高维数据映射为低维数据
降维函数
def transform(self, X):
"""将给定的X,映射到各个主成分分量中"""
assert X.shape[1] == self.components_.shape[1]
return X.dot(self.components_.T)
将降维后的数据恢复成高维
def inverse_transform(self, X):
"""将给定的X,反向映射回原来的特征空间"""
assert X.shape[1] == self.components_.shape[0]
return X.dot(self.components_)
scikit-learn中的PCA
引入数据集
In [6]: import numpy as np
...: import matplotlib.pyplot as plt
...: from sklearn import datasets
In [7]: digits = datasets.load_digits()
...: X = digits.data
...: y = digits.target
划分训练数据集、测试数据集
In [14]: from sklearn.model_selection import train_test_split
...: X_train,X_test,y_train,y_test = train_test_split(X,y,random_state=666)
In [15]: X_train.shape
Out[15]: (1347, 64)
对原始样本进行训练,得到相应的识别率
In [16]: %%time
...: from sklearn.neighbors import KNeighborsClassifier
...: knn_clf = KNeighborsClassifier()
...: knn_clf.fit(X_train,y_train)
...:
Wall time: 137 ms
In [17]: knn_clf.score(X_test,y_test)
Out[17]: 0.9866666666666667
使用PCA进行降维后再训练数据集
降到2维。时间减少,但精度也下降了
[18]: from sklearn.decomposition import PCA
...: pca = PCA(n_components=2)
...: pca.fit(X_train)
...: X_train_reduction = pca.transform(X_train)
#不能使用测试数据集再重新训练一个PCA
#应使用用训练数据集X_train训练过的PCA
...: X_test_reduction = pca.transform(X_test)
#降到2维。时间减少,但精度也下降了
In [19]: %%time
...: knn_clf = KNeighborsClassifier()
...: knn_clf.fit(X_train_reduction,y_train)
Wall time: 964 µs
In [20]: knn_clf.score(X_test_reduction,y_test)
Out[20]: 0.6066666666666667
如何选取最合适的维数
pca.explained_variance_ratio_
每个主成分可以解释的方差是多少
In [23]: pca = PCA(n_components=X_train.shape[1])
...: pca.fit(X_train)
...: pca.explained_variance_ratio_
#第二个参数是前i个特征可解释的方差的和
In [24]: plt.plot([i for i in range(X_train.shape[1])],[np.sum(pca.explained_variance_ratio_[:i+1]) for i in range(X_train.shape[1])])
通过确定的训练精度得到维数
In [25]: pca = PCA(0.95)
...: pca.fit(X_train)
In [26]: pca.n_components_
Out[26]: 28
In [27]: X_train_reduction = pca.transform(X_train)
...: X_test_reduction = pca.transform(X_test)
In [28]: %%time
...: knn_clf = KNeighborsClassifier()
...: knn_clf.fit(X_train_reduction,y_train)
Wall time: 2.05 ms
In [29]: knn_clf.score(X_test_reduction,y_test)
Out[29]: 0.98