专栏 | 基于 Jupyter 的特征工程手册：数据预处理（一）

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作者：Yingxiang Chen & Zihan Yang

编辑：红色石头

特征工程在机器学习中的重要性不言而喻，恰当的特征工程能显著提升机器学习模型性能。我们在 Github 上整理编写了一份系统的特征工程教程，供大家参考学习。

项目地址：

https://github.com/YC-Coder-Chen/feature-engineering-handbook

本文将探讨数据预处理部分：介绍了如何利用 scikit-learn 处理静态的连续变量，利用 Category Encoders 处理静态的类别变量以及利用 Featuretools 处理常见的时间序列变量。

目录

特征工程的数据预处理我们将分为三大部分来介绍：

• 静态连续变量

• 静态类别变量

• 时间序列变量

本文将介绍 1.1 静态连续变量的数据预处理。下面将结合 Jupyter，使用 sklearn，进行详解。

1.1 静态连续变量

1.1.1 离散化

离散化连续变量可以使模型更加稳健。例如，当预测客户的购买行为时，一个已有 30 次购买行为的客户可能与一个已有 32 次购买行为的客户具有非常相似的行为。有时特征中的过精度可能是噪声，这就是为什么在 LightGBM 中，模型采用直方图算法来防止过拟合。离散连续变量有两种方法。

1.1.1.1 二值化

将数值特征二值化。

# load the sample data from sklearn.datasets import fetch_california_housing dataset = fetch_california_housing() X, y = dataset.data, dataset.target # we will take the first column as the example later %matplotlib inline import seaborn as sns import matplotlib.pyplot as pltfig, ax = plt.subplots() sns.distplot(X[:,0], hist = True, kde=True) ax.set_title('Histogram', fontsize=12) ax.set_xlabel('Value', fontsize=12) ax.set_ylabel('Frequency', fontsize=12); # this feature has long-tail distribution

from sklearn.preprocessing import Binarizersample_columns = X[0:10,0] # select the top 10 samples # return array([8.3252, 8.3014, 7.2574, 5.6431, 3.8462, 4.0368, 3.6591, 3.12, 2.0804, 3.6912])model = Binarizer(threshold=6) # set 6 to be the threshold # if value <= 6, then return 0 else return 1 result = model.fit_transform(sample_columns.reshape(-1,1)).reshape(-1) # return array([1., 1., 1., 0., 0., 0., 0., 0., 0., 0.])

1.1.1.2 分箱

将数值特征分箱。

均匀分箱：

from sklearn.preprocessing import KBinsDiscretizer# in order to mimic the operation in real-world, we shall fit the KBinsDiscretizer # on the trainset and transform the testset # we take the top ten samples in the first column as test set # take the rest samples in the first column as train settest_set = X[0:10,0] # return array([8.3252, 8.3014, 7.2574, 5.6431, 3.8462, 4.0368, 3.6591, 3.12, 2.0804, 3.6912]) train_set = X[10:,0]model = KBinsDiscretizer(n_bins=5, encode='ordinal', strategy='uniform') # set 5 bins # return oridinal bin number, set all bins to have identical widthsmodel.fit(train_set.reshape(-1,1)) result = model.transform(test_set.reshape(-1,1)).reshape(-1) # return array([2., 2., 2., 1., 1., 1., 1., 0., 0., 1.]) bin_edge = model.bin_edges_[0] # return array([ 0.4999 , 3.39994, 6.29998, 9.20002, 12.10006, 15.0001 ]), the bin edges # visualiza the bin edges fig, ax = plt.subplots() sns.distplot(train_set, hist = True, kde=True)for edge in bin_edge: # uniform binsline = plt.axvline(edge, color='b') ax.legend([line], ['Uniform Bin Edges'], fontsize=10) ax.set_title('Histogram', fontsize=12) ax.set_xlabel('Value', fontsize=12) ax.set_ylabel('Frequency', fontsize=12);

