.NET: Convert a Softmax Loss to a Percentage Accuracy

1. For perfect predictions, the loss approaches 0, which should correspond to 100% accuracy (1.0)
2. For random predictions, the loss approaches -ln(1/numClasses), which should correspond to 0% accuracy (0.0)
3. The function performs a linear interpolation between these points

/// <summary>
/// Converts a softmax cross-entropy loss value to an accuracy percentage (0-1).
/// For a perfectly predicted class, the loss approaches 0, resulting in accuracy approaching 1.
/// For completely random predictions, the loss approaches -ln(1/numClasses), resulting in accuracy approaching 0.
/// </summary>
/// <param name="loss">The cross-entropy loss value from the network</param>
/// <param name="numClasses">Number of possible classes in the classification task</param>
/// <returns>Accuracy value between 0 and 1 (0% to 100%)</returns>
public static float ConvertSoftmaxLossToAccuracy(float loss, int numClasses)
{
	// Calculate the maximum possible loss (random prediction)
	// For random prediction, probability is 1/numClasses for each class
	float maxLoss = -MathF.Log(1.0f / numClasses);

	// Calculate accuracy: 1 when loss is 0, 0 when loss equals maxLoss
	// Using a linear interpolation between these points
	float accuracy = 1.0f - (loss / maxLoss);

	// Clamp the result between 0 and 1 to handle edge cases
	return Math.Max(0.0f, Math.Min(1.0f, accuracy));
}

How to use this static method:

float loss = 0.01f;
int numberOfClasses = 10;
float percentAccuracy = ConvertSoftmaxLossToAccuracy(loss, numberOfClasses);
Console.WriteLine($"{percentAccuracy:P2}");
// --- OR ---
float percentAccuracy2 = ConvertSoftmaxLossToAccuracy(loss, numberOfClasses) * 100;
Console.WriteLine($"{percentAccuracy2:F2}%");

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