从头开始实施主成分分析以不同于 scikit-learn 的方式定位数据
Implementation of Principal Component Analysis from Scratch Orients the Data Differently than scikit-learn
根据指南Implementing PCA in Python, by Sebastian Raschka我正在从头构建 PCA 算法用于我的研究目的。 class 定义为:
import numpy as np
class PCA(object):
"""Dimension Reduction using Principal Component Analysis (PCA)
It is the procces of computing principal components which explains the
maximum variation of the dataset using fewer components.
:type n_components: int, optional
:param n_components: Number of components to consider, if not set then
`n_components = min(n_samples, n_features)`, where
`n_samples` is the number of samples, and
`n_features` is the number of features (i.e.,
dimension of the dataset).
Attributes
==========
:type covariance_: np.ndarray
:param covariance_: Coviarance Matrix
:type eig_vals_: np.ndarray
:param eig_vals_: Calculated Eigen Values
:type eig_vecs_: np.ndarray
:param eig_vecs_: Calculated Eigen Vectors
:type explained_variance_: np.ndarray
:param explained_variance_: Explained Variance of Each Principal Components
:type cum_explained_variance_: np.ndarray
:param cum_explained_variance_: Cumulative Explained Variables
"""
def __init__(self, n_components : int = None):
"""Default Constructor for Initialization"""
self.n_components = n_components
def fit_transform(self, X : np.ndarray):
"""Fit the PCA algorithm into the Dataset"""
if not self.n_components:
self.n_components = min(X.shape)
self.covariance_ = np.cov(X.T)
# calculate eigens
self.eig_vals_, self.eig_vecs_ = np.linalg.eig(self.covariance_)
# explained variance
_tot_eig_vals = sum(self.eig_vals_)
self.explained_variance_ = np.array([(i / _tot_eig_vals) * 100 for i in sorted(self.eig_vals_, reverse = True)])
self.cum_explained_variance_ = np.cumsum(self.explained_variance_)
# define `W` as `d x k`-dimension
self.W_ = self.eig_vecs_[:, :self.n_components]
print(X.shape, self.W_.shape)
return X.dot(self.W_)
以iris-dataset作为测试用例,实现PCA并可视化如下:
import numpy as np
import pandas as pd
import seaborn as sns
import matplotlib.pyplot as plt
# loading iris data, and normalize
from sklearn.datasets import load_iris
iris = load_iris()
from sklearn.preprocessing import MinMaxScaler
X, y = iris.data, iris.target
X = MinMaxScaler().fit_transform(X)
# using the PCA function (defined above)
# to fit_transform the X value
# naming the PCA object as dPCA (d = defined)
dPCA = PCA()
principalComponents = dPCA.fit_transform(X)
# creating a pandas dataframe for the principal components
# and visualize the data using scatter plot
PCAResult = pd.DataFrame(principalComponents, columns = [f"PCA-{i}" for i in range(1, dPCA.n_components + 1)])
PCAResult["target"] = y # possible as original order does not change
sns.scatterplot(x = "PCA-1", y = "PCA-2", data = PCAResult, hue = "target", s = 50)
plt.show()
输出如下:
现在,我想验证输出,为此我使用了sklearn
库,输出如下:
from sklearn.decomposition import PCA # note the same name
sPCA = PCA() # consider all the components
principalComponents_ = sPCA.fit_transform(X)
PCAResult_ = pd.DataFrame(principalComponents_, columns = [f"PCA-{i}" for i in range(1, 5)])
PCAResult_["target"] = y # possible as original order does not change
sns.scatterplot(x = "PCA-1", y = "PCA-2", data = PCAResult_, hue = "target", s = 50)
plt.show()
我不明白为什么输出的方向不同,值略有不同。我研究了很多代码 [1, 2, 3],所有代码都有同样的问题。我的问题:
sklearn
有什么不同,就是剧情不同?我也尝试过使用不同的数据集 - 同样的问题。
- 有办法解决这个问题吗?
我无法研究 sklearn.decompose.PCA
算法,因为我对 python 的 OOP 概念不熟悉。
Sebastian Raschka 的博客 post 中的输出也有细微差别。下图:
计算特征向量时,您可能 change its sign 并且解也是有效的。
因此任何 PCA 轴都可以反转,解决方案将有效。
不过,您可能希望 PCA 轴与数据集中的原始变量之一强加正相关,并在需要时反转轴。
值的差异来自使用 svd 分解的 sklearn 的 PCA。在 sklearn 中有一个函数 svd_flip
用于翻转 PC,这解释了为什么你会看到这个翻转
有关帮助的更多详细信息 page:
It uses the LAPACK implementation of the full SVD or a randomized
truncated SVD by the method of Halko et al. 2009, depending on the
shape of the input data and the number of components to extract.
