调试神经网络

时间:2017-02-09 15:39:40

标签: python machine-learning neural-network backpropagation mnist

TLDR

我一直试图在MNIST上安装一个简单的神经网络,它适用于一个小的调试设置,但是当我把它带到MNIST的一个子集时,它训练速度超快,梯度很快接近0 ,但随后它为任何给定的输入输出相同的值,最终成本相当高。我一直试图有目的地过度装备,以确保它实际上工作,但它不会这样做MNIST表明在设置中的深层问题。我已经使用渐变检查检查了我的反向传播实现,它似乎匹配,所以不确定错误在哪里,或者现在要处理什么!

非常感谢您提供的任何帮助,我一直在努力解决这个问题!

解释

根据这个解释,我一直在努力在Numpy建立一个神经网络: http://ufldl.stanford.edu/wiki/index.php/Neural_Networks http://ufldl.stanford.edu/wiki/index.php/Backpropagation_Algorithm

反向传播似乎与渐变检查匹配:

Backpropagation: [ 0.01168585,  0.06629858, -0.00112408, -0.00642625, -0.01339408,
   -0.07580145,  0.00285868,  0.01628148,  0.00365659,  0.0208475 ,
    0.11194151,  0.16696139,  0.10999967,  0.13873069,  0.13049299,
   -0.09012582, -0.1344335 , -0.08857648, -0.11168955, -0.10506167]
Gradient Checking: [-0.01168585 -0.06629858  0.00112408  0.00642625  0.01339408  
    0.07580145 -0.00285868 -0.01628148 -0.00365659 -0.0208475  
   -0.11194151 -0.16696139 -0.10999967 -0.13873069 -0.13049299  
    0.09012582  0.1344335   0.08857648  0.11168955  0.10506167]

当我在这个简单的调试设置上训练时:

a is a neural net w/ 2 inputs -> 5 hidden -> 2 outputs, and learning rate 0.5
a.gradDesc(np.array([[0.1,0.9],[0.2,0.8]]),np.array([[0,1],[0,1]]))
ie. x1 = [0.1, 0.9] and y1 = [0,1]

我得到了这些可爱的训练曲线 error on iterations gradient on iterations

不可否认,这显然是一种愚蠢的,非常容易适应的功能。 但是,只要我将它带到MNIST,就可以使用以下设置:

    # Number of input, hidden and ouput nodes
    # Input = 28 x 28 pixels
    input_nodes=784
    # Arbitrary number of hidden nodes, experiment to improve
    hidden_nodes=200
    # Output = one of the digits [0,1,2,3,4,5,6,7,8,9]
    output_nodes=10

    # Learning rate
    learning_rate=0.4

    # Regularisation parameter
    lambd=0.0

在下面的代码中运行此设置,进行100次迭代,它似乎首先训练然后只是“扁平线”很快并且没有实现非常好的模型:

Initial ===== Cost (unregularised):  2.09203670985  /// Cost (regularised):      2.09203670985    Mean Gradient:  0.0321241229793
Iteration  100  Cost (unregularised):  0.980999805477   /// Cost (regularised):  0.980999805477    Mean Gradient:  -5.29639499854e-09
TRAINED IN  26.45932364463806

这样可以提供非常差的测试精度并预测相同的输出,即使在所有输入为0.1或全部为0.9的情况下进行测试时我也得到相同的输出(尽管它输出的确切数字根据初始随机权重而变化):

Test accuracy:  8.92
Targets    2  2  1  7  2  2  0  2  3  
Hypothesis 5  5  5  5  5  5  5  5  5 

MNIST培训的曲线: enter image description here enter image description here

代码转储:

# Import dependencies
import numpy as np
import time
import csv
import matplotlib.pyplot
import random
import math

# Read in training data
with open('MNIST/mnist_train_100.csv') as file:
    train_data=np.array([list(map(int,line.strip().split(','))) for line in file.readlines()])


# In[197]:

# Plot a sample of training data to visualise
displayData(train_data[:,1:], 25)


# In[198]:

# Read in test data
with open('MNIST/mnist_test.csv') as file:
    test_data=np.array([list(map(int,line.strip().split(','))) for line in file.readlines()])

# Main neural network class
class neuralNetwork:
    # Define the architecture
    def __init__(self, i, h, o, lr, lda):
        # Number of nodes in each layer
        self.i=i
        self.h=h
        self.o=o
        # Learning rate
        self.lr=lr
        # Lambda for regularisation
        self.lda=lda

