trim Array in JS

Question

I Have an array say

['(','C','1','2',')']

Now I want to trim data from array beginning from indexOf('C') + 2 ,If it is a digit I need to remove it from array.. So for the above example final array has to be

 ['(','C','1',')']

For example if I have ['(','C','1','2','3','*',')'] I want it to be trimmed to ['(','C','1','*',')'] , After element 'C' only one numeral is allowed.

I know I can traverse the array by getting the indexOf('C') and then checking each element for numeric.. but help me with some efficient and better way. like using splice or something.

Traversing the array is not anymore inefficient than using .splice() or something.... Since you would have to have used .indexOf() to get the indices to do the splice. Another method is joining the array back into a String and performing regex matching. — Amy
– Amy, Commented Mar 25, 2013 at 5:36
@sweetamylase—joining and using a regular expression may work in trivial cases (per the OP), but where the data is a random string it's very difficult. Regular expressions aren't good for parsing irregular data (even fairly ordered irregular data, like markup). — RobG
– RobG, Commented Mar 25, 2013 at 5:53

KooiInc · Accepted Answer · 2013-03-26 06:22:36Z

3

If the position from where you want to 'trim' is known, you could use filter here, like.:

var a = ['(','C','1','2','3','*',')']
        .filter( function(a){
                   this.x += 1; 
                   return this.x<=2 ? a : isNaN(+a);}, {x:-1} 
         );

Which could lead to this Array extension:

Array.prototype.trimNumbersFrom = function(n){ 
    return this.filter( function(a){
                         this.x += 1; 
                         return this.x<=n ? a : isNaN(+a);
                        }, {x:-1} 
           );
};
//=> usage
var a = ['(','C','1','2','3','*',')'].trimNumbersFrom(2);
    //=> ["(", "C", "1", "*", ")"]

See also ...

edited Mar 26, 2013 at 6:22

answered Mar 25, 2013 at 6:19

KooiInc

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2 Comments

Garry Over a year ago

i tried Array.prototype.trimNumbersFrom but its not workin fine for just one case ... ["(", "C", "1", "0"] ...

KooiInc Over a year ago

Yep, !+a returns true for '0'. IsNaN(+a) doesn"t, edited my answer.

gpasci · Accepted Answer · 2013-03-25 05:59:38Z

2

var a = ['(','C', 'a', '8','2',')'].join("").split("C");
var nPos = a[1].search(/[0-9]/g);
var firstNumber = a[1][nPos];
var b = a[1].split(n);

// rebuild
a = a[0] + "C" + b[0] + firstNumber + b[1].replace(/[0-9]/g, "");

Not tested thoroughly but for your case it works.

answered Mar 25, 2013 at 5:59

gpasci

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Comments

Prasath K · Accepted Answer · 2013-03-25 06:10:33Z

1

You can use of isNaN() function which returns false if its a valid number or true if it's not a number

var str = ['(','C','1','2','3','4','*',')']; // Your Input
var temp = [],count = 0;

for(var i=0;i<str.length;i++)
{
   if(i<str.indexOf('C') + 2)
   {
      temp[count] = str[i];
      count++;
   }
   else
   {
      if(isNaN(str[i]) == true)
      {
        temp[count] = str[i];
        count++;
      }      
   }
 }
str = temp;
alert(str);

LIVE DEMO

answered Mar 25, 2013 at 6:10

Prasath K

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1 Comment

Garry Over a year ago

ur solution is perfect .. even i was doin the same ... was jst lukin for some more efficient way.. thanks ....

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