Eric Lippert's challenge "comma-quibbling", best answer?

Question

I wanted to bring this challenge to the attention of the stackoverflow community. The original problem and answers are here. BTW, if you did not follow it before, you should try to read Eric's blog, it is pure wisdom.

Summary:

Write a function that takes a non-null IEnumerable and returns a string with the following characteristics:

1. If the sequence is empty the resulting string is "{}".
2. If the sequence is a single item "ABC" then the resulting string is "{ABC}".
3. If the sequence is the two item sequence "ABC", "DEF" then the resulting string is "{ABC and DEF}".
4. If the sequence has more than two items, say, "ABC", "DEF", "G", "H" then the resulting string is "{ABC, DEF, G and H}". (Note: no Oxford comma!)

As you can see even our very own Jon Skeet (yes, it is well known that he can be in two places at the same time) has posted a solution but his (IMHO) is not the most elegant although probably you can not beat its performance.

What do you think? There are pretty good options there. I really like one of the solutions that involves the select and aggregate methods (from Fernando Nicolet). Linq is very powerful and dedicating some time to challenges like this make you learn a lot. I twisted it a bit so it is a bit more performant and clear (by using Count and avoiding Reverse):

public static string CommaQuibbling(IEnumerable<string> items)
{
    int last = items.Count() - 1;
    Func<int, string> getSeparator = (i) => i == 0 ? string.Empty : (i == last ? " and " : ", ");
    string answer = string.Empty;

    return "{" + items.Select((s, i) => new { Index = i, Value = s })
                      .Aggregate(answer, (s, a) => s + getSeparator(a.Index) + a.Value) + "}";
}

Answer 1

Inefficient, but I think clear.

public static string CommaQuibbling(IEnumerable<string> items)
{
    List<String> list = new List<String>(items);
    if (list.Count == 0) { return "{}"; }
    if (list.Count == 1) { return "{" + list[0] + "}"; }

    String[] initial = list.GetRange(0, list.Count - 1).ToArray();
    return "{" + String.Join(", ", initial) + " and " + list[list.Count - 1] + "}";
}

If I was maintaining the code, I'd prefer this to more clever versions.

Answer 2

How about this approach? Purely cumulative - no back-tracking, and only iterates once. For raw performance, I'm not sure you'll do better with LINQ etc, regardless of how "pretty" a LINQ answer might be.

using System;
using System.Collections.Generic;
using System.Text;

static class Program
{
    public static string CommaQuibbling(IEnumerable<string> items)
    {
        StringBuilder sb = new StringBuilder('{');
        using (var iter = items.GetEnumerator())
        {
            if (iter.MoveNext())
            { // first item can be appended directly
                sb.Append(iter.Current);
                if (iter.MoveNext())
                { // more than one; only add each
                  // term when we know there is another
                    string lastItem = iter.Current;
                    while (iter.MoveNext())
                    { // middle term; use ", "
                        sb.Append(", ").Append(lastItem);
                        lastItem = iter.Current;
                    }
                    // add the final term; since we are on at least the
                    // second term, always use " and "
                    sb.Append(" and ").Append(lastItem);
                }
            }
        }
        return sb.Append('}').ToString();
    }
    static void Main()
    {
        Console.WriteLine(CommaQuibbling(new string[] { }));
        Console.WriteLine(CommaQuibbling(new string[] { "ABC" }));
        Console.WriteLine(CommaQuibbling(new string[] { "ABC", "DEF" }));
        Console.WriteLine(CommaQuibbling(new string[] {
             "ABC", "DEF", "G", "H" }));
    }
}

Answer 3

If I was doing a lot with streams which required first/last information, I'd have thid extension:

[Flags]
public enum StreamPosition
{
   First = 1, Last = 2
}

public static IEnumerable<R> MapWithPositions<T, R> (this IEnumerable<T> stream, 
    Func<StreamPosition, T, R> map)
{
    using (var enumerator = stream.GetEnumerator ())
    {
        if (!enumerator.MoveNext ()) yield break ;

        var cur   = enumerator.Current   ;
        var flags = StreamPosition.First ;
        while (true)
        {
            if (!enumerator.MoveNext ()) flags |= StreamPosition.Last ;
            yield return map (flags, cur) ;
            if ((flags & StreamPosition.Last) != 0) yield break ;
            cur   = enumerator.Current ;
            flags = 0 ;
        }
    }
}

Then the simplest (not the quickest, that would need a couple more handy extension methods) solution will be:

public static string Quibble (IEnumerable<string> strings)
{
    return "{" + String.Join ("", strings.MapWithPositions ((pos, item) => (
       (pos &  StreamPosition.First) != 0      ? "" : 
        pos == StreamPosition.Last   ? " and " : ", ") + item)) + "}" ;
}

Answer 4

Here as a Python one liner

>>> f=lambda s:"{%s}"%", ".join(s)[::-1].replace(',','dna ',1)[::-1]
>>> f([])
'{}'
>>> f(["ABC"])
'{ABC}'
>>> f(["ABC","DEF"])
'{ABC and DEF}'
>>> f(["ABC","DEF","G","H"])
'{ABC, DEF, G and H}'

This version might be easier to understand

>>> f=lambda s:"{%s}"%" and ".join(s).replace(' and',',',len(s)-2)
>>> f([])
'{}'
>>> f(["ABC"])
'{ABC}'
>>> f(["ABC","DEF"])
'{ABC and DEF}'
>>> f(["ABC","DEF","G","H"])
'{ABC, DEF, G and H}'

Answer 5

Here's a simple F# solution, that only does one forward iteration:

let CommaQuibble items =
    let sb = System.Text.StringBuilder("{")
    // pp is 2 previous, p is previous
    let pp,p = items |> Seq.fold (fun (pp:string option,p) s -> 
        if pp <> None then
            sb.Append(pp.Value).Append(", ") |> ignore
        (p, Some(s))) (None,None)
    if pp <> None then
        sb.Append(pp.Value).Append(" and ") |> ignore
    if p <> None then
        sb.Append(p.Value) |> ignore
    sb.Append("}").ToString()

(EDIT: Turns out this is very similar to Skeet's.)

The test code:

let Test l =
    printfn "%s" (CommaQuibble l)

Test []
Test ["ABC"]        
Test ["ABC";"DEF"]        
Test ["ABC";"DEF";"G"]        
Test ["ABC";"DEF";"G";"H"]        
Test ["ABC";null;"G";"H"]        

Answer 6

I'm a fan of the serial comma: I eat, shoot, and leave.

I continually need a solution to this problem and have solved it in 3 languages (though not C#). I would adapt the following solution (in Lua, does not wrap answer in curly braces) by writing a concat method that works on any IEnumerable:

function commafy(t, andword)
  andword = andword or 'and'
  local n = #t -- number of elements in the numeration
  if n == 1 then
    return t[1]
  elseif n == 2 then
    return concat { t[1], ' ', andword, ' ', t[2] }
  else
    local last = t[n]
    t[n] = andword .. ' ' .. t[n]
    local answer = concat(t, ', ')
    t[n] = last
    return answer
  end
end