How to sort a numbered list

Question

So I have this numbered list that I need to sort numerically:

1
1.1.
1.1.1.
1.1.2.
1.1.3.
1.1.4.
1.1.5.
1.1.6.
1.1.7.
1.2.
1.2.1.
1.2.2.
1.2.3.
1.2.4.
1.2.5.
1.2.6.
1.3.
1.3.1.
1.3.2.
1.3.3.
1.3.4.
1.4.
1.4.1.
1.4.2.
1.4.3.
1.5.
1.5.1.
1.5.2.
1.5.3.
1.6.
1.6.1.
1.6.1.1.
1.6.1.2.
1.6.2.
1.6.2.1.
1.6.2.2.
1.6.2.3.
1.6.2.4.
1.6.2.5.
1.6.2.6.
1.6.3.
1.6.3.1.
1.6.3.2.
1.6.3.3.
1.6.3.4.
1.6.3.5.
1.6.4.
1.6.5.
1.7.
1.7.1.
1.7.1.1.
1.7.1.2.
1.7.1.3.
1.7.2.
1.7.3.
1.7.4.
1.7.5.
1.7.5.1.
1.7.5.2.
1.7.6.
1.7.7.
1.7.8.
1.7.9.
1.8.
1.9.
1.10.

Issue is when trying to sort 1.10. will come right before 1.2. instead of after 1.9.

Any Idea how I can achieve this using C# or linq?

Thanks.

You're going to need to parse them into some kind of tuple or array of integers and sort those. — 15ee8f99-57ff-4f92-890c-b56153, Oct 09 '19 at 18:56
Assuming they're strings, split them into components and sort by component. — squillman, Oct 09 '19 at 18:56
First split on period, then parse each to an `int`, then order by the first, then second, and so on. — juharr, Oct 09 '19 at 18:56
best way is to sort integers, but if you must deal with strings then rewrite each entry as 2 digit strings such that `1.10` becomes `01.10`, `1.2` becomes `01.02` etc — fnostro, Oct 09 '19 at 18:59
`when trying to sort 1.10. will come right before 1.2` that hapens when they are not numbers but strings (numerals). There are several great posts here on 'Natural Sort' and even how to call/use the one built into Windows — Ňɏssa Pøngjǣrdenlarp, Oct 09 '19 at 19:01

itsme86 · Answer 1 · 2019-10-09T19:17:53.183

1

You can create an IComparer<T> class to help the sorter do what you want:

class MyComparer : IComparer<string>
{
    public int Compare(string x, string y)
    {
        string[] xParts = x.Split(new[] { '.' }, StringSplitOptions.RemoveEmptyEntries);
        string[] yParts = y.Split(new[] { '.' }, StringSplitOptions.RemoveEmptyEntries);

        for (int i = 0;;i++)
        {
            if (xParts.Length >= i && yParts.Length < i)
                return 1;
            if (xParts.Length < i && yParts.Length >= i)
                return -1;
            if (xParts.Length < i && yParts.Length < i)
                return 0;

            int compared = int.Parse(xParts[i]).CompareTo(int.Parse(yParts[i]));
            if (compared != 0)
                return compared;
        }
    }
}

Then you can sort your values like so:

var sorted = values.OrderBy(s => s, new MyComparer()).ToList();

edited Oct 09 '19 at 19:17

answered Oct 09 '19 at 19:08

itsme86

19,266
4
41
57

That doesn't even compile. What if `compared` is 0? – juharr Oct 09 '19 at 19:10
2

@juharr What doesn't compile about it? It does on my computer. If `compared` is 0 it loops again and checks the next value. – itsme86 Oct 09 '19 at 19:10
Not all paths have a return. – juharr Oct 09 '19 at 19:11
2

@juharr Did you *try* to compile it? All paths have a return value. – itsme86 Oct 09 '19 at 19:11
Did not realize the compiler would be smart enough to figure that out. – juharr Oct 09 '19 at 19:13
1

Although this will crash when comparing "1.2." to "1.2.1." since you cannot parse an empty string into an `int`. – juharr Oct 09 '19 at 19:16
@juharr Thanks, I've edited my answer to address that. – itsme86 Oct 09 '19 at 19:18
This will still crash when comparing "1" to "1.5.7". – Selim Yildiz Oct 09 '19 at 19:22
@SelimYıldız It shouldn't. When `i` is 1, the `if (xParts.Length < i && yParts.Length >= i)` will evaluate to true and -1 will be returned. What are you seeing that I'm not? – itsme86 Oct 09 '19 at 20:48

score -1 · Answer 2 · answered Oct 09 '19 at 19:07

-1

you can use Sort method of the List

 List<int> lista = new List<int>() {2,6,1,2,4,9,0,3};
                lista.Sort();

                foreach (var item in lista)
                {
                    Console.WriteLine(item);
                }

                Console.ReadKey();

answered Oct 09 '19 at 19:07

Hairon Chaviano

435
2
5

1

This doesn't address the issue in the question: There is a collection of strings in the form "1.10.1.", "1.2.1.", etc. OP knows how to perform a default sort. The question is how to make "1.10.1." sort ***after*** "1.2.1." – 15ee8f99-57ff-4f92-890c-b56153 Oct 09 '19 at 19:10

How to sort a numbered list

2 Answers2