Dictionary foreach loop with multiple occurences

Question

A little look on what are Dictionaries:

dictWordCounter<string, int> 
dictWordPercent<string, double>
topTwentySeven<string, double>

• WordCounter is the number of words. For example, if the word frog is 10 times it will be {frog, 10}.

• WordPercent is the percent after some math in the program. For example, frog could be {frog, 0,55}.

• topTwentySeven is the top 27 values from the WordPercent dictionary that I want to operate on.

So, I need to count something like this:

• If a word is just one, count it normally
• If a word is twice or more, count it x2
• Total number of words (after x2) must be still 27, not more

For example if it was "topThree" and words:

{Frog, 2}, {Monkey, 1}, {Giraffee, 1}, and their values from Percent Dictionary would be {Frog, 0,1}, {Giraffee, 0,2}, {Monkey, 0,45}, I want to count: 0,1*0,1*0,2 and (1-0,1) * (1-0,1)*(1-0,2).

Could someone look on this and check if it can be improved?

 var topTwentySeven = dictWordPercent.OrderByDescending(kvp => Math.Abs(kvp.Value - 0.5))
 .Take(27)
 .ToDictionary(kvp => kvp.Key, kvp => kvp.Value);
 int compare = 0;
 double temp2 = 1.0, A = 1.0, B = 1.0;
 for (int counter = 0; counter < 27; )
 { 
 foreach (var word in topTwentySeven)
 {
 dictWordCounter.TryGetValue(word.Key, out compare);
 if (compare >= 2 && counter < 26)
 {
 temp = word.Value;
 temp2 = temp * temp;
 A *= temp2;
 B *= (1 - temp2);
 counter = counter + 2;
 }
 else if (compare == 1 && counter < 27)
 {
 temp = word.Value;
 A *= temp;
 B *= (1 - temp);
 counter = counter + 1;
 }
 else if (counter >= 27)
 {
 //Do nothing
 }
 }
 }
int top27C = topTwentySeven.Count();
double aN = 0.0, bN = 0.0;
aN = Math.Pow(A, (1 / top27C));
bN = Math.Pow(B, (1 / top27C));

I need both A and B.

totalSum = Math.Round(((aN / (aN + bN)) * 100), 2);

• Ewan: I knew there was a reason to use . as decimal point
• RubberDuck: Where's the method signature? Can you include that please?
• Ken'ichi Matsuyama: Which method? The one converting WordCounter to WordPercentage?
• Ewan: i'm confused. you say count twice but then square the value? You should be able to reduce this down to a single line of linq. But I dont understand the requirement
• Ken'ichi Matsuyama: @Ewan twice in the meaning of square, I am multiplying ( * ) all the values from dictionary. So if all the values would be single in WordCount I would multiple all of them, but if one of the values would be 2,3,4..n times in the wordCount dictionary I would multiply by this value twice, but keep in mind to count only 27 total values.

Answer 1

Updated. You have some tricky bits to do in linq, but you can see the code is much clearer (although less efficient) without the loop. You could further neaten, by creating a class WordCountAndValue with methods rather than using the anon type and inline ifs

using System;
using System.Linq;
using System.Collections.Generic;
using Microsoft.VisualStudio.TestTools.UnitTesting;
namespace UnitTestProject3
{
 [TestClass]
 public class UnitTest1
 {
 [TestMethod]
 public void TestMethod1()
 {
 Dictionary<string, int> wordCounts = new Dictionary<string, int>();
 wordCounts.Add("t1", 1);
 wordCounts.Add("t2", 2);
 wordCounts.Add("t3", 3);
 wordCounts.Add("t4", 4);
 wordCounts.Add("t5", 5);
 Dictionary<string, double> wordPercent = new Dictionary<string, double>();
 wordPercent.Add("t1", 0.3);
 wordPercent.Add("t2", 0.4);
 wordPercent.Add("t3", 0.5);
 wordPercent.Add("t4", 0.6);
 wordPercent.Add("t5", 0.7);
 var wordsCombined = wordPercent
 .Select(kvp => new { word = kvp.Key, percent = kvp.Value, count = GetCount(wordCounts, kvp.Key) })
 .OrderByDescending(i => i.percent);
 //find number to get
 int target = 5;
 int count = 0;
 var sample = wordsCombined.TakeWhile(i=> target > (count += (i.count >= 2 ? 2 : i.count))).ToList();
 Assert.AreEqual(2, sample.Count());
 //find A
 double A = sample
 .Select(i => i.count >= 2 ? i.percent * i.percent : i.percent)
 .Aggregate(1.0, (t, n) => t * n);
 //find B
 double B = sample
 .Select(i => i.count >= 2 ? i.percent * i.percent : i.percent)
 .Aggregate(1.0, (t, n) => t * (1-n));
 Assert.AreEqual(2, sample.Count());
 Assert.AreEqual((double)0.1764, Math.Round(A,4), "A is incorrect");
 Assert.AreEqual((double)0.3264, Math.Round(B, 4), "B is incorrect");
 }
 public int GetCount(Dictionary<string, int> counts, string key)
 {
 if (!counts.ContainsKey(key))
 {
 return 0;
 }
 return counts[key];
 }
 }
}

Answer 2

Don't write dead code - it rots like a corpse and makes it hard to maintain your code:

else if (counter >= 27)
{
 //Do nothing
}

Just delete this.

---

You can use the ++ increment operator here:

counter = counter + 1;

Like this:

++counter;

---

There is a += operator you can use here:

counter = counter + 2;

Like this:

counter += 2;

---

There is a Math.Pow() method you can use:

temp2 = temp * temp;

Like this:

temp2 = Math.Pow(temp, 2);

• aydjay: I would have written counter++. It's a little easier to read and you don't care if it's incremented before or after the value is accessed.