How to get only Numbers from a string and store it in an Array using Javascript

Question

I have a string that contains a mix of characters and numbers. I want to separate out the numbers from the string and store them in an Array. So far I have been able to separate the strings and numbers and put them in Array and then able to get only numbers but the string type data is NaN. I am unable to get rid of NaN values from the Array.

So far my code looks like this:

arrayList = "abcd_456_yasdb_382_h_83_2";
new_arrayList = arrayList.split('_');
console.log(new_arrayList); // Output : [ 'abcd', '456', 'yasdb', '382', 'h', '83', '2' ]


new_arrayList = new_arrayList.map(Number);
console.log(new_arrayList); // Output :  [ NaN, 456, NaN, 382, NaN, 83, 2 ]

onlyNumArray = []

for(i =0; i<new_arrayList.length; i++) {
    // console.log(new_arrayList[i]);
    if(typeof (new_arrayList[i]) === "number" && typeof(new_arrayList[i]) != NaN) {
        onlyNumArray.push(new_arrayList[i])
    }
    else {
        console.log("Not a number")
    }
}
console.log(onlyNumArray)  // Output : [ NaN, 456, NaN, 382, NaN, 83, 2 ]

Output = [ NaN, 456, NaN, 382, NaN, 83, 2 ]
Expected Output = [456, 382,83,2]

typeof NaN returns 'number', so you'll want to replace that condition with !isNaN(new_arrayList[i]) — samuei
– samuei, Commented Jan 26, 2022 at 14:52

D M · Accepted Answer · 2022-01-26 15:12:18Z

4

Is there a reason you can't use Array#filter? Also, you can't check for NaN by equality (except that NaN != NaN as Passerby points out). You have to use isNaN (or Number#isNaN).

const input = 'abcd_456_yasdb_382_h_83_2';

const tokens = input
  .split('_')
  .map(Number)
  .filter(i => !isNaN(i));

console.log(tokens);

edited Jan 26, 2022 at 15:12

answered Jan 26, 2022 at 14:52

D M

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2 Comments

Passerby Over a year ago

Beat me for a minute :) Technically you can check NaN by equality: .filter((i)=>{return i==i;}) can do the trick too.

D M Over a year ago

True, I hadn't considered checking for the absence of self-equality. Not sure I'd immediately pick up on what was being checked if I came across it in someone else's code.

Sandrow · Accepted Answer · 2022-01-26 14:54:50Z

1

You could use the array.filter() method and do something like this :

const unfilteredArray = [ NaN, 456, NaN, 382, NaN, 83, 2 ]
const numbersArray = unfilteredArray.filter(Number)
// Output : [ 456, 382, 83, 2 ]

answered Jan 26, 2022 at 14:54

Sandrow

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1 Comment

Passerby Over a year ago

.filter(Number) could be problematic when 0 is involved: [83,2,0].filter(Number) -> [83,2]

Alexandru Oriean · Accepted Answer · 2022-01-26 15:02:18Z

0

const newArrayList = arrayList.split('_').filter(element => {
  return Number.parseInt(element);
}).map(element => {
  return Number.parseInt(element)
});

answered Jan 26, 2022 at 15:02

Alexandru Oriean

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1 Comment

Luca Kiebel Over a year ago

While this code snippet may solve the problem, it doesn't explain why or how it answers the question. Please include an explanation for your code, as that really helps to improve the quality of your post. Remember that you are answering the question for readers in the future, and those people might not know the reasons for your code suggestion.

sqram · Accepted Answer · 2022-01-27 21:13:21Z

0

input = 'abcd_456_yasdb_382_h_83_2';
array = [...input.matchAll(/\d+/g)].map(a => parseInt(a[0]));

console.log(array); // [456, 382, 83, 2]

Explanation:

matchAll returns an array of all the regex matches. in our case, it matches sequence of number(s) and return them in an array
The array returned contains extra elements in it. The first element is the actual matched string. So we must run a .map against our returned array, and convert the first value (our number string) into an int with parseInt

edited Jan 27, 2022 at 21:13

answered Jan 26, 2022 at 15:07

sqram

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1 Comment

Luca Kiebel Over a year ago

While this code snippet may solve the problem, it doesn't explain why or how it answers the question. Please include an explanation for your code, as that really helps to improve the quality of your post. Remember that you are answering the question for readers in the future, and those people might not know the reasons for your code suggestion.

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