Ex 2: Restricted Boltzmann Machine features for digit classification

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http://scikit-learn.org/stable/auto_examples/neural_networks/plot_rbm_logistic_classification.html#sphx-glr-auto-examples-neural-networks-plot-rbm-logistic-classification-py

此範例將使用BernoulliRBM特徵選取方法,提升手寫數字識別的精確率,伯努利限制玻爾茲曼機器模型(`BernoulliRBM

`)將可以對數據做有效的非線性 特徵提取的處理。 為了讓此模型訓練出來更為強健,將輸入的圖檔,分別做上左右下,一像素的平移,用以增加更多訓練資料, 訓練網路的參數是使用grid search演算法,但此訓練太耗費時間,因此不再這重現,。 此範例結果將比較, 1.使用原本的像素值做的邏輯回歸 2.使用BernoulliRBM做特徵選取的邏輯回歸 結果將顯示:使用BernoulliRBM將可以提升分類的準確度。

(一)引入函式庫與資料

from __future__ import print_function
print(__doc__)
# Authors: Yann N. Dauphin, Vlad Niculae, Gabriel Synnaeve
# License: BSD
import numpy as np
import matplotlib.pyplot as plt
from scipy.ndimage import convolve
from sklearn import linear_model, datasets, metrics
from sklearn.model_selection import train_test_split
from sklearn.neural_network import BernoulliRBM
from sklearn.pipeline import Pipeline

(二)資料前處理、讀取資料、選取模型

def nudge_dataset(X, Y):
"""
此副函式是用來將輸入資料的數字圖形,分別做上左右下一像素的平移,目的是製造更多的訓練資料讓模型訓練出來更強健
"""
direction_vectors = [
[[0, 1, 0],
[0, 0, 0],
[0, 0, 0]],
[[0, 0, 0],
[1, 0, 0],
[0, 0, 0]],
[[0, 0, 0],
[0, 0, 1],
[0, 0, 0]],
[[0, 0, 0],
[0, 0, 0],
[0, 1, 0]]]
shift = lambda x, w: convolve(x.reshape((8, 8)), mode='constant',
weights=w).ravel()
X = np.concatenate([X] +
[np.apply_along_axis(shift, 1, X, vector)
for vector in direction_vectors])
Y = np.concatenate([Y for _ in range(5)], axis=0)
return X, Y
# Load Data
digits = datasets.load_digits()
X = np.asarray(digits.data, 'float32')
X, Y = nudge_dataset(X, digits.target)
X = (X - np.min(X, 0)) / (np.max(X, 0) + 0.0001) # 將灰階影像降尺度降到[0,1]
# 將資料切割成訓練集與測試集
X_train, X_test, Y_train, Y_test = train_test_split(X, Y,
test_size=0.2,
random_state=0)
# Models we will use
logistic = linear_model.LogisticRegression()
rbm = BernoulliRBM(random_state=0, verbose=True)
classifier = Pipeline(steps=[('rbm', rbm), ('logistic', logistic)])

(三)設定模型參數與訓練模型

# 參數選擇需使用cross-validation去比較
# 此參數是使用GridSearchCV找出來的. Here we are not performing cross-validation to save time.
#GridSratch 就是將參數設定好,跑過全部參數後去找結果最好的一組參數
rbm.learning_rate = 0.06
rbm.n_iter = 20
#.n_components = 100 表示隱藏層單元為100,即表示萃取出100個特徵,特徵萃取的越多準確率會越高,但越耗時間
rbm.n_components = 100
logistic.C = 6000.0
# Training RBM-Logistic Pipeline
classifier.fit(X_train, Y_train)
# Training Logistic regression
logistic_classifier = linear_model.LogisticRegression(C=100.0)
logistic_classifier.fit(X_train, Y_train)

(四)評估模型的分辨準確率

print()
print("Logistic regression using RBM features:\n%s\n" % (
metrics.classification_report(
Y_test,
classifier.predict(X_test))))
print("Logistic regression using raw pixel features:\n%s\n" % (
metrics.classification_report(
Y_test,
logistic_classifier.predict(X_test))))

