How to convert a String[] to an IDictionary<String, String>?

问题

How to convert a String[] to an IDictionary<String, String>?

The values at the indices 0,2,4,... shall be keys, and consequently values at the indices 1,3,5,... shall be values.

Example:

new[] { "^BI", "connectORCL", "^CR", "connectCR" }

=>

new Dictionary<String, String> {{"^BI", "connectORCL"}, {"^CR", "connectCR"}};

回答1:

Dictionary<string,string> ArrayToDict(string[] arr)
{
    if(arr.Length%2!=0)
        throw new ArgumentException("Array doesn't contain an even number of entries");
    Dictionary<string,string> dict=new Dictionary<string,string>();
    for(int i=0;i<arr.Length/2;i++)
    {
      string key=arr[2*i];
      string value=arr[2*i+1];
      dict.Add(key,value);
    }
    return dict;
}

回答2:

I'd recommend a good old for loop for clarity. But if you insist on a LINQ query, this should work:

var dictionary = Enumerable.Range(0, array.Length/2)
                           .ToDictionary(i => array[2*i], i => array[2*i+1])

回答3:

There's really no easy way to do this in LINQ (And even if there were, it's certainly not going to be clear as to the intent). It's easily accomplished by a simple loop though:

// This code assumes you can guarantee your array to always have an even number
// of elements.

var array = new[] { "^BI", "connectORCL", "^CR", "connectCR" };
var dict = new Dictionary<string, string>();

for(int i=0; i < array.Length; i+=2)
{
    dict.Add(array[i], array[i+1]);
}

回答4:

Something like this maybe:

        string[] keyValues = new string[20];
        Dictionary<string, string> dict = new Dictionary<string, string>();
        for (int i = 0; i < keyValues.Length; i+=2)
        {
            dict.Add(keyValues[i], keyValues[i + 1]);
        }

Edit: People in the C# tag are damn fast...

回答5:

If you have Rx as a dependency you can do:

strings
    .BufferWithCount(2)
    .ToDictionary(
         buffer => buffer.First(), // key selector
         buffer => buffer.Last()); // value selector

BufferWithCount(int count) takes the first count values from the input sequence and yield them as a list, then it takes the next count values and so on. I.e. from your input sequence you will get the pairs as lists: {"^BI", "connectORCL"}, {"^CR", "connectCR"}, the ToDictionary then takes the first list item as key and the last ( == second for lists of two items) as value.

However, if you don't use Rx, you can use this implementation of BufferWithCount:

static class EnumerableX
{
    public static IEnumerable<IList<T>> BufferWithCount<T>(this IEnumerable<T> source, int count)
    {
        if (source == null)
        {
            throw new ArgumentNullException("source");
        }

        if (count <= 0)
        {
            throw new ArgumentOutOfRangeException("count");
        }

        var buffer = new List<T>();

        foreach (var t in source)
        {
            buffer.Add(t);

            if (buffer.Count == count)
            {
                yield return buffer;
                buffer = new List<T>();
            }
        }

        if (buffer.Count > 0)
        {
            yield return buffer;
        }
    }
}

回答6:

It looks like other people have already beaten me to it and/or have more efficient answers but I'm posting 2 ways:

A for loop might be the clearest way to accomplish in this case...

var words = new[] { "^BI", "connectORCL", "^CR", "connectCR" };

var final = words.Where((w, i) => i % 2 == 0)
                 .Select((w, i) => new[] { w, words[(i * 2) + 1] })
                 .ToDictionary(arr => arr[0], arr => arr[1])
                 ;

final.Dump();

//alternate way using zip

var As = words.Where((w, i) => i % 2 == 0);
var Bs = words.Where((w, i) => i % 2 == 1);

var dictionary = new Dictionary<string, string>(As.Count());

var pairs = As.Zip(Bs, (first, second) => new[] {first, second})
                .ToDictionary(arr => arr[0], arr => arr[1])
                ;

pairs.Dump();

回答7:

FYI, this is what I ended up with using a loop and implementing it as an extension method:

internal static Boolean IsEven(this Int32 @this)
{
    return @this % 2 == 0;
}

internal static IDictionary<String, String> ToDictionary(this String[] @this)
{
    if (!@this.Length.IsEven())
        throw new ArgumentException( "Array doesn't contain an even number of entries" );

    var dictionary = new Dictionary<String, String>();

    for (var i = 0; i < @this.Length; i += 2)
    {
        var key = @this[i];
        var value = @this[i + 1];

        dictionary.Add(key, value);
    }

    return dictionary;
}

回答8:

Pure Linq

1. Select : Project original string value and its index.
2. GroupBy : Group adjacent pairs.
3. Convert each group into dictionary entry.

string[] arr = new string[] { "^BI", "connectORCL", "^CR", "connectCR" };

var dictionary = arr.Select((value,i) => new {Value = value,Index = i})
                .GroupBy(value => value.Index / 2)
                .ToDictionary(g => g.FirstOrDefault().Value, 
                                   g => g.Skip(1).FirstOrDefault().Value);

来源：https://stackoverflow.com/questions/4670279/how-to-convert-a-string-to-an-idictionarystring-string