Add separator to string at every N characters?

Question

I have a string which contains binary digits. How to separate string after each 8 digit?

Suppose the string is:

string x = \"111111110000000011111111         


        

Answer 1

Ugly but less garbage:

private string InsertStrings(string s, int insertEvery, char insert)
{
    char[] ins = s.ToCharArray();
    int length = s.Length + (s.Length / insertEvery);
    if (ins.Length % insertEvery == 0)
    {
        length--;
    }
    var outs = new char[length];
    long di = 0;
    long si = 0;
    while (si < s.Length - insertEvery)
    {
        Array.Copy(ins, si, outs, di, insertEvery);
        si += insertEvery;
        di += insertEvery;
        outs[di] = insert;
        di ++;
    }
    Array.Copy(ins, si, outs, di, ins.Length - si);
    return new string(outs);
}

String overload:

private string InsertStrings(string s, int insertEvery, string insert)
{
    char[] ins = s.ToCharArray();
    char[] inserts = insert.ToCharArray();
    int insertLength = inserts.Length;
    int length = s.Length + (s.Length / insertEvery) * insert.Length;
    if (ins.Length % insertEvery == 0)
    {
        length -= insert.Length;
    }
    var outs = new char[length];
    long di = 0;
    long si = 0;
    while (si < s.Length - insertEvery)
    {
        Array.Copy(ins, si, outs, di, insertEvery);
        si += insertEvery;
        di += insertEvery;
        Array.Copy(inserts, 0, outs, di, insertLength);
        di += insertLength;
    }
    Array.Copy(ins, si, outs, di, ins.Length - si);
    return new string(outs);
}

Answer 2

One way using LINQ:

string data = "111111110000000011111111000000001111111100000000";
const int separateOnLength = 8;

string separated = new string(
    data.Select((x,i) => i > 0 && i % separateOnLength == 0 ? new [] { ',', x } : new [] { x })
        .SelectMany(x => x)
        .ToArray()
    );

Answer 3

A little late to the party, but here's a simplified LINQ expression to break an input string x into groups of n separated by another string sep:

string sep = ",";
int n = 8;
string result = String.Join(sep, x.InSetsOf(n).Select(g => new String(g.ToArray())));

A quick rundown of what's happening here:

• x is being treated as an IEnumberable<char>, which is where the InSetsOf extension method comes in.
• InSetsOf(n) groups characters into an IEnumerable of IEnumerable -- each entry in the outer grouping contains an inner group of n characters.
• Inside the Select method, each group of n characters is turned back into a string by using the String() constructor that takes an array of chars.
• The result of Select is now an IEnumerable<string>, which is passed into String.Join to interleave the sep string, just like any other example.

Answer 4

If I understand your last requirement correctly (it's not clear to me if you need the intermediate comma-delimited string or not), you could do this:

var enumerable = "111111110000000011111111000000001111111100000000".Batch(8).Reverse();

By utilizing morelinq.

Answer 5

I am more than late with my answer but you can use this one:

    static string PutLineBreak(string str, int split)
    {
        for (int a = 1; a <= str.Length; a++)
        {
            if (a % split == 0)
                str = str.Insert(a, "\n");
        }

        return str;
    }

Answer 6

I did it using Pattern & Matcher as following way:

fun addAnyCharacter(input: String, insertion: String, interval: Int): String {
  val pattern = Pattern.compile("(.{$interval})", Pattern.DOTALL)
  val matcher = pattern.matcher(input)
  return matcher.replaceAll("$1$insertion")
}

Where:

input indicates Input string. Check results section.

insertion indicates Insert string between those characters. For example comma (,), start(*), hash(#).

interval indicates at which interval you want to add insertion character.

input indicates Input string. Check results section. Check results section; here I've added insertion at every 4th character.

Results:

I/P: 1234XXXXXXXX5678 O/P: 1234 XXXX XXXX 5678

I/P: 1234567812345678 O/P: 1234 5678 1234 5678

I/P: ABCDEFGHIJKLMNOP O/P: ABCD EFGH IJKL MNOP

Hope this helps.