圖1:使用RBM演算法後準確率為0.95

圖2:不使用任何特徵選取方法做的做的邏輯回歸準確率0.77

(五)畫出100個RBM萃取出的特徵

plt.figure(figsize=(4.2, 4))
for i, comp in enumerate(rbm.components_):
plt.subplot(10, 10, i + 1)
plt.imshow(comp.reshape((8, 8)), cmap=plt.cm.gray_r,
interpolation='nearest')
plt.xticks(())
plt.yticks(())
plt.suptitle('100 components extracted by RBM', fontsize=16)
plt.subplots_adjust(0.08, 0.02, 0.92, 0.85, 0.08, 0.23)
plt.show()

圖3:使用RBM演算法,尋找出來的特徵

(六)完整程式碼

from __future__ import print_function
print(__doc__)
# Authors: Yann N. Dauphin, Vlad Niculae, Gabriel Synnaeve
# License: BSD
import numpy as np
import matplotlib.pyplot as plt
from scipy.ndimage import convolve
from sklearn import linear_model, datasets, metrics
from sklearn.model_selection import train_test_split
from sklearn.neural_network import BernoulliRBM
from sklearn.pipeline import Pipeline
###############################################################################
# Setting up
def nudge_dataset(X, Y):
"""
This produces a dataset 5 times bigger than the original one,
by moving the 8x8 images in X around by 1px to left, right, down, up
"""
direction_vectors = [
[[0, 1, 0],
[0, 0, 0],
[0, 0, 0]],
[[0, 0, 0],
[1, 0, 0],
[0, 0, 0]],
[[0, 0, 0],
[0, 0, 1],
[0, 0, 0]],
[[0, 0, 0],
[0, 0, 0],
[0, 1, 0]]]
shift = lambda x, w: convolve(x.reshape((8, 8)), mode='constant',
weights=w).ravel()
X = np.concatenate([X] +
[np.apply_along_axis(shift, 1, X, vector)
for vector in direction_vectors])
Y = np.concatenate([Y for _ in range(5)], axis=0)
return X, Y
# Load Data
digits = datasets.load_digits()
X = np.asarray(digits.data, 'float32')
X, Y = nudge_dataset(X, digits.target)
X = (X - np.min(X, 0)) / (np.max(X, 0) + 0.0001) # 0-1 scaling
X_train, X_test, Y_train, Y_test = train_test_split(X, Y,
test_size=0.2,
random_state=0)
# Models we will use
logistic = linear_model.LogisticRegression()
rbm = BernoulliRBM(random_state=0, verbose=True)
classifier = Pipeline(steps=[('rbm', rbm), ('logistic', logistic)])
###############################################################################
# Training
# Hyper-parameters. These were set by cross-validation,
# using a GridSearchCV. Here we are not performing cross-validation to
# save time.
rbm.learning_rate = 0.06
rbm.n_iter = 20
# More components tend to give better prediction performance, but larger
# fitting time
rbm.n_components = 100
logistic.C = 6000.0
# Training RBM-Logistic Pipeline
classifier.fit(X_train, Y_train)
# Training Logistic regression
logistic_classifier = linear_model.LogisticRegression(C=100.0)
logistic_classifier.fit(X_train, Y_train)
###############################################################################
# Evaluation
print()
print("Logistic regression using RBM features:\n%s\n" % (
metrics.classification_report(
Y_test,
classifier.predict(X_test))))
print("Logistic regression using raw pixel features:\n%s\n" % (
metrics.classification_report(
Y_test,
logistic_classifier.predict(X_test))))
###############################################################################
# Plotting
plt.figure(figsize=(4.2, 4))
for i, comp in enumerate(rbm.components_):
plt.subplot(10, 10, i + 1)
plt.imshow(comp.reshape((8, 8)), cmap=plt.cm.gray_r,
interpolation='nearest')
plt.xticks(())
plt.yticks(())
plt.suptitle('100 components extracted by RBM', fontsize=16)
plt.subplots_adjust(0.08, 0.02, 0.92, 0.85, 0.08, 0.23)
plt.show()