Question 193244 in Javascript Question posted in
A very good W3school tutorial can be found here.

In JavaScript, how to easily remove boolean values from an array consisting of numbers and booleans?

In JavaScript, I have an array containing boolean values and integer numbers, eg

[1, true, 2, false, 3]

I want to get an array with the boolean values removed. I tried

> [1, true, 2, false, 3].filter(Number)
[ 1, true, 2, 3 ]

Strangely, JavaScript thinks that false is not a Number, but true is. How so?

If my array contains strings, it works.

> [1, '', 2, 'foo', 3].filter(Number)
[ 1, 2, 3 ]

I know I can write the following, I was hoping for an easier way.

> [1, true, 2, false, 3].filter(x => typeof x === 'number')
[ 1, 2, 3 ]

What’s the easiest way to achieve what I want?

Thanks

Tags:

3

Answers


  1. 0

    The .filter() method interprets the return value of the callback as boolean. If you use Number() as the filter callback, everything that evaluates to 0 or NaN will be excluded from the result.

    What you really want to do is check the type of each value, which you can do with typeof.

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  2. 0

    In case of just targeting integer or even safe integer values one could do …

    [1, true, 2, false, 3].filter(Number.isInteger);

    … respectively …

    [1, true, 2, false, 3].filter(Number.isSafeInteger);

    … thats pretty much the easiest to go with.

    And from another of my above comments …

    The "easiest" way is also the one that expresses exactly what is going to be achieved. Thus, the very expressive / readable / comprehensible last variant presented by the OP is the one to go with … [1, true, 2, false, 3].filter(x => typeof x === 'number') …

    … Of cause one could be more expressive by writing an isNumber function which implements a number validation. Then the code becomes even more expressive …

    function isNumber(value) {
  return Object
    .prototype
    .toString
    .call(value) === '[object Number]';
}

function isNumberValue(value) {
  return typeof value === 'number';
}
function isFiniteNumberValue(value) {
  return isNumberValue(value) && Number.isFinite(value);
}
const testData = [1, true, 2, '', 3, 0/0, new Number(1111)];

console.log(
  'any number ...',
  testData.filter(isNumber)
);
console.log(
  'any number value ...',
  testData.filter(isNumberValue)
);
console.log(
  'any finite number value ...',
  testData.filter(isFiniteNumberValue)
);
    .as-console-wrapper { min-height: 100%!important; top: 0; }
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  3. 0

    Using the input provided in your question, try this:
    [1, true, 2, false, 3].filter(el => (el === 0 || parseInt(el)));
    Or try this simpler one [1, true, 2, false, 3].filter(Number.isInteger);

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