,'').split(''); // ['5','5','6','7','8'];

Ooops

you want to keep sequences of digits, so match is better:

str.match(/\d+/g); // ["55", "6", "78"]

If you want to sum the numbers, use reduce:

function numberAddition(s) {
  return s.match(/\d+/g).reduce(function(prev, curr){return prev + +curr}, 0);
}

console.log(numberAddition(str)); // 139

noting that a for loop will usually be faster, but not enough to matter unless you're doing this a lot or with very big arrays of numbers.

PS. Don't forget to declare variables, and function names starting with a capital letter are, by convention, reserved for constructors.

Upvotes: 1

thefourtheye
thefourtheye

Reputation: 239463

  1. If you want only the numbers, then there are so many ways to do it. But the simplest would be using a RegEx,

    console.log("55abcdef6ghij78".match(/\d+/g).join(""));
// 55678

    This finds all the consecutive digits (one or more digits) in the string and returns them as an array and we finally join everything together to get a single string.

  2. If you want all the numbers to be separate elements in the array, then you can tweak the RegEx a little bit, like this

    console.log("55abcdef6ghij78".match(/\d/g));
// [ '5', '5', '6', '7', '8' ]

  3. If you want to group the numbers, as you asked for in the comments then just don't join them

    console.log("55abcdef6ghij78".match(/\d+/g));
// [ '55', '6', '78' ]

  4. If you want to convert them to numbers as soon as you group them, you can map the result array to Number function, like this

    console.log("55abcdef6ghij78".match(/\d+/g).map(Number));
// [ 55, 6, 78 ]

Apart from that, if you want to fix your program, then there two major mistakes.

  1. typeof str[i] === 'number'.

    The elements in str will already be of type string only. So, they will never be equal to number. To fix this, you can use the classic trick,

    if (+str[i] == str[i])

    This converts the current string to a number and compares it with the string value of it. If they are equal, then it means that the current element is a digit.

  2. Apart from that, the return statement is within the for loop itself.

So, your program can be fixed like this

function NumberAddition(str) {
    var numArr = [];
    str = str.split("");

    for (var i = 0; i < str.length; i++) {
        if (+str[i] == str[i]) {
            numArr.push(str[i]);
        }
    }
    return numArr;                           // Outside the for loop
}

console.log(NumberAddition("55abcdef6ghij78"));
// [ '5', '5', '6', '7', '8' ]

But the idiomatic way to do this would be to use Array.prototype.filter, like this

function NumberAddition(str) {
    return str.split("").filter(function(currentChar) {
        return +currentChar == currentChar;
    })
}

console.log(NumberAddition("55abcdef6ghij78"));

But, the shorter version would be using Number, like this

console.log("55abcdef6ghij78".split("").filter(Number));
// [ '5', '5', '6', '7', '8' ]

Upvotes: 3

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