提问人:Richard Knop 提问时间:11/9/2009 最后编辑:Kevin Ford The SubmarinerRichard Knop 更新时间:10/9/2022 访问量:28300
感知器学习算法未收敛到 0
Perceptron learning algorithm not converging to 0
问:
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:数据集
我已经删除了所有不相关的代码。基本上它现在所做的是读取文件并将值从中加载到三个数组中:、、。test1.txt
x
y
outputs
然后有一个感知器学习算法,由于某种原因,它没有收敛到 0(应该收敛到 0),因此我得到了一个无限的 do while 循环。globalError
当我使用较小的训练集(如 5 分)时,效果很好。任何想法可能出在哪里?
我编写的这个算法与这个 C# 感知器算法非常相似:
编辑:
下面是一个具有较小训练集的示例:
#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;
}
答:
如果你把随机生成器的种子放在main的开头,而不是在每次调用时重新播种,这可能会有所帮助,即randomFloat
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];
评论
我在您的源代码中发现的一些小错误:
int patternCount = sizeof(x) / sizeof(int);
最好将其更改为
int patternCount = i;
因此,您不必依赖 X 数组来获得合适的大小。
您可以在 p 循环中增加迭代次数,而原始 C# 代码在 p 循环之外执行此操作。最好将 printf 和 iteration++ 移到 PAUSE 语句之前的 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,从而产生在 56 到 60 之间振荡的误差。
您可以尝试的最后一件事是在此数据集上测试原始 C# 程序,如果它也没有终止,则算法有问题(因为您的数据集看起来正确,请参阅我的可视化评论)。
评论
globalError
不会变成零,它会像你说的那样收敛到零,即它会变得非常小。
像这样更改您的循环:
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));
给予和适用于您的问题的价值观。maxIterations
maxError
评论
在当前代码中,感知器成功学习了决策边界的方向,但无法转换它。
y y ^ ^ | - + \\ + | - \\ + + | - +\\ + + | - \\ + + + | - - \\ + | - - \\ + | - - + \\ + | - - \\ + + ---------------------> x --------------------> x stuck like this need to get like this
(正如有人指出的那样,这是一个更准确的版本)
问题在于你的感知器没有偏置项,即连接到值 1 的输入的第三个权重分量。
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;
}
...输出如下:
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
下面是上面使用 MATLAB 的代码的简短动画,显示了每次迭代时的决策边界:
评论
y = ax + c
a
c
w0*x + w1*y + w2 = 0
w_i
if (outputs[i] == 0) outputs[i] = -1;
calculateOutput
评论