感知器学习算法不收敛到0

时间:2009-11-08 17:25:51

标签: c algorithm machine-learning neural-network perceptron

这是我在ANSI C中的感知器实现:

#include <stdio.h>
#include <stdlib.h>
#include <math.h>

float randomFloat()
{
    srand(time(NULL));
    float r = (float)rand() / (float)RAND_MAX;
    return r;
}

int calculateOutput(float weights[], float x, float y)
{
    float sum = x * weights[0] + y * weights[1];
    return (sum >= 0) ? 1 : -1;
}

int main(int argc, char *argv[])
{
    // X, Y coordinates of the training set.
    float x[208], y[208];

    // Training set outputs.
    int outputs[208];

    int i = 0; // iterator

    FILE *fp;

    if ((fp = fopen("test1.txt", "r")) == NULL)
    {
        printf("Cannot open file.\n");
    }
    else
    {
        while (fscanf(fp, "%f %f %d", &x[i], &y[i], &outputs[i]) != EOF)
        {
            if (outputs[i] == 0)
            {
                outputs[i] = -1;
            }
            printf("%f   %f   %d\n", x[i], y[i], outputs[i]);
            i++;
        }
    }

    system("PAUSE");

    int patternCount = sizeof(x) / sizeof(int);

    float weights[2];
    weights[0] = randomFloat();
    weights[1] = randomFloat();

    float learningRate = 0.1;

    int iteration = 0;
    float globalError;

    do {
        globalError = 0;
        int p = 0; // iterator
        for (p = 0; p < patternCount; p++)
        {
            // Calculate output.
            int output = calculateOutput(weights, x[p], y[p]);

            // Calculate error.
            float localError = outputs[p] - output;

            if (localError != 0)
            {
                // Update weights.
                for (i = 0; i < 2; i++)
                {
                    float add = learningRate * localError;
                    if (i == 0)
                    {
                        add *= x[p];
                    }
                    else if (i == 1)
                    {
                        add *= y[p];
                    }
                    weights[i] +=  add;
                }
            }

            // Convert error to absolute value.
            globalError += fabs(localError);

            printf("Iteration %d Error %.2f %.2f\n", iteration, globalError, localError);

            iteration++;
        }

        system("PAUSE");

    } while (globalError != 0);

    system("PAUSE");
    return 0;
}

我正在使用的训练集:Data Set

我删除了所有不相关的代码。基本上它现在所做的就是读取test1.txt文件并将其中的值加载到三个数组:xyoutputs

然后有一个perceptron learning algorithm由于某种原因,它没有收敛到0(globalError应该收敛到0),因此我得到一个无限的while while循环。

当我使用较小的训练集(如5分)时,它的效果非常好。任何想法可能是问题?

我写的这个算法与这个C# Perceptron algorithm非常相似:


修改

以下是一个较小训练集的示例:

#include <stdio.h>
#include <stdlib.h>
#include <math.h>

float randomFloat()
{
    float r = (float)rand() / (float)RAND_MAX;
    return r;
}

int calculateOutput(float weights[], float x, float y)
{
    float sum = x * weights[0] + y * weights[1];
    return (sum >= 0) ? 1 : -1;
}

int main(int argc, char *argv[])
{
    srand(time(NULL));

    // X coordinates of the training set.
    float x[] = { -3.2, 1.1, 2.7, -1 };

    // Y coordinates of the training set.
    float y[] = { 1.5, 3.3, 5.12, 2.1 };

    // The training set outputs.
    int outputs[] = { 1, -1, -1, 1 };

    int i = 0; // iterator

    FILE *fp;

    system("PAUSE");

    int patternCount = sizeof(x) / sizeof(int);

    float weights[2];
    weights[0] = randomFloat();
    weights[1] = randomFloat();

    float learningRate = 0.1;

    int iteration = 0;
    float globalError;

    do {
        globalError = 0;
        int p = 0; // iterator
        for (p = 0; p < patternCount; p++)
        {
            // Calculate output.
            int output = calculateOutput(weights, x[p], y[p]);

            // Calculate error.
            float localError = outputs[p] - output;

            if (localError != 0)
            {
                // Update weights.
                for (i = 0; i < 2; i++)
                {
                    float add = learningRate * localError;
                    if (i == 0)
                    {
                        add *= x[p];
                    }
                    else if (i == 1)
                    {
                        add *= y[p];
                    }
                    weights[i] +=  add;
                }
            }

            // Convert error to absolute value.
            globalError += fabs(localError);

            printf("Iteration %d Error %.2f\n", iteration, globalError);          
        }

        iteration++;

    } while (globalError != 0);

    // Display network generalisation.
    printf("X       Y     Output\n");
    float j, k;
    for (j = -1; j <= 1; j += .5)
    {
        for (j = -1; j <= 1; j += .5)
        {
            // Calculate output.
            int output = calculateOutput(weights, j, k);
            printf("%.2f  %.2f  %s\n", j, k, (output == 1) ? "Blue" : "Red");
        }
    }

    // Display modified weights.
    printf("Modified weights: %.2f %.2f\n", weights[0], weights[1]);

    system("PAUSE");
    return 0;
}

4 个答案:

