javascript filter missing array

Question

I try make a reject function find a missing values in a array:

var store = [0,1,2,3,4,5,6,7,8,15,16,18,20,21];
var fullArray= [0,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20,21];
function reject(array, iteratorFunction) {
    return array.filter(function(element) {
        return !iteratorFunction(element)});
}
var missingValues = reject(fullArray, function(number){
  return store.find(function(item){return number === item});
}); 
console.log(missingValues);

but my result shows:

[0,9,10,11,12,13,14,17,19]

showing me a 0 in the first value. why this occurs and how to fix it?

Ry- · Accepted Answer · 2017-08-12 05:00:36Z

2

store.find returns the element it finds. If that’s 0, your iteratorFunction will return 0, and !0 is true because 0 is falsy.

A correct presence check using Array#includes is even more concise, though:

var missingValues = reject(fullArray, function (number) {
  return store.includes(number);
});

answered Aug 12, 2017 at 5:00

Ry-♦

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Redu · Accepted Answer · 2017-08-12 13:47:23Z

0

You may also do as follows in a more concise way;

var store     = [0,1,2,3,4,5,6,7,8,15,16,18,20,21],
    fullArray = [0,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20,21],
    result    = fullArray.filter(n => !store.includes(n));
console.log(result);

However this would run in O(n.m) so a more efficient way would be;

var store     = [0,1,2,3,4,5,6,7,8,15,16,18,20,21],
    fullArray = [0,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20,21],
    result    = fullArray.reduce((t,m) => (!t[m] && t.r.push(m),t), store.reduce((h,n) => (h[n] = true,h),{r:[]})).r;
console.log(result);

which runs in O(n + m)

answered Aug 12, 2017 at 13:47

Redu

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