我來尋找我正在寫的程序的一般提示。神經網絡[ocr]
目標是: 使用神經網絡程序識別3個字母[D,O,M](或者如果我輸入除3之外的任何內容則顯示「什麼都不識別」)。
這是我到目前爲止有:
一類爲我的單神經元
public class neuron
{
double[] weights;
public neuron()
{
weights = null;
}
public neuron(int size)
{
weights = new double[size + 1];
Random r = new Random();
for (int i = 0; i <= size; i++)
{
weights[i] = r.NextDouble()/5 - 0.1;
}
}
public double output(double[] wej)
{
double s = 0.0;
for (int i = 0; i < weights.Length; i++) s += weights[i] * wej[i];
s = 1/(1 + Math.Exp(s));
return s;
}
}
一種用於層類:
public class layer
{
neuron[] tab;
public layer()
{
tab = null;
}
public layer(int numNeurons, int numInputs)
{
tab = new neuron[numNeurons];
for (int i = 0; i < numNeurons; i++)
{
tab[i] = new neuron(numInputs);
}
}
public double[] compute(double[] wejscia)
{
double[] output = new double[tab.Length + 1];
output[0] = 1;
for (int i = 1; i <= tab.Length; i++)
{
output[i] = tab[i - 1].output(wejscia);
}
return output;
}
}
最後一類的網絡
public class network
{
layer[] layers = null;
public network(int numLayers, int numInputs, int[] npl)
{
layers = new layer[numLayers];
for (int i = 0; i < numLayers; i++)
{
layers[i] = new layer(npl[i], (i == 0) ? numInputs : (npl[i - 1]));
}
}
double[] compute(double[] inputs)
{
double[] output = layers[0].compute(inputs);
for (int i = 1; i < layers.Length; i++)
{
output = layers[i].compute(output);
}
return output;
}
}
現在爲algorythm我選擇了:
我有一個圖片框,大小200x200,您可以在其中繪製一個字母(或從jpg文件中讀取一個字母)。
我然後將其轉換爲我的第一個陣列(讓整個畫面)和2一(切周圍的非相關背景),像這樣:
Bitmap bmp2 = new Bitmap(this.pictureBox1.Image);
int[,] binaryfrom = new int[bmp2.Width, bmp2.Height];
int minrow=0, maxrow=0, mincol=0, maxcol=0;
for (int i = 0; i < bmp2.Height; i++)
{
for (int j = 0; j < bmp2.Width; j++)
{
if (bmp2.GetPixel(j, i).R == 0)
{
binaryfrom[i, j] = 1;
if (minrow == 0) minrow = i;
if (maxrow < i) maxrow = i;
if (mincol == 0) mincol = j;
else if (mincol > j) mincol = j;
if (maxcol < j) maxcol = j;
}
else
{
binaryfrom[i, j] = 0;
}
}
}
int[,] boundaries = new int[binaryfrom.GetLength(0)-minrow-(binaryfrom.GetLength(0)-(maxrow+1)),binaryfrom.GetLength(1)-mincol-(binaryfrom.GetLength(1)-(maxcol+1))];
for(int i = 0; i < boundaries.GetLength(0); i++)
{
for(int j = 0; j < boundaries.GetLength(1); j++)
{
boundaries[i, j] = binaryfrom[i + minrow, j + mincol];
}
}
並將其轉換爲12X8像我的最後一個數組所以(我知道我可以這樣縮短公平一點,但希望在不同循環的每一步,所以我可以看到哪裏出了問題更容易[如果有的話確實]):
int[,] finalnet = new int[12, 8];
int k = 1;
int l = 1;
for (int i = 0; i < finalnet.GetLength(0); i++)
{
for (int j = 0; j < finalnet.GetLength(1); j++)
{
finalnet[i, j] = 0;
}
}
while (k <= finalnet.GetLength(0))
{
while (l <= finalnet.GetLength(1))
{
for (int i = (int)(boundaries.GetLength(0)/finalnet.GetLength(0)) * (k - 1); i < (int)(boundaries.GetLength(0)/finalnet.GetLength(0)) * k; i++)
{
for (int j = (int)(boundaries.GetLength(1)/finalnet.GetLength(1)) * (l - 1); j < (int)(boundaries.GetLength(1)/finalnet.GetLength(1)) * l; j++)
{
if (boundaries[i, j] == 1) finalnet[k-1, l-1] = 1;
}
}
l++;
}
l = 1;
k++;
}
int a = boundaries.GetLength(0);
int b = finalnet.GetLength(1);
if((a%b) != 0){
k = 1;
while (k <= finalnet.GetLength(1))
{
for (int i = (int)(boundaries.GetLength(0)/finalnet.GetLength(0)) * finalnet.GetLength(0); i < boundaries.GetLength(0); i++)
{
for (int j = (int)(boundaries.GetLength(1)/finalnet.GetLength(1)) * (k - 1); j < (int)(boundaries.GetLength(1)/finalnet.GetLength(1)) * k; j++)
{
if (boundaries[i, j] == 1) finalnet[finalnet.GetLength(0) - 1, k - 1] = 1;
}
}
k++;
}
}
if (boundaries.GetLength(1) % finalnet.GetLength(1) != 0)
{
k = 1;
while (k <= finalnet.GetLength(0))
{
for (int i = (int)(boundaries.GetLength(0)/finalnet.GetLength(0)) * (k - 1); i < (int)(boundaries.GetLength(0)/finalnet.GetLength(0)) * k; i++)
{
for (int j = (int)(boundaries.GetLength(1)/finalnet.GetLength(1)) * finalnet.GetLength(1); j < boundaries.GetLength(1); j++)
{
if (boundaries[i, j] == 1) finalnet[k - 1, finalnet.GetLength(1) - 1] = 1;
}
}
k++;
}
for (int i = (int)(boundaries.GetLength(0)/finalnet.GetLength(0)) * finalnet.GetLength(0); i < boundaries.GetLength(0); i++)
{
for (int j = (int)(boundaries.GetLength(1)/finalnet.GetLength(1)) * finalnet.GetLength(1); j < boundaries.GetLength(1); j++)
{
if (boundaries[i, j] == 1) finalnet[finalnet.GetLength(0) - 1, finalnet.GetLength(1) - 1] = 1;
}
}
}
結果是12X8(我可以在代碼中更改它以從一些窗體控件中獲取它)0和1的數組,其中1構成粗糙的shap e你畫的一封信。
現在我的問題是: 這是一個正確的algorythm? 是我的功能
1/(1+Math.Exp(x))
好一個在這裏使用? 什麼應該是拓撲結構? 2或3層,如果3,隱藏層中有多少個神經元?我有96個輸入(finalnet數組的每個字段),所以我應該在第一層還需要96個神經元?我應該在最後一層還是4層有3個神經元(考慮到「不承認」的情況),還是沒有必要?
謝謝你的幫助。
編輯:哦,我忘了補充,我要訓練我的網絡使用Backpropagation algorythm。
那麼,在我的演講中,沒有說明是使用雙極性還是單極性功能更好,所以我在網上進行了搜索,但我仍然不確定這一點。 3個輸出神經元看起來更容易是的,但是閾值讓我感到困惑,所以我們假設所有的神經元都返回接近0的值,如0,1; 0,2; 0,15是否可以安全地假設它不識別? – NagashTDN 2014-11-08 10:46:35
@NagashTDN yes – BlackBear 2014-11-08 10:47:08
alrgiht,thank you :) – NagashTDN 2014-11-08 10:53:14