Visit the parallel-letter-frequency exercise on Exercism to read the full instructions and download the exercise files.
Dig Deeper
Use AsParallel()
AsParallel
using System.Collections.Generic;
using System.Linq;
public static class ParallelLetterFrequency
{
public static Dictionary<char, int> Calculate(IEnumerable<string> texts)
{
return texts.AsParallel().Aggregate(new Dictionary<char, int>(), AddCount);
}
private static Dictionary<char, int> AddCount(Dictionary<char, int> letterCounts, string text)
{
foreach (var letterCount in text.ToLower().Where(char.IsLetter).GroupBy(c => c))
{
if (letterCounts.TryGetValue(letterCount.Key, out var count))
letterCounts[letterCount.Key] = letterCount.Count() + count;
else
letterCounts[letterCount.Key] = letterCount.Count();
}
return letterCounts;
}
}
The key to this approach is to use the AsParallel() method, which enables parallel LINQ query execution via PLINQ.
We thus start our solution with calling AsParallel() on the IEnumerable<string> representing the texts:
texts.AsParallel()
This will return a ParallelEnumerable instance, which has a lot of methods that look like regular LINQ methods, but are subtly different due to them being executed in parallel.
Having enable parallel processing of each text, we then need to count the letters in that text and then combine all those text counts in the end to end up with the global letter count (with all letters lowercased).
The method that is intended for exactly this purpose is the Aggregate() method.
We’ll use the overload that takes an accumulator value (which is a Dictionary<char, int> representing the total letter count) and a function that takes the accumulator value and a single text, and returns a new, updated accumulator value:
texts.AsParallel().Aggregate(new Dictionary<char, int>(), AddCount)
Let’s move on to the AddCount() method, which signature looks like this:
private static Dictionary<char, int> AddCount(Dictionary<char, int> letterCounts, string text)
Let’s start by counting each letter, and worry about updating the accumulator after that:
text.ToLower().Where(char.IsLetter).GroupBy(c => c)
What this code does is it first converts the text to lowercase, then only keeps letter characters after which we group each letter by its own value.
This will return an IGrouping<char, char> instance, that we then iterate over:
foreach (var letterCount in text.ToLower().Where(char.IsLetter).GroupBy(c => c))
We can get the letter we group on via letterCount.Key and the count via letterCount.Count().
The next step is to update the accumulator value, taking into account the fact that the accumulator might already have a count for the letter we’re processing.
if (letterCounts.TryGetValue(letterCount.Key, out var count))
letterCounts[letterCount.Key] = letterCount.Count() + count;
else
letterCounts[letterCount.Key] = letterCount.Count();
We use TryGetValue() to check if there already is an existing count associated with the key, and if so, we increment the existing value, otherwise we’ll assign a new value.
You don't have to worry about any concurrent updates to the dictionary, PLINQ will handle this for you.
Finally, we’ll return the updated accumulator value:
return letterCounts;
Shortening
When the body of a function is a single expression, the function can be implemented as an expression-bodied member, like so
public static Dictionary<char, int> Calculate(IEnumerable<string> texts) =>
texts.AsParallel().Aggregate(new Dictionary<char, int>(), AddCount);
View Community Solutions
The following are the top 3 community solutions by stars:
Source: Exercism csharp/parallel-letter-frequency