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I have a simple bitmask, 3 ("011" in base 2) which denotes that I should extract array[0] and array[1] but not array[2]

What is an efficient way to do this?

Ultimately, I'm generating a new array with values that passed a .filter

Something like this:

var bitmask = 37, // "100101"
    array = ["a", "b", "c", "d", "e", "f"];

var array2 = array.filter((value, index) => {
    // do something with bitmask and index to return true
});

// array2 should be ["a", "c", "f"];

I have a simple bitmask, 3 ("011" in base 2) which denotes that I should extract array[0] and array[1] but not array[2]

What is an efficient way to do this?

Ultimately, I'm generating a new array with values that passed a .filter

Something like this:

var bitmask = 37, // "100101"
    array = ["a", "b", "c", "d", "e", "f"];

var array2 = array.filter((value, index) => {
    // do something with bitmask and index to return true
});

// array2 should be ["a", "c", "f"];

• javascript
• algorithm
• bitmask

Share Follow edited Jun 30, 2016 at 2:15 neaumusic asked Jun 30, 2016 at 2:10 neaumusicneaumusic 10.5k1111 gold badges5959 silver badges8686 bronze badges 6

• That's not a bitmask, it's a string. – RJM Commented Jun 30, 2016 at 2:14
• you're right, but you know what i mean, it's actually a number – neaumusic Commented Jun 30, 2016 at 2:14
• Actually covering it to binary then a string would be usefully, that way you can iterate through it and check for "1". – Spencer Wieczorek Commented Jun 30, 2016 at 2:16
• the thing is, I'm iterating from 000001 to 111111 so that i can get all possible selections to fill up to 6 slots, not sure i want to loop through a string as well as the array – neaumusic Commented Jun 30, 2016 at 2:18
• Isn't that supposed to return ['a', 'd', 'f']? Or I get your question wrong.. And what if the bitmask length is higher than the array length? – choz Commented Jun 30, 2016 at 2:20

 |  Show 1 more ment

3 Answers 3

Sorted by: Reset to default 10

Expanding on your original example you can do this:

var bitmask = 37, // "100101"
    array = ["a", "b", "c", "d", "e", "f"];

var array2 = array.filter((value, index) => {
    // do something with bitmask and index to return true
    return bitmask & (1 << index);
});

// array2 should be ["a", "c", "f"];
console.log(array2);

var bitmask = 5, idx=0;
// Loop till bitmask reach 0, works when bitmask >= 0
// If you want to sure instead of implicit type converting (from number to boolean)
// Just change it to while(bitmask >= 0)
while(bitmask){
   // If the rightmost bit is 1, take the array[index]
   if(bitmask & 1) console.log("take arr["+idx+"]");
   // Shift right by 1 bit, say 5 = 101,  this will make the number bee 2 = 10
   bitmask>>=1; 
   // Increase the index 
   idx++;
}

Using your own example, here is the code works:

var bitmask = 37, // "100101"
    array = ["a", "b", "c", "d", "e", "f"],
    idx = 0;
var array2 = [];
while(bitmask){
   if(bitmask & 1) array2.push(array[idx]);
   bitmask>>=1; 
   idx++;
}

Simply use some bit operation to loop it. As it is looping bit by bit, I think it is the fastest you can get

One way to do this is cast your number into a binary string, then check if the index of the bitmask is "1" inside your filter.

    var bitmask = (37).toString(2), // "100101"
        array = ["a", "b", "c", "d", "e", "f"];
    
    var array2 = array.filter((value, index) => {
        if(bitmask[index] == "1") return value;
    });
    
    console.log(array2);

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