        # Randomly initialise the parameters, input-> hidden and hidden-> output
        self.ih=np.random.normal(0.0,pow(self.h,-0.5),(self.h,self.i))
        self.ho=np.random.normal(0.0,pow(self.o,-0.5),(self.o,self.h))

    def predict(self, X):
        # GET HYPOTHESIS ESTIMATES/ OUTPUTS
        # Add bias node x(0)=1 for all training examples, X is now m x n+1
        # Then compute activation to hidden node
        z2=np.dot(X,self.ih.T) + 1
        #print(a1.shape)
        a2=sigmoid(z2)
        #print(ha)
        # Add bias node h(0)=1 for all training examples, H is now m x h+1
        # Then compute activation to output node
        z3=np.dot(a2,self.ho.T) + 1
        h=sigmoid(z3)
        outputs=np.argmax(h.T,axis=0)

        return outputs

    def backprop (self, X, y):
        try:
            m = X.shape[0]
        except:
            m=1

        # GET HYPOTHESIS ESTIMATES/ OUTPUTS
        # Add bias node x(0)=1 for all training examples, X is now m x n+1
        # Then compute activation to hidden node
        z2=np.dot(X,self.ih.T)
        #print(a1.shape)
        a2=sigmoid(z2)
        #print(ha)
        # Add bias node h(0)=1 for all training examples, H is now m x h+1
        # Then compute activation to output node
        z3=np.dot(a2,self.ho.T)
        h=sigmoid(z3)

        # Compute error/ cost for this setup (unregularised and regularise)
        costReg=self.costFunc(h,y)
        costUn=self.costFuncReg(h,y)

        # Output error term
        d3=-(y-h)*sigmoidGradient(z3)

        # Hidden error term
        d2=np.dot(d3,self.ho)*sigmoidGradient(z2)

        # Partial derivatives for weights
        D2=np.dot(d3.T,a2)
        D1=np.dot(d2.T,X)

        # Partial derivatives of theta with regularisation
        T2Grad=(D2/m)+(self.lda/m)*(self.ho)
        T1Grad=(D1/m)+(self.lda/m)*(self.ih)

        # Update weights
        # Hidden layer (weights 1)
        self.ih-=self.lr*(((D1)/m) + (self.lda/m)*self.ih)
        # Output layer (weights 2)
        self.ho-=self.lr*(((D2)/m) + (self.lda/m)*self.ho)

        # Unroll gradients to one long vector
        grad=np.concatenate(((T1Grad).ravel(),(T2Grad).ravel()))

        return costReg, costUn, grad

    def backpropIter (self, X, y):
        try:
            m = X.shape[0]
        except:
            m=1

        # GET HYPOTHESIS ESTIMATES/ OUTPUTS
        # Add bias node x(0)=1 for all training examples, X is now m x n+1
        # Then compute activation to hidden node
        z2=np.dot(X,self.ih.T)
        #print(a1.shape)
        a2=sigmoid(z2)
        #print(ha)
        # Add bias node h(0)=1 for all training examples, H is now m x h+1
        # Then compute activation to output node
        z3=np.dot(a2,self.ho.T)
        h=sigmoid(z3)

        # Compute error/ cost for this setup (unregularised and regularise)
        costUn=self.costFunc(h,y)
        costReg=self.costFuncReg(h,y)

        gradW1=np.zeros(self.ih.shape)
        gradW2=np.zeros(self.ho.shape)
        for i in range(m):
            delta3 = -(y[i,:]-h[i,:])*sigmoidGradient(z3[i,:])
            delta2 = np.dot(self.ho.T,delta3)*sigmoidGradient(z2[i,:])

            gradW2= gradW2 + np.outer(delta3,a2[i,:])
            gradW1 = gradW1 + np.outer(delta2,X[i,:])

        # Update weights
        # Hidden layer (weights 1)
        #self.ih-=self.lr*(((gradW1)/m) + (self.lda/m)*self.ih)
        # Output layer (weights 2)
        #self.ho-=self.lr*(((gradW2)/m) + (self.lda/m)*self.ho)

        # Unroll gradients to one long vector
        grad=np.concatenate(((gradW1).ravel(),(gradW2).ravel()))

        return costUn, costReg, grad

    def gradDesc(self, X, y):
        # Backpropagate to get updates
        cost,costreg,grad=self.backpropIter(X,y)

        # Unroll parameters
        deltaW1=np.reshape(grad[0:self.h*self.i],(self.h,self.i))
        deltaW2=np.reshape(grad[self.h*self.i:],(self.o,self.h))

        # m = no. training examples
        m=X.shape[0]
        #print (self.ih)
        self.ih -= self.lr * ((deltaW1))#/m) + (self.lda * self.ih)) 
        self.ho -= self.lr * ((deltaW2))#/m) + (self.lda * self.ho)) 
        #print(deltaW1)
        #print(self.ih)
        return cost,costreg,grad


    # Gradient checking to compute the gradient numerically to debug backpropagation
    def gradCheck(self, X, y):
        # Unroll theta
        theta=np.concatenate(((self.ih).ravel(),(self.ho).ravel()))
        # perturb will add and subtract epsilon, numgrad will store answers
        perturb=np.zeros(len(theta))
        numgrad=np.zeros(len(theta))
        # epsilon, e is a small number
        e = 0.00001
        # Loop over all theta
        for i in range(len(theta)):
            # Perturb is zeros with one index being e
            perturb[i]=e
            loss1=self.costFuncGradientCheck(theta-perturb, X, y)
            loss2=self.costFuncGradientCheck(theta+perturb, X, y)
            # Compute numerical gradient and update vectors
            numgrad[i]=(loss1-loss2)/(2*e)
            perturb[i]=0
        return numgrad

    def costFuncGradientCheck(self,theta,X,y):
        T1=np.reshape(theta[0:self.h*self.i],(self.h,self.i))
        T2=np.reshape(theta[self.h*self.i:],(self.o,self.h))
        m=X.shape[0]
        # GET HYPOTHESIS ESTIMATES/ OUTPUTS
        # Compute activation to hidden node
        z2=np.dot(X,T1.T)
        a2=sigmoid(z2)
        # Compute activation to output node
        z3=np.dot(a2,T2.T)
        h=sigmoid(z3)

        cost=self.costFunc(h, y)
        return cost #+ ((self.lda/2)*(np.sum(pow(T1,2)) + np.sum(pow(T2,2))))

    def costFunc(self, h, y):
        m=h.shape[0]
        return np.sum(pow((h-y),2))/m

    def costFuncReg(self, h, y):
        cost=self.costFunc(h, y)
        return cost #+ ((self.lda/2)*(np.sum(pow(self.ih,2)) + np.sum(pow(self.ho,2))))

# Helper functions to compute sigmoid and gradient for an input number or matrix
def sigmoid(Z):
    return np.divide(1,np.add(1,np.exp(-Z)))
def sigmoidGradient(Z):
    return sigmoid(Z)*(1-sigmoid(Z))

# Pre=processing helper functions
# Normalise data to 0.1-1 as 0 inputs kills the weights and changes
def scaleDataVec(data):
    return (np.asfarray(data[1:]) / 255.0 * 0.99) + 0.1

def scaleData(data):
    return (np.asfarray(data[:,1:]) / 255.0 * 0.99) + 0.1

# DISPLAY DATA
# plot_data will be what to plot, num_ex must be a square number of how many examples to plot, random examples will then be plotted
def displayData(plot_data, num_ex, rand=1):
    if rand==0:
        data=plot_data
    else:
        rand_indexes=random.sample(range(plot_data.shape[0]),num_ex)
        data=plot_data[rand_indexes,:]
    # Useful variables, m= no. train ex, n= no. features
    m=data.shape[0]
    n=data.shape[1]
    # Shape for one example
    example_width=math.ceil(math.sqrt(n))
    example_height=math.ceil(n/example_width)
    # No. of items to display
    display_rows=math.floor(math.sqrt(m))
    display_cols=math.ceil(m/display_rows)
    # Padding between images
    pad=1
    # Setup blank display
    display_array = -np.ones((pad + display_rows * (example_height + pad), (pad + display_cols * (example_width + pad))))
    curr_ex=0
    for i in range(1,display_rows+1):
        for j in range(1,display_cols+1):
            if curr_ex>m:
                break
            # Max value of this patch
            max_val=max(abs(data[curr_ex, :]))
            display_array[pad + (j-1) * (example_height + pad) : j*(example_height+1), pad + (i-1) * (example_width + pad) :                           i*(example_width+1)] = data[curr_ex, :].reshape(example_height, example_width)/max_val
            curr_ex+=1

    matplotlib.pyplot.imshow(display_array, cmap='Greys', interpolation='None')