您可以阅读有关关系 here
我们首先运行您的示例数据集:
from sklearn.preprocessing import MinMaxScaler
from sklearn.decomposition import PCA
from sklearn.datasets import load_iris
from sklearn.utils.extmath import svd_flip
import pandas as pd
import numpy as np
import scipy
iris = load_iris()
X, y = iris.data, iris.target
X = MinMaxScaler().fit_transform(X)
n_components = 4
sPCA = PCA(n_components,svd_solver="full")
sklearnPCs = pd.DataFrame(sPCA.fit_transform(X))
我们现在对您的居中矩阵执行 SVD:
U,S,Vt = scipy.linalg.svd(X - X.mean(axis=0))
U = U[:,:n_components]
U, Vt = svd_flip(U, Vt)
svdPCs = pd.DataFrame(U*S)
结果:
0 1 2 3
0 -0.630703 0.107578 -0.018719 -0.007307
1 -0.622905 -0.104260 -0.049142 -0.032359
2 -0.669520 -0.051417 0.019644 -0.007434
3 -0.654153 -0.102885 0.023219 0.020114
4 -0.648788 0.133488 0.015116 0.011786
.. ... ... ... ...
145 0.551462 0.059841 0.086283 -0.110092
146 0.407146 -0.171821 -0.004102 -0.065241
147 0.447143 0.037560 0.049546 -0.032743
148 0.488208 0.149678 0.239209 0.002864
149 0.312066 -0.031130 0.118672 0.052505
svdPCs
0 1 2 3
0 -0.630703 0.107578 -0.018719 -0.007307
1 -0.622905 -0.104260 -0.049142 -0.032359
2 -0.669520 -0.051417 0.019644 -0.007434
3 -0.654153 -0.102885 0.023219 0.020114
4 -0.648788 0.133488 0.015116 0.011786
.. ... ... ... ...
145 0.551462 0.059841 0.086283 -0.110092
146 0.407146 -0.171821 -0.004102 -0.065241
147 0.447143 0.037560 0.049546 -0.032743
148 0.488208 0.149678 0.239209 0.002864
149 0.312066 -0.031130 0.118672 0.052505
您可以在不翻转的情况下实现。值将相同,您的 PCA 将如其他答案中所述有效。
根据指南Implementing PCA in Python, by Sebastian Raschka我正在从头构建 PCA 算法用于我的研究目的。 class 定义为:
import numpy as np
class PCA(object):
"""Dimension Reduction using Principal Component Analysis (PCA)
It is the procces of computing principal components which explains the
maximum variation of the dataset using fewer components.
:type n_components: int, optional
:param n_components: Number of components to consider, if not set then
`n_components = min(n_samples, n_features)`, where
`n_samples` is the number of samples, and
`n_features` is the number of features (i.e.,
dimension of the dataset).
Attributes
==========
:type covariance_: np.ndarray
:param covariance_: Coviarance Matrix
:type eig_vals_: np.ndarray
:param eig_vals_: Calculated Eigen Values
:type eig_vecs_: np.ndarray
:param eig_vecs_: Calculated Eigen Vectors
:type explained_variance_: np.ndarray
:param explained_variance_: Explained Variance of Each Principal Components
:type cum_explained_variance_: np.ndarray
:param cum_explained_variance_: Cumulative Explained Variables
"""
def __init__(self, n_components : int = None):
"""Default Constructor for Initialization"""
self.n_components = n_components
def fit_transform(self, X : np.ndarray):
"""Fit the PCA algorithm into the Dataset"""
if not self.n_components:
self.n_components = min(X.shape)
self.covariance_ = np.cov(X.T)
# calculate eigens
self.eig_vals_, self.eig_vecs_ = np.linalg.eig(self.covariance_)
# explained variance
_tot_eig_vals = sum(self.eig_vals_)
self.explained_variance_ = np.array([(i / _tot_eig_vals) * 100 for i in sorted(self.eig_vals_, reverse = True)])
self.cum_explained_variance_ = np.cumsum(self.explained_variance_)
# define `W` as `d x k`-dimension
self.W_ = self.eig_vecs_[:, :self.n_components]
print(X.shape, self.W_.shape)
return X.dot(self.W_)
以iris-dataset作为测试用例,实现PCA并可视化如下:
import numpy as np
import pandas as pd
