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Question 1
Description: identification number on the picture.
. 1 Import numpy AS NP 2 Import scipy.io AS SCIO . 3 Import matplotlib.pyplot AS PLT . 4 Import scipy.optimize AS OP . 5 . 6 # display image data . 7 DEF displayData (X-): . 8 m = np.size (X-, 0) # number of rows of X, i.e., the number of samples . 9 n-np.size = (X,. 1) # number of columns of X, i.e., a single sample size 10 example_width = int (np.round (np.sqrt (n-))) # single image width . 11 example_height = int (np.floor (n-/ example_width)) # single image height 12 int = display_rows (np.floor (np.sqrt (m))) # shown in the figure, how many line in FIG. 13 is display_cols = int (np.ceil (m / display_rows)) # shown in the figure, a number pictures 14 PAD = 1 # interval between image 15 display_array = - np.ones ((+ display_rows * PAD (example_height + PAD), 16 PAD * + display_cols (example_width + PAD))) # initialization image matrix . 17 curr_ex = 0 # current picture count 18 # to each thumbnail image is inserted into the array . 19 for J in Range (0, display_rows): 20 is for I in range(0, display_cols): 21 if curr_ex >= m: 22 break 23 max_val = np.max(abs(X[curr_ex, :])) 24 jstart = pad + j * (example_height + pad) 25 istart = pad + i * (example_width + pad) 26 display_array[jstart: (jstart + example_height), istart: (istart + example_width)] = \ 27 np.array(X[curr_ex, :]).reshape(example_height, example_width) / max_val 28 curr_ex = curr_ex + 1 29 if curr_ex >= m: 30 break 31 display_array = display_array.T 32 plt.imshow(display_array,cmap=plt.cm.gray) 33 plt.axis('off') 34 plt.show() 35 36 #计算hθ(z) 37 def sigmoid(z): 38 g = 1.0 / (1.0 + np.exp(-z)) 39 return g 40 41 #计算cost 42 def lrCostFunction(theta, X, y, lamb): 43 theta = np.array(theta).reshape((np.size(theta), 1)) 44 m = np.size(y) 45 h = sigmoid(np.dot(X, theta)) 46 J = 1 / m * (-np.dot(y.T, np.log(h)) - np.dot((1 - y.T), np.log(1 - h))) 47 theta2 = theta[1:, 0] 48 Jadd = lamb / (2 * m) * np.sum(theta2 ** 2) 49 J = J + Jadd 50 return J.flatten() 51 52 #计算梯度 53 def gradient(theta, X, y, lamb): 54 theta = np.array(theta).reshape((np.size(theta), 1)) 55 m = np.size(y) 56 h = sigmoid(np.dot(X, theta)) 57 grad = 1/m*np.dot(X.T, h - y) 58 theta[0,0] = 0 59 gradadd = lamb/m*theta 60 grad = grad + gradadd 61 return grad.flatten() 62 63 #θ计算 64 def oneVsAll(X, y, num_labels, lamb): 65 m = np.size(X, 0) 66 n = np.size(X, 1) 67 all_theta = np.zeros((num_labels, n+1)) 68 one = np.ones(m) 69 X = np.insert(X, 0, values=one, axis=1) 70 for c in range(0, num_labels): 71 initial_theta = np.zeros(n+1) 72 y_t = (y==c) 73 result = op.minimize(fun=lrCostFunction, x0=initial_theta, args=(X, y_t, lamb), method='TNC', jac=gradient) 74 all_theta[c, :] = result.x 75 return all_theta 76 77 #计算准确率 78 def predictOneVsAll(all_theta, X): 79 m =np.size (X-, 0) 80 num_labels = np.size (all_theta, 0) 81 P = np.zeros ((m,. 1)) # is used to save the maximum of each row 82 G = np.zeros ((NP .size (X-, 0), num_labels)) # is used to save the results of each classification (classification total of 10 times, each time to save a) 83 One = np.ones (m) 84 X-= np.insert (X-, 0, values One =, = Axis. 1 ) 85 for C in Range (0, num_labels): 86 Theta = all_theta [C,:] 87 G [:, C] = Sigmoid (np.dot (X-, Theta .T)) 88 P = g.argmax (Axis =. 