Question

我正在研究需要识别面部特征的面部特征的项目。我将此表示为回归问题，并希望从简单的conv网络开始，并在下面定义网络。

我注意到预测的输出总是相同的，之后会有更多的调试，我看到得分层的权重和梯度在迭代时不会改变。我使用~5e-2的固定学习率来生成以下示例。随着迭代的进展，训练损失似乎减少但我无法理解为什么。我还记录了其他图层：'conv1'，'conv2'，'fc1'，并查看了在迭代中保持不变的相同行为。由于损失似乎在减少，所以必须改变一些东西，我的猜测是我在下面记录的方式可能不正确。

你可以给我一些检查指示吗？如果您需要更多信息，请告诉我

修改了lenet：

# Modified LeNet. Added relu1, relu2 and, dropout. 
# Loss function is an Euclidean distance
def lenet(hdf5_list, batch_size=64, dropout_ratio=0.5, train=True):
    # our version of LeNet: a series of linear and simple nonlinear transformations
    n = caffe.NetSpec()

    n.data, n.label = L.HDF5Data(batch_size=batch_size, source=hdf5_list, ntop=2)

    n.conv1 = L.Convolution(n.data, kernel_size=5, num_output=20, weight_filler=dict(type='xavier'), bias_filler=dict(type='constant', value=0.1))
    n.relu1 = L.ReLU(n.conv1, in_place=False, relu_param=dict(negative_slope=0.1))
    n.pool1 = L.Pooling(n.relu1, kernel_size=2, stride=2, pool=P.Pooling.MAX)

    n.conv2 = L.Convolution(n.pool1, kernel_size=5, num_output=50, weight_filler=dict(type='xavier'), bias_filler=dict(type='constant', value=0.1))
    n.relu2 = L.ReLU(n.conv2, in_place=False, relu_param=dict(negative_slope=0.1))
    n.pool2 = L.Pooling(n.relu2, kernel_size=2, stride=2, pool=P.Pooling.MAX)

    if train:
        n.drop3 = fc1_input = L.Dropout(n.pool2, in_place=True, dropout_param = dict(dropout_ratio=dropout_ratio) )
    else:
        fc1_input = n.pool2

    n.fc1 =   L.InnerProduct(fc1_input, num_output=500, weight_filler=dict(type='xavier'), bias_filler=dict(type='constant', value=0.1))
    n.relu3 = L.ReLU(n.fc1, in_place=True, relu_param=dict(negative_slope=0.1))
    n.score = L.InnerProduct(n.relu3, num_output=30, weight_filler=dict(type='xavier'))
    n.loss =  L.EuclideanLoss(n.score, n.label)

    return n.to_proto()

求解器循环：

#custom solver loop
for it in range(niter):
    solver.step(1)

    train_loss[it] = solver.net.blobs['loss'].data

    score_weights.append(solver.net.params['score'][0].data)
    score_biases.append(solver.net.params['score'][1].data)
    score_weights_diff.append(solver.net.params['score'][0].diff)
    score_biases_diff.append(solver.net.params['score'][1].diff)


    if (it % val_interval) == 0 or (it == niter - 1):

        val_error_this = 0
        for test_it in range(niter_val_error):
            solver.test_nets[0].forward()
            val_error_this += euclidean_loss(solver.test_nets[0].blobs['score'].data , 
                                             solver.test_nets[0].blobs['label'].data) / niter_val_error
        val_error[it // val_interval] = val_error_this

打印分数：

print score_weights_diff[0].shape
for i in range(10):
    score_weights_i = score_weights_diff[i]
    print score_weights_i[0:30:10,0]


print score_biases_diff[0].shape
for i in range(5):
    score_biases_i = score_biases_diff[i]
    print score_biases_i[0:30:6]

输出：

(30, 500)
[ -3.71852257e-05   7.34565838e-05   2.61445384e-04]
[ -3.71852257e-05   7.34565838e-05   2.61445384e-04]
[ -3.71852257e-05   7.34565838e-05   2.61445384e-04]
[ -3.71852257e-05   7.34565838e-05   2.61445384e-04]
[ -3.71852257e-05   7.34565838e-05   2.61445384e-04]
[ -3.71852257e-05   7.34565838e-05   2.61445384e-04]
[ -3.71852257e-05   7.34565838e-05   2.61445384e-04]
[ -3.71852257e-05   7.34565838e-05   2.61445384e-04]
[ -3.71852257e-05   7.34565838e-05   2.61445384e-04]
[ -3.71852257e-05   7.34565838e-05   2.61445384e-04]
131
(30,)
[  3.22921231e-04   5.66378840e-05  -5.15143370e-07  -1.51118627e-04
   2.30352176e-04]
[  3.22921231e-04   5.66378840e-05  -5.15143370e-07  -1.51118627e-04
   2.30352176e-04]
[  3.22921231e-04   5.66378840e-05  -5.15143370e-07  -1.51118627e-04
   2.30352176e-04]
[  3.22921231e-04   5.66378840e-05  -5.15143370e-07  -1.51118627e-04
   2.30352176e-04]
[  3.22921231e-04   5.66378840e-05  -5.15143370e-07  -1.51118627e-04
   2.30352176e-04]

Answer 1

从您的代码中看起来有点困难，但score_weights_diff，score_biases_diff和其他列表可能会将引用存储到solver.net.params['score'][0].diff和因此，列表中的所有条目实际上都是相同的，并且在每次迭代时一起变化。

尝试并保存副本：

score_weights_diff.append(solver.net.params['score'][0].diff[...].copy())

在每次迭代后尝试并打印权重/偏差，看它们是否会发生变化。

记录咖啡的重量和偏差

1 个答案: