【Python第三方包】scikit-learn
KFold与StratifiedKFold的区别
class sklearn.model_selection.StratifiedKFold(n_splits=3, shuffle=False, random_state=None)
Stratified K-Folds cross-validator Provides train/test indices to split data in train/test sets.This cross-validation object is a variation of KFold that returns stratified folds. The folds are made by preserving the percentage of samples for each class
意思就是Stra~是按着不同类别标签的相对占比来进行的分折
看看代码
import sklearn.model_selection as skmodsel
print('假设有100个样本，X为对应特征矩阵，y为对应的类标签')
X = [[i] for i in range(100)]
y = ['a'] * 30 + ['b'] * 30 + ['c'] * 30 + ['d'] * 10
print('用KFold进行分折')
K_folds = skmodsel.KFold(n_splits=10)
for train_indices, test_indices in K_folds.split(X):
print('test set:', test_indices)
d = {}
for i in train_indices:
d[y[i]] = d.setdefault(y[i], 0) + 1
print(d)
K_strafold = skmodsel.StratifiedKFold(n_splits=10)
print('用StratifiedKFold进行分折')
for train_indices, test_indices in K_strafold.split(X, y):
print('test set:', test_indices)
d = {}
for i in train_indices:
d[y[i]] = d.setdefault(y[i], 0) + 1
print(d)
结果如下
假设有100个样本，X为对应特征矩阵，y为对应的类标签
用KFold进行分折
test set: [0 1 2 3 4 5 6 7 8 9]
{'a': 20, 'b': 30, 'c': 30, 'd': 10}
test set: [10 11 12 13 14 15 16 17 18 19]
{'a': 20, 'b': 30, 'c': 30, 'd': 10}
test set: [20 21 22 23 24 25 26 27 28 29]
{'a': 20, 'b': 30, 'c': 30, 'd': 10}
test set: [30 31 32 33 34 35 36 37 38 39]
{'a': 30, 'b': 20, 'c': 30, 'd': 10}
test set: [40 41 42 43 44 45 46 47 48 49]
{'a': 30, 'b': 20, 'c': 30, 'd': 10}
test set: [50 51 52 53 54 55 56 57 58 59]
{'a': 30, 'b': 20, 'c': 30, 'd': 10}
test set: [60 61 62 63 64 65 66 67 68 69]
{'a': 30, 'b': 30, 'c': 20, 'd': 10}
test set: [70 71 72 73 74 75 76 77 78 79]
{'a': 30, 'b': 30, 'c': 20, 'd': 10}
test set: [80 81 82 83 84 85 86 87 88 89]
{'a': 30, 'b': 30, 'c': 20, 'd': 10}
test set: [90 91 92 93 94 95 96 97 98 99]
{'a': 30, 'b': 30, 'c': 30}
用StratifiedKFold进行分折
test set: [ 0 1 2 30 31 32 60 61 62 90]
{'a': 27, 'b': 27, 'c': 27, 'd': 9}
test set: [ 3 4 5 33 34 35 63 64 65 91]
{'a': 27, 'b': 27, 'c': 27, 'd': 9}
test set: [ 6 7 8 36 37 38 66 67 68 92]
{'a': 27, 'b': 27, 'c': 27, 'd': 9}
test set: [ 9 10 11 39 40 41 69 70 71 93]
{'a': 27, 'b': 27, 'c': 27, 'd': 9}
test set: [12 13 14 42 43 44 72 73 74 94]
{'a': 27, 'b': 27, 'c': 27, 'd': 9}
test set: [15 16 17 45 46 47 75 76 77 95]
{'a': 27, 'b': 27, 'c': 27, 'd': 9}
test set: [18 19 20 48 49 50 78 79 80 96]
{'a': 27, 'b': 27, 'c': 27, 'd': 9}
test set: [21 22 23 51 52 53 81 82 83 97]
{'a': 27, 'b': 27, 'c': 27, 'd': 9}
test set: [24 25 26 54 55 56 84 85 86 98]
{'a': 27, 'b': 27, 'c': 27, 'd': 9}
test set: [27 28 29 57 58 59 87 88 89 99]
{'a': 27, 'b': 27, 'c': 27, 'd': 9}
我们设置的是十折，也就是说将a分为十份，每次取一份来做test集
结果很明显，假设现在有100个样本，类别标签为a,b,c的各30个，和10个标签为d的样本，他们的占比就是3:3:3:1
当直接使用KFold的时候，第0~9个样本作为第一折，第二次取10-19做第二折，以此类推，不考虑每一折内的样本分布情况
但当使用Stra~的时候，分折的时候是要考虑样本分布情况的，以保证每一折内的各类样本比例与全部数据中的一致
SVM的decision_func和predict_proba
默认情况下，sklearn包中的svm方法是没有predict_proba方法的，也就是说默认情况下，没法获得对某个样本预测的后验概率，svm有自己的decision_func方法，如下
import numpy as np
from sklearn import svm
X = np.array([[-1, -1], [-2, -1], [1, 1], [2, 1]])
y = np.array([1, 1, 2, 2])
svm1 = svm.SVC(kernel='linear', probability=False, random_state=random_state)
svm1.fit(X, y)
print(svm1.decision_function(X))
#output
#[-1. -1.5 1. 1.5]
#如果是multi-class问题
X = np.array([[-1, -1], [-2, -1], [1, 1], [2, 1], [0, 0], [0, 0.1]])
y = np.array([1, 1, 2, 2, 3, 3])
svm1 = svm.SVC(kernel='linear', probability=True, random_state=random_state)
svm1.fit(X, y)
print(svm1.decision_function(X))
#输出
#[[ 2.18181818 -0.35863636 1.17681818]
#[ 2.31818182 -0.495 1.17681818]
#[-0.36363636 2.16863636 1.195 ]
#[-0.5 2.305 1.195 ]
#[ 0.90909091 -0.095 2.18590909]
#[-0.10454545 0.91772727 2.18681818]]
可以看到，如果是两类问题，那么decision_function输出的就只是一维的，而如果是多类问题，那么输出的维度是与decision_function_shape这个参数有关，详情见SVM-api
而且，如果是两类问题，如上例，如果某个样本的decision_function输出如果为负，那么则说明，这个样本应该分到类标号较小的那个类中去，也就是1
而如果把SVC的参数probability=True，那么svm也会带有predict_proba方法，如下
from sklearn import svm
X = np.array([[-1, -1], [-2, -1], [1, 1], [2, 1]])
y = np.array([1, 1, 2, 2])
svm = svm.SVC(kernel='linear', probability=True, random_state=random_state)
svm.fit(X, y)
print(svm.predict_proba(X))
#输出
# [[0.71303947 0.28696053]
# [0.79661758 0.20338242]
# [0.28696001 0.71303999]
# [0.20335122 0.79664878]]
predict_proba这个方法的输出是基于decision_func的输出、运用Platt scaling计算出来的，详情见SVM-api中的Attributes中的probaA_和probaB_
PCA
explained_variance_即为特征值
import numpy as np
from sklearn.decomposition import PCA
X = np.array([(2.5,2.4), (0.5,0.7), (2.2,2.9), (1.9,2.2), (3.1,3.0), (2.3, 2.7), (2, 1.6), (1, 1.1), (1.5, 1.6), (1.1, 0.9)])
pca = PCA(n_components=1)
pca.fit(X)
print(pca.components_)
print(pca.explained_variance_)
#[[-0.6778734 -0.73517866]]
#[1.28402771]