• 笔记:CS231n+assignment2(作业二)(三)


    终于来到了最终的大BOSS,卷积神经网络~

    这里我想还是主要关注代码的实现,具体的CNN的知识点想以后在好好写一写,CNN的代码关键就是要加上卷积层和池话层.

    一、卷积层

      卷积层的前向传播还是比较容易的,我们主要关注的是反向传播,看下图就知道了:

      

    def conv_forward_naive(x, w, b, conv_param):
        stride, pad = conv_param['stride'], conv_param['pad']
        N, C, H, W = x.shape
        F, C, HH, WW = w.shape
        x_padded = np.pad(x, ((0, 0), (0, 0), (pad, pad), (pad, pad)), mode='constant') #补零
        H_new = 1 + (H + 2 * pad - HH) / stride
        W_new = 1 + (W + 2 * pad - WW) / stride
        s = stride
        out = np.zeros((N, F, H_new, W_new))
    
        for i in xrange(N):       # ith image    
            for f in xrange(F):   # fth filter        
                for j in xrange(H_new):            
                    for k in xrange(W_new):                
                        out[i, f, j, k] = np.sum(x_padded[i, :, j*s:HH+j*s, k*s:WW+k*s] * w[f]) + b[f]#对应位相乘
    
        cache = (x, w, b, conv_param)
    
        return out, cache
    
    
    def conv_backward_naive(dout, cache):
        x, w, b, conv_param = cache
        pad = conv_param['pad']
        stride = conv_param['stride']
        F, C, HH, WW = w.shape
        N, C, H, W = x.shape
        H_new = 1 + (H + 2 * pad - HH) / stride
        W_new = 1 + (W + 2 * pad - WW) / stride
    
        dx = np.zeros_like(x)
        dw = np.zeros_like(w)
        db = np.zeros_like(b)
    
        s = stride
        x_padded = np.pad(x, ((0, 0), (0, 0), (pad, pad), (pad, pad)), 'constant')
        dx_padded = np.pad(dx, ((0, 0), (0, 0), (pad, pad), (pad, pad)), 'constant')
    
        for i in xrange(N):       # ith image    
            for f in xrange(F):   # fth filter        
                for j in xrange(H_new):            
                    for k in xrange(W_new):                
                        window = x_padded[i, :, j*s:HH+j*s, k*s:WW+k*s]
                        db[f] += dout[i, f, j, k]                
                        dw[f] += window * dout[i, f, j, k]                
                        dx_padded[i, :, j*s:HH+j*s, k*s:WW+k*s] += w[f] * dout[i, f, j, k]#上面的式子,关键就在于+号
    
        # Unpad
        dx = dx_padded[:, :, pad:pad+H, pad:pad+W]
    
        return dx, dw, db

      和http://www.cnblogs.com/tornadomeet/p/3468450.html中提到的一样,卷积层的BP算法就是这么计算的,也就是一个正统的卷积操作

    二、pooling层

      

    def max_pool_forward_naive(x, pool_param):
        HH, WW = pool_param['pool_height'], pool_param['pool_width']
        s = pool_param['stride']
        N, C, H, W = x.shape
        H_new = 1 + (H - HH) / s
        W_new = 1 + (W - WW) / s
        out = np.zeros((N, C, H_new, W_new))
        for i in xrange(N):    
            for j in xrange(C):        
                for k in xrange(H_new):            
                    for l in xrange(W_new):                
                        window = x[i, j, k*s:HH+k*s, l*s:WW+l*s] 
                        out[i, j, k, l] = np.max(window)
    
        cache = (x, pool_param)
    
        return out, cache
    
    
    def max_pool_backward_naive(dout, cache):
        x, pool_param = cache
        HH, WW = pool_param['pool_height'], pool_param['pool_width']
        s = pool_param['stride']
        N, C, H, W = x.shape
        H_new = 1 + (H - HH) / s
        W_new = 1 + (W - WW) / s
        dx = np.zeros_like(x)
        for i in xrange(N):    
            for j in xrange(C):        
                for k in xrange(H_new):            
                    for l in xrange(W_new):                
                        window = x[i, j, k*s:HH+k*s, l*s:WW+l*s]                
                        m = np.max(window)               #获得之前的那个值,这样下面只要windows==m就能得到相应的位置
                        dx[i, j, k*s:HH+k*s, l*s:WW+l*s] = (window == m) * dout[i, j, k, l]
    
        return dx

    三、与之前的区别

      这里BN算法与之前是不太一样的,因为网络的输入变成了saptail的

      

    def spatial_batchnorm_forward(x, gamma, beta, bn_param):
        N, C, H, W = x.shape
        x_new = x.transpose(0, 2, 3, 1).reshape(N*H*W, C)#分成不同的channel来算,所以可以直接用之前的代码
        out, cache = batchnorm_forward(x_new, gamma, beta, bn_param)
        out = out.reshape(N, H, W, C).transpose(0, 3, 1, 2)
    
        return out, cache
    
    
    def spatial_batchnorm_backward(dout, cache):
        N, C, H, W = dout.shape
        dout_new = dout.transpose(0, 2, 3, 1).reshape(N*H*W, C)
        dx, dgamma, dbeta = batchnorm_backward(dout_new, cache)
        dx = dx.reshape(N, H, W, C).transpose(0, 3, 1, 2)
    
        return dx, dgamma, dbeta

    四、总结

      assignment2终于弄完了,总的来说..numpy还是要多熟悉,具体的操作也要熟悉。卷积层的前向传播很好理解,反向传播和之前的区别不大,只不过需要做一个卷积的操作。

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  • 原文地址:https://www.cnblogs.com/daihengchen/p/5770129.html
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