# In[312]:

a=neuralNetwork(2,5,2,0.5,0.0)
print(a.backpropIter(np.array([[0.1,0.9],[0.2,0.8]]),np.array([[0,1],[0,1]])))
print(a.gradCheck(np.array([[0.1,0.9],[0.2,0.8]]),np.array([[0,1],[0,1]])))
D=[]
C=[]
for i in range(100):
    c,b,d=a.gradDesc(np.array([[0.1,0.9],[0.2,0.8]]),np.array([[0,1],[0,1]]))
    C.append(c)
    D.append(np.mean(d))
    #print(c)

print(a.predict(np.array([[0.1,0.9]])))
# Debugging plot
matplotlib.pyplot.figure()
matplotlib.pyplot.plot(C)
matplotlib.pyplot.ylabel("Error")
matplotlib.pyplot.xlabel("Iterations")
matplotlib.pyplot.figure()
matplotlib.pyplot.plot(D)
matplotlib.pyplot.ylabel("Gradient")
matplotlib.pyplot.xlabel("Iterations")
#print(J)


# In[313]:

# Class instance

# Number of input, hidden and ouput nodes
# Input = 28 x 28 pixels
input_nodes=784
# Arbitrary number of hidden nodes, experiment to improve
hidden_nodes=200
# Output = one of the digits [0,1,2,3,4,5,6,7,8,9]
output_nodes=10

# Learning rate
learning_rate=0.4

# Regularisation parameter
lambd=0.0

# Create instance of Nnet class
nn=neuralNetwork(input_nodes,hidden_nodes,output_nodes,learning_rate,lambd)


# In[314]:

time1=time.time()
# Scale inputs
inputs=scaleData(train_data)
# 0.01-0.99 range as the sigmoid function can't reach 0 or 1, 0.01 for all except 0.99 for target
targets=(np.identity(output_nodes)*0.98)[train_data[:,0],:]+0.01
J=[]
JR=[]
Grad=[]
iterations=100
for i in range(iterations):
    j,jr,grad=nn.gradDesc(inputs, targets)
    grad=np.mean(grad)
    if i == 0:
        print("Initial ===== Cost (unregularised): ", j, "\t///", "Cost (regularised): ",jr,"   Mean Gradient: ",grad)
    print("\r", end="")
    print("Iteration ", i+1, "\tCost (unregularised): ", j, "\t///", "Cost (regularised): ", jr,"   Mean Gradient: ",grad,end="")
    J.append(j)
    JR.append(jr)
    Grad.append(grad)
time2 = time.time()
print ("\nTRAINED IN ",time2-time1)


# In[315]:

# Debugging plot
matplotlib.pyplot.figure()
matplotlib.pyplot.plot(J)
matplotlib.pyplot.plot(JR)
matplotlib.pyplot.ylabel("Error")
matplotlib.pyplot.xlabel("Iterations")
matplotlib.pyplot.figure()
matplotlib.pyplot.plot(Grad)
matplotlib.pyplot.ylabel("Gradient")
matplotlib.pyplot.xlabel("Iterations")
#print(J)


# In[316]:

# Scale inputs
inputs=scaleData(test_data)
# 0.01-0.99 range as the sigmoid function can't reach 0 or 1, 0.01 for all except 0.99 for target
targets=test_data[:,0]
h=nn.predict(inputs)
score=[]
targ=[]
hyp=[]
for i,line in enumerate(targets):
    if line == h[i]:
        score.append(1)
    else:
        score.append(0)
    hyp.append(h[i])
    targ.append(line)
print("Test accuracy: ", sum(score)/len(score)*100)
indexes=random.sample(range(len(hyp)),9)
print("Targets    ",end="")
for j in indexes:
    print (targ[j]," ",end="")
print("\nHypothesis ",end="")
for j in indexes:
    print (hyp[j]," ",end="")
displayData(test_data[indexes, 1:], 9, rand=0)


# In[277]:

nn.predict(0.9*np.ones((784,)))

编辑1

建议使用不同的学习率,但不幸的是,它们都得到了类似的结果,这里是使用MNIST 100子集进行30次迭代的图:

enter image description here enter image description here

具体而言,以下是他们开始和结束的数字:

