IGrouping <int, Keypair> 到字典

Question

I seperated my main dictonary into a group of multiple dictonaries so that I can shoot threads off with the smaller set of dictonaries.

Here is the code that seperates my main Dictonary into multiple smaller ones:

int numberOfGroups = 10;
int counter = 0;
var result = offenderWorkload.GroupBy(x => counter++ % numberOfGroups);

Now I am having trouble to obtain one dictonary from the set of 10 within the result.

I need to seperate result into 10 different dictonaries and shoot off threads so something like...

foreach(var something in result)
{
    Dictionary<String, int> workLoad = (Dictionary<String, int>)something.ToDictionary();
    Console.WriteLine("workload: " + something.Key + " has " + workLoad.Keys.Count);
}

So that something is keypair that gets casted to a dictonary and ignoring the int in the Igroup.

Answer 1

Dictionary<String, int> workLoad = something.ToDictionary(x => x.Key, x => x.Value);

因为在进行分组之后，每个分组成员都不是字典，而是KeyValuePair的集合。

Answer 2

This will give you a Group of dictionaries. each dictionary have n elements.

int n = (int)Math.Ceiling((double)offenderWorkload.Count/numberOfGroups);

IEnumerable<Dictionary<string, int>> result =
    offenderWorkload.GroupBy(x => counter++/n)
        .Select(x => x.ToDictionary(d => d.Key, d => d.Value));

Note that if your dictionary have 12 keys and you want to group them in to 5 dictionaries you will get 4 dictionaries instead, each one have 3 keys.

Because 12/5 = 2.4 . you can have 5 dictionaries with 2 keys plus one dictionary with two keys 2*5+0.4*5 . In total 6 dictionaries which is more than maximum number of groups. Or better way, by taking its ceiling you will get maximum possible (or less) than amount of specified groups. Here [2.4] = 3 . in total 4 dictionaries each have 3 keys.

Another example if you want to group 12 into 7 dictionaries you will get 5 dictionaries instead each one have 2 keys. If you group 13 into 3 you will get 3 dictionaries, Two first dictionaries will have 5 keys and Last one have 3 keys.

In order to get maximum numberOfGroups that you have specified choose something that offenderWorkload.Count can be divided to. for example If you group 12 into 6 you will get 6 dictionaries each one have 2 keys ( 12 / 6 = 2 ).

int numberOfGroups = 10;
int counter = 0;
int n = (int)Math.Ceiling((double)offenderWorkload.Count/numberOfGroups);

var result =
    offenderWorkload.GroupBy(x => counter++/n)
        .Select(x => x.ToDictionary(d => d.Key, d => d.Value));

int i = 0;
foreach (var workLoad in result)
{
    Console.WriteLine("workload: " + i++ + " has " + workLoad.Keys.Count);
}