import seaborn as sns
import matplotlib.pyplot as plt
# loading iris data, and normalize
from sklearn.datasets import load_iris
iris = load_iris()
from sklearn.preprocessing import MinMaxScaler
X, y = iris.data, iris.target
X = MinMaxScaler().fit_transform(X)
# using the PCA function (defined above)
# to fit_transform the X value
# naming the PCA object as dPCA (d = defined)
dPCA = PCA()
principalComponents = dPCA.fit_transform(X)
# creating a pandas dataframe for the principal components
# and visualize the data using scatter plot
PCAResult = pd.DataFrame(principalComponents, columns = [f"PCA-{i}" for i in range(1, dPCA.n_components + 1)])
PCAResult["target"] = y # possible as original order does not change
sns.scatterplot(x = "PCA-1", y = "PCA-2", data = PCAResult, hue = "target", s = 50)
plt.show()
输出如下:
现在,我想验证输出,为此我使用了sklearn
库,输出如下:
from sklearn.decomposition import PCA # note the same name
sPCA = PCA() # consider all the components
principalComponents_ = sPCA.fit_transform(X)
PCAResult_ = pd.DataFrame(principalComponents_, columns = [f"PCA-{i}" for i in range(1, 5)])
PCAResult_["target"] = y # possible as original order does not change
sns.scatterplot(x = "PCA-1", y = "PCA-2", data = PCAResult_, hue = "target", s = 50)
plt.show()
我不明白为什么输出的方向不同,值略有不同。我研究了很多代码 [1, 2, 3],所有代码都有同样的问题。我的问题:
sklearn
有什么不同,就是剧情不同?我也尝试过使用不同的数据集 - 同样的问题。- 有办法解决这个问题吗?
我无法研究 sklearn.decompose.PCA
算法,因为我对 python 的 OOP 概念不熟悉。
Sebastian Raschka 的博客 post 中的输出也有细微差别。下图:
计算特征向量时,您可能 change its sign 并且解也是有效的。
因此任何 PCA 轴都可以反转,解决方案将有效。
不过,您可能希望 PCA 轴与数据集中的原始变量之一强加正相关,并在需要时反转轴。
值的差异来自使用 svd 分解的 sklearn 的 PCA。在 sklearn 中有一个函数 svd_flip
用于翻转 PC,这解释了为什么你会看到这个翻转
有关帮助的更多详细信息 page:
It uses the LAPACK implementation of the full SVD or a randomized truncated SVD by the method of Halko et al. 2009, depending on the shape of the input data and the number of components to extract.
您可以阅读有关关系 here
我们首先运行您的示例数据集:
from sklearn.preprocessing import MinMaxScaler
from sklearn.decomposition import PCA
from sklearn.datasets import load_iris
from sklearn.utils.extmath import svd_flip
import pandas as pd
import numpy as np
import scipy
iris = load_iris()
X, y = iris.data, iris.target
X = MinMaxScaler().fit_transform(X)
n_components = 4
sPCA = PCA(n_components,svd_solver="full")
sklearnPCs = pd.DataFrame(sPCA.fit_transform(X))
我们现在对您的居中矩阵执行 SVD:
U,S,Vt = scipy.linalg.svd(X - X.mean(axis=0))
U = U[:,:n_components]
U, Vt = svd_flip(U, Vt)
svdPCs = pd.DataFrame(U*S)
结果:
0 1 2 3
0 -0.630703 0.107578 -0.018719 -0.007307
1 -0.622905 -0.104260 -0.049142 -0.032359
2 -0.669520 -0.051417 0.019644 -0.007434
3 -0.654153 -0.102885 0.023219 0.020114
4 -0.648788 0.133488 0.015116 0.011786
.. ... ... ... ...
145 0.551462 0.059841 0.086283 -0.110092
146 0.407146 -0.171821 -0.004102 -0.065241
147 0.447143 0.037560 0.049546 -0.032743
148 0.488208 0.149678 0.239209 0.002864
149 0.312066 -0.031130 0.118672 0.052505
svdPCs
0 1 2 3
0 -0.630703 0.107578 -0.018719 -0.007307
1 -0.622905 -0.104260 -0.049142 -0.032359
2 -0.669520 -0.051417 0.019644 -0.007434
3 -0.654153 -0.102885 0.023219 0.020114
4 -0.648788 0.133488 0.015116 0.011786
.. ... ... ... ...
145 0.551462 0.059841 0.086283 -0.110092
146 0.407146 -0.171821 -0.004102 -0.065241
147 0.447143 0.037560 0.049546 -0.032743
148 0.488208 0.149678 0.239209 0.002864
149 0.312066 -0.031130 0.118672 0.052505
您可以在不翻转的情况下实现。值将相同,您的 PCA 将如其他答案中所述有效。