答案 0 :(得分:156)

在您当前的代码中,perceptron成功了解了决策边界的方向,但是无法翻译

    y                              y
    ^                              ^
    |  - + \\  +                   |  - \\ +   +
    | -    +\\ +   +               | -   \\  + +   +
    | - -    \\ +                  | - -  \\    +
    | -  -  + \\  +                | -  -  \\ +   +
    ---------------------> x       --------------------> x
        stuck like this            need to get like this

(有人指出,这里是more accurate version

问题在于你的感知器没有偏差项,即连接到值为1的输入的第三个权重组件。

       w0   -----
    x ---->|     |
           |  f  |----> output (+1/-1)
    y ---->|     |
       w1   -----
               ^ w2
    1(bias) ---|

以下是我纠正问题的方法:

#include <stdio.h>
#include <stdlib.h>
#include <math.h>
#include <time.h>

#define LEARNING_RATE    0.1
#define MAX_ITERATION    100

float randomFloat()
{
    return (float)rand() / (float)RAND_MAX;
}

int calculateOutput(float weights[], float x, float y)
{
    float sum = x * weights[0] + y * weights[1] + weights[2];
    return (sum >= 0) ? 1 : -1;
}

int main(int argc, char *argv[])
{
    srand(time(NULL));

    float x[208], y[208], weights[3], localError, globalError;
    int outputs[208], patternCount, i, p, iteration, output;

    FILE *fp;
    if ((fp = fopen("test1.txt", "r")) == NULL) {
        printf("Cannot open file.\n");
        exit(1);
    }

    i = 0;
    while (fscanf(fp, "%f %f %d", &x[i], &y[i], &outputs[i]) != EOF) {
        if (outputs[i] == 0) {
            outputs[i] = -1;
        }
        i++;
    }
    patternCount = i;

    weights[0] = randomFloat();
    weights[1] = randomFloat();
    weights[2] = randomFloat();

    iteration = 0;
    do {
        iteration++;
        globalError = 0;
        for (p = 0; p < patternCount; p++) {
            output = calculateOutput(weights, x[p], y[p]);

            localError = outputs[p] - output;
            weights[0] += LEARNING_RATE * localError * x[p];
            weights[1] += LEARNING_RATE * localError * y[p];
            weights[2] += LEARNING_RATE * localError;

            globalError += (localError*localError);
        }

        /* Root Mean Squared Error */
        printf("Iteration %d : RMSE = %.4f\n",
            iteration, sqrt(globalError/patternCount));
    } while (globalError > 0 && iteration <= MAX_ITERATION);

    printf("\nDecision boundary (line) equation: %.2f*x + %.2f*y + %.2f = 0\n",
        weights[0], weights[1], weights[2]);

    return 0;
}

...使用以下输出:

Iteration 1 : RMSE = 0.7206
Iteration 2 : RMSE = 0.5189
Iteration 3 : RMSE = 0.4804
Iteration 4 : RMSE = 0.4804
Iteration 5 : RMSE = 0.3101
Iteration 6 : RMSE = 0.4160
Iteration 7 : RMSE = 0.4599
Iteration 8 : RMSE = 0.3922
Iteration 9 : RMSE = 0.0000

Decision boundary (line) equation: -2.37*x + -2.51*y + -7.55 = 0

以下是使用MATLAB的上述代码的简短动画,在每次迭代时显示decision boundary

screenshot

答案 1 :(得分:6)

如果您将随机生成器的播种放在主播的开头而不是每次调用randomFloat时重新播种,即

,这可能会有所帮助。
float randomFloat()
{
    float r = (float)rand() / (float)RAND_MAX;
    return r;
}

// ...

int main(int argc, char *argv[])
{
    srand(time(NULL));

    // X, Y coordinates of the training set.
    float x[208], y[208];

答案 2 :(得分:3)

我在源代码中发现的一些小错误:

int patternCount = sizeof(x) / sizeof(int);

最好将此更改为

int patternCount = i;

因此您不必依赖x数组来获得正确的大小。

您在p循环内增加迭代,而原始C#代码在p循环外执行此操作。最好在PAUSE语句之前将printf和迭代++移到p循环之外 - 我也会删除PAUSE语句或将其更改为

if ((iteration % 25) == 0) system("PAUSE");

即使进行了所有这些更改,您的程序仍然不会使用您的数据集终止,但输出更加一致,错误会在56到60之间振荡。

您可以尝试的最后一件事是测试此数据集上的原始C#程序,如果它也没有终止,则算法出现问题(因为您的数据集看起来正确,请参阅我的可视化注释)。

答案 3 :(得分:1)

globalError不会变为零,如你所说的那样 会收敛到 零,即它会变得非常小。

改变你的循环:

int maxIterations = 1000000; //stop after one million iterations regardless
float maxError = 0.001; //one in thousand points in wrong class

do {
    //loop stuff here

    //convert to fractional error
    globalError = globalError/((float)patternCount);

} while ((globalError > maxError) && (i<maxIterations));

提供适用于您问题的maxIterationsmaxError值。