    Initial ===== Cost (unregularised):  4.07208963507  /// Cost (regularised):  4.07208963507    Mean Gradient:  0.0540251381858
    Iteration  50   Cost (unregularised):  0.613310215166   /// Cost (regularised):  0.613310215166    Mean Gradient:  -0.000133981500849Initial ===== Cost (unregularised):  5.67535252616     /// Cost (regularised):  5.67535252616    Mean Gradient:  0.0644797515914
    Iteration  50   Cost (unregularised):  0.381080434935   /// Cost (regularised):  0.381080434935    Mean Gradient:  0.000427866902699Initial ===== Cost (unregularised):  3.54658422176  /// Cost (regularised):  3.54658422176    Mean Gradient:  0.0672211732868
    Iteration  50   Cost (unregularised):  0.981    /// Cost (regularised):  0.981    Mean Gradient:  2.34515341943e-20Initial ===== Cost (unregularised):  4.05269658215   /// Cost (regularised):  4.05269658215    Mean Gradient:  0.0469666696193
    Iteration  50   Cost (unregularised):  0.980999999999   /// Cost (regularised):  0.980999999999    Mean Gradient:  -1.0582706063e-14Initial ===== Cost (unregularised):  2.40881492228  /// Cost (regularised):  2.40881492228    Mean Gradient:  0.0516056901574
    Iteration  50   Cost (unregularised):  1.74539997258    /// Cost (regularised):  1.74539997258    Mean Gradient:  1.01955789614e-09Initial ===== Cost (unregularised):  2.58498876008   /// Cost (regularised):  2.58498876008    Mean Gradient:  0.0388768685257
    Iteration  3    Cost (unregularised):  1.72520399313    /// Cost (regularised):  1.72520399313    Mean Gradient:  0.0134040908157
    Iteration  50   Cost (unregularised):  0.981    /// Cost (regularised):  0.981    Mean Gradient:  -4.49319474346e-43Initial ===== Cost (unregularised):  4.40141352357  /// Cost (regularised):  4.40141352357    Mean Gradient:  0.0689167742968
    Iteration  50   Cost (unregularised):  0.981    /// Cost (regularised):  0.981    Mean Gradient:  -1.01563966458e-22

学习率0.01,相当低,有最好的结果,但是在这个地区探索学习率,我的准确率只有30-40%,比我有8%甚至0%的大幅提升以前见过,但不是真的应该实现的目标!

编辑2

我现在已经完成并添加了针对矩阵而不是迭代公式优化的反向传播函数,所以现在我可以在大的纪元/迭代上运行它而不会非常缓慢。所以类的“backprop”函数与渐变检查相匹配(实际上它是1/2的大小,但我认为这是梯度检查中的一个问题,所以我们将保留bc它应该不成比例,我试过添加分区来解决这个问题)。随着大量的时代我获得了更好的准确性,但仍然有一个问题,因为当我以前编程一个略有不同风格的简单3层神经网络作为一本书的一部分,在相同的数据集csvs,我获得更好的训练结果。以下是大型时期的一些情节和数据。

enter image description here enter image description here

看起来很不错但是,我们仍然有一个相当差的测试集精度,这是通过数据集进行2,500次运行,应该用更少的数据获得好结果!

    Test accuracy:  61.150000000000006
    Targets    6  9  8  2  2  2  4  3  8  
    Hypothesis 6  9  8  4  7  1  4  3  8

编辑3,什么数据集?

http://makeyourownneuralnetwork.blogspot.co.uk/2015/03/the-mnist-dataset-of-handwitten-digits.html?m=1

使用train.csv和test.csv来尝试更多数据,没有更好的只需要更长时间,所以我在调试时使用了train_100和test_10子集。

编辑4

似乎在大量时期(如14,000)之后学习一些东西,因为整个数据集用于backprop函数(不是backpropiter),每个循环实际上都是一个时代,并且在子集上有一些荒谬的时期100列车和10个测试样品,测试精度相当不错。然而,通过这个小样本,这可能很容易归因于偶然的机会,即便如此,即使在小数据集上也不会达到70%的目标。但它确实表明它似乎在学习,我正在非常广泛地尝试参数来排除这种情况。

1 个答案:

答案 0 :(得分:1)

解决

我解决了我的神经网络。如果它有助于其他任何人,请简要说明。感谢所有帮助提出建议的人。 基本上,我用完全矩阵方法实现了它,即。反向传播每次都使用所有示例。我后来尝试将其作为矢量方法实现,即。每个例子的反向传播。这是当我意识到矩阵方法不会更新每个示例的参数时,因此通过这种方式运行的方法与依次遍历每个示例的方法不同,实际上整个训练集反向传播为一个示例。因此,我的矩阵实现确实有效,但经过多次迭代后,最终需要比矢量方法更长的时间!已经开了一个新问题来了解这个特定部分的更多信息,但是我们去了,它只需要使用矩阵方法进行大量迭代,或者通过示例方法进行更加渐进的示例。