1 ) 89 #Print (P) 90 return p.flatten () 91 is 92 # loading the data file 93 Data = scio.loadmat ( ' ex3data1.mat ' ) 94 X-Data = [ ' X- ' ] 95 Y = Data [ ' Y ' ] 96 Y 10% Y = # because the data set is considered from the start of the matlab 1, save the results to 10 0, modulo here, the back 10 0 97 m = np.size (X-, 0) 98 rand_indices = NP. the random.randint (0, m, 100 ) 99 SEL = X-[rand_indices,:] 100 displayData (SEL) 101 102 # calculating [theta] 103 Lamb = 0.1 104 num_labels = 10 105 all_theta = oneVsAll (X-, Y, num_labels, Lamb) 106 # Print (all_theta) 107 108 # calculating a predicted accuracy of 109 Pred = predictOneVsAll (all_theta, X-) 110 # np.set_printoptions (threshold = np.inf) 111 # encountered a pit on this calculation: 112 # Pred output is [[...]] form, into 1-dimensional vectors need to flatten, y becomes the same with flatten into a one-dimensional vector 113 ACC = np.mean (Pred y.flatten == ()) * 100 114 Print ( ' Training the Set the Accuracy: ' , ACC,'%')
operation result:
Problem 2
Introduction: Using neural network digital identification (Θ provided)
. 1 Import numpy AS NP 2 Import scipy.io AS SCIO . 3 Import matplotlib.pyplot AS PLT . 4 Import scipy.optimize AS OP . 5 . 6 # display image data . 7 DEF displayData (X-): . 8 m = np.size (X-, 0) # number of rows of X, i.e., the number of samples . 9 n-np.size = (X,. 1) # number of columns of X, i.e., a single sample size 10 example_width = int (np.round (np.sqrt (n-))) # single image width . 11 example_height = int (np.floor (n-/ example_width)) # single image height 12 int = display_rows (np.floor (np.sqrt (m))) # shown in the figure, how many line in FIG. 13 is display_cols = int (np.ceil (m / display_rows)) # shown in the figure, a number pictures 14 PAD = 1 # interval between image 15 display_array = - np.ones ((+ display_rows * PAD (example_height + PAD), 16 PAD * + display_cols (example_width + PAD))) # initialization image matrix . 17 curr_ex = 0 # current picture count 18 # to each thumbnail image is inserted into the array . 19 for J in Range (0, display_rows): 20 is for I in range(0, display_cols): 21 if curr_ex >= m: 22 break 23 max_val = np.max(abs(X[curr_ex, :])) 24 jstart = pad + j * (example_height + pad) 25 istart = pad + i * (example_width + pad) 26 display_array[jstart: (jstart + example_height), istart: (istart + example_width)] = \ 27 np.array(X[curr_ex, :]).reshape(example_height, example_width) / max_val 28 curr_ex = curr_ex + 1 29 if curr_ex >= m: 30 break 31 display_array = display_array.T 32 plt.imshow(display_array,cmap=plt.cm.gray) 33 plt.axis('off') 34 plt.show() 35 36 #计算hθ(z) 37 def sigmoid(z): 38 g = 1.0 / (1.0 + np.exp(-z)) 39 return g 40 41 #实现神经网络 42 def predict(theta1, theta2, X): 43 m = np.size(X,0) 44 p = np.zeros((np.size(X, 0), 1)) 45 # of the second layer is calculated 46 is One = np.ones (m) 47 X-np.insert = (X-, 0, values One =, = Axis. 1 ) 48 A2 = Sigmoid (np.dot (X-, theta1.T)) 49 # third layer calculated 50 One = np.ones (np.size (a2,0)) 51 is A2 = np.insert (A2, 0, values One =, = Axis. 1 ) 52 is A3 = Sigmoid (np.dot ( A2, theta2.T)) 53 is P = a3.argmax (Axis =. 1). 1 + # Y value of 1-10, 0-9 to add it here. 1 54 is return p.flatten () 55 56 is # read data file 57 is data = scio.loadmat ( 'ex3data1.mat') 58 X = data['X'] 59 y = data['y'] 60 m = np.size(X, 0) 61 rand_indices = np.random.randint(0,m,100) 62 sel = X[rand_indices, :] 63 # displayData(sel) 64 65 theta = scio.loadmat('ex3weights.mat') 66 theta1 = theta['Theta1'] 67 theta2 = theta['Theta2'] 68 69 # forecast accuracy 70 Pred = Predict (Theta1, Theta2, X-) 71 is ACC = np.mean (Pred y.flatten == ()) * 100 72 Print ( ' Training the Set the Accuracy: ' , ACC, ' % ' ) 73 is 74 # to identify a single image 75 for I in Range (0, m): 76 IT np.random.randint = (0, m,. 1 ) 77 IT = IT [0] 78 displayData (X-[IT: IT + 1'd ,:]) 79 Pred = Predict (Theta1, Theta2, X-[IT: IT +. 1 ,:]) 80 print('Neural Network Prediction:', pred) 81 print('input q to exit:') 82 cin = input() 83 if cin == 'q': 84 break
Neural network analysis matrix representation:
operation result:
