Why does Array.filter(Number) filter zero out in JavaScript?

Because 0 is one of the many falsy values in javascript

All these conditions will be sent to else blocks:

if (false)
if (null)
if (undefined)
if (0)
if (NaN)
if ('')
if ("")
if (``)

From the Array.prototype.filter() documentation:

filter() calls a provided callback function once for each element in an array, and constructs a new array of all the values for which callback returns a value that coerces to true

In your case the callback function is the Number. So your code is equivalent to:

[-1, 0, 1, 2, 3, 4, Number(0), '', 'test'].filter(a => Number(a)) 

// Number(0) -> 0
// Number(Number(0)) -> 0
// Number('') -> 0
// Number('test') -> NaN

When filter function picks truthy values (or values that coerces to true), the items which return 0 and NaN are ignored. So, it returns [-1, 1, 2, 3, 4]

To prevent a falsy zero from filtering, you could use another callback for getting only numerical values: Number.isFinite

console.log([-1, 0, 1, 2, 3, 4, Number(0), '', 'test'].filter(Number.isFinite))

Expected behavior

This behavior isn't unique to using Number as the filter function. A filter function that simple returns the 0 value would also remove it from the list.

var a = [-1, 0, 1, 2, 3, 4, Number(0), '', 'test'].filter(v => v)
console.log(a); // [-1, 1, 2, 3, 4, "test"]

This is because Number isn't specifically a filter function, it's primarily a type-casting function (and a class constructor, but not a very useful one). So when a number (like 0) is passed to Number, it just returns that number.

Array.prototype.filter removes values that are falsy. In JavaScript, the following are falsy and thus removed by filter.

false
null
undefined
0
NaN
''
""
``

(For complicated backwards compatibility reasons MDN goes into, document.all is also falsy in many browsers despite being an object, but that's a side-note)