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:x
,y
,outputs
。
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;
}
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:
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
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maxIterations
和maxError
值。