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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.

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.

• javascript
• jquery
• arrays

Share Improve this question Follow edited Mar 25, 2013 at 5:50 000 27.3k1010 gold badges7474 silver badges103103 bronze badges asked Mar 25, 2013 at 5:33 GarryGarry 30611 gold badge77 silver badges2424 bronze badges 3

• Sounds like you're looking for Array.slice() ?? – adeneo Commented Mar 25, 2013 at 5:35
• 4 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 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 Commented Mar 25, 2013 at 5:53

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3 Answers 3

Sorted by: Reset to default 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 ...

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.

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);

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