Filter object properties by key in ES6

If you have a list of allowed values, you can easily retain them in an object using:

const raw = {
  item1: { key: 'sdfd', value:'sdfd' },
  item2: { key: 'sdfd', value:'sdfd' },
  item3: { key: 'sdfd', value:'sdfd' }
};

const allowed = ['item1', 'item3'];

const filtered = Object.keys(raw)
  .filter(key => allowed.includes(key))
  .reduce((obj, key) => {
    obj[key] = raw[key];
    return obj;
  }, {});

console.log(filtered);

This uses:

1. Object.keys to list all properties in raw (the original data), then
2. Array.prototype.filter to select keys that are present in the allowed list, using
   
   
   1. Array.prototype.includes to make sure they are present
3. Array.prototype.reduce to build a new object with only the allowed properties.

This will make a shallow copy with the allowed properties (but won't copy the properties themselves).

You can also use the object spread operator to create a series of objects without mutating them (thanks to rjerue for mentioning this):

const raw = {
  item1: { key: 'sdfd', value:'sdfd' },
  item2: { key: 'sdfd', value:'sdfd' },
  item3: { key: 'sdfd', value:'sdfd' }
};

const allowed = ['item1', 'item3'];

const filtered = Object.keys(raw)
  .filter(key => allowed.includes(key))
  .reduce((obj, key) => {
    return {
      ...obj,
      [key]: raw[key]
    };
  }, {});

console.log(filtered);

For purposes of trivia, if you wanted to remove the unwanted fields from the original data (which I would not recommend doing, since it involves some ugly mutations), you could invert the includes check like so:

const raw = {
  item1: { key: 'sdfd', value:'sdfd' },
  item2: { key: 'sdfd', value:'sdfd' },
  item3: { key: 'sdfd', value:'sdfd' }
};

const allowed = ['item1', 'item3'];

Object.keys(raw)
  .filter(key => !allowed.includes(key))
  .forEach(key => delete raw[key]);

console.log(raw);

I'm including this example to show a mutation-based solution, but I don't suggest using it.

This expression,

(.*"[^,]*),([^,]*".*)

with a replacement of $1 $2 might work.

Demo

Example

using System;
using System.Text.RegularExpressions;

public class Example
{
    public static void Main()
    {
        string pattern = @"(.*""[^,]*),([^,]*"".*)";
        string substitution = @"\1 \2";
        string input = @"-1299-5,""XXX,XXXX"",trft,4,0,10800";
        RegexOptions options = RegexOptions.Multiline;

        Regex regex = new Regex(pattern, options);
        string result = regex.Replace(input, substitution);
    }
}