I have very long array containing numbers. I need to remove trailing zeros from that array.
if my array will look like this:
var arr = [1,2,0,1,0,1,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0];
I want to remove everything except [1, 2, 0, 1, 0, 1]
.
I have created function that is doing what is expected, but I'm wondering if there is a build in function I could use.
var arr = [1,2,0,1,0,1,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0];
for(i=arr.length-1;i>=0;i--)
{
if(arr[i]==0)
{
arr.pop();
} else {
break;
}
}
console.log(arr);
Can this be done better/faster?
Step 1: Get the string Step 2: Count number of trailing zeros n Step 3: Remove n characters from the beginning Step 4: return remaining string.
x = floor((x * 100) + 0.5)/100; and then print using printf to truncate any trailing zeros.. It is better and simpler to let printf do the rounding by specifying the precision in the format string: %. 2g (or %.
rtrim(0) method if you wish. it breaks if len(items) == 0 .
Assuming:
var arr = [1,2,0,1,0,1,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0];
You can use this shorter code:
while(arr[arr.length-1] === 0){ // While the last element is a 0,
arr.pop(); // Remove that last element
}
Result:
arr == [1,2,0,1,0,1]
var arr = [1,2,0,1,0,1,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0];
var copy = arr.slice(0);
var len = Number(copy.reverse().join('')).toString().length;
arr.length = len;
arr -> [1, 2, 0, 1, 0, 1]
how it works
copy.reverse().join('')
becomes "00000000000000000101021"
when you convert a numerical string to number all the preceding zeroes are kicked off
var len = Number(copy.reverse().join('')) becomes 101021
now by just counting the number i know from where i have to remove the trailing zeroes and the fastest way to delete traling elements is by resetting the length of the array.
arr.length = len;
DEMO
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