分位数分箱：

from sklearn.preprocessing import KBinsDiscretizer# in order to mimic the operation in real-world, we shall fit the KBinsDiscretizer # on the trainset and transform the testset # we take the top ten samples in the first column as test set # take the rest samples in the first column as train settest_set = X[0:10,0] # return array([8.3252, 8.3014, 7.2574, 5.6431, 3.8462, 4.0368, 3.6591, 3.12, 2.0804, 3.6912]) train_set = X[10:,0]model = KBinsDiscretizer(n_bins=5, encode='ordinal', strategy='quantile') # set 3 bins # return oridinal bin number, set all bins based on quantilemodel.fit(train_set.reshape(-1,1)) result = model.transform(test_set.reshape(-1,1)).reshape(-1) # return array([4., 4., 4., 4., 2., 3., 2., 1., 0., 2.]) bin_edge = model.bin_edges_[0] # return array([ 0.4999 , 2.3523 , 3.1406 , 3.9667 , 5.10824, 15.0001 ]), the bin edges # 2.3523 is the 20% quantile # 3.1406 is the 40% quantile, etc.. # visualiza the bin edges fig, ax = plt.subplots() sns.distplot(train_set, hist = True, kde=True)for edge in bin_edge: # quantile based binsline = plt.axvline(edge, color='b') ax.legend([line], ['Quantiles Bin Edges'], fontsize=10) ax.set_title('Histogram', fontsize=12) ax.set_xlabel('Value', fontsize=12) ax.set_ylabel('Frequency', fontsize=12);

1.1.2 缩放

不同尺度的特征之间难以比较，特别是在线性回归和逻辑回归等线性模型中。在基于欧氏距离的 k-means 聚类或 KNN 模型中，就需要进行特征缩放，否则距离的测量是无用的。而对于任何使用梯度下降的算法，缩放也会加快收敛速度。

一些常用的模型：

注：偏度影响 PCA 模型，因此最好使用幂变换来消除偏度。

1.1.2.1 标准缩放（Z 分数标准化）

公式：

其中，X 是变量（特征），???? 是 X 的均值，???? 是 X 的标准差。此方法对异常值非常敏感，因为异常值同时影响到 ???? 和 ????。

from sklearn.preprocessing import StandardScaler# in order to mimic the operation in real-world, we shall fit the StandardScaler # on the trainset and transform the testset # we take the top ten samples in the first column as test set # take the rest samples in the first column as train settest_set = X[0:10,0] # return array([8.3252, 8.3014, 7.2574, 5.6431, 3.8462, 4.0368, 3.6591, 3.12, 2.0804, 3.6912]) train_set = X[10:,0]model = StandardScaler()model.fit(train_set.reshape(-1,1)) # fit on the train set and transform the test set # top ten numbers for simplification result = model.transform(test_set.reshape(-1,1)).reshape(-1) # return array([ 2.34539745, 2.33286782, 1.78324852, 0.93339178, -0.0125957 , # 0.08774668, -0.11109548, -0.39490751, -0.94221041, -0.09419626]) # result is the same as ((X[0:10,0] - X[10:,0].mean())/X[10:,0].std()) # visualize the distribution after the scaling # fit and transform the entire first featureimport seaborn as sns import matplotlib.pyplot as pltfig, ax = plt.subplots(2,1, figsize = (13,9)) sns.distplot(X[:,0], hist = True, kde=True, ax=ax[0]) ax[0].set_title('Histogram of the Original Distribution', fontsize=12) ax[0].set_xlabel('Value', fontsize=12) ax[0].set_ylabel('Frequency', fontsize=12); # this feature has long-tail distributionmodel = StandardScaler() model.fit(X[:,0].reshape(-1,1)) result = model.transform(X[:,0].reshape(-1,1)).reshape(-1)# show the distribution of the entire feature sns.distplot(result, hist = True, kde=True, ax=ax[1]) ax[1].set_title('Histogram of the Transformed Distribution', fontsize=12) ax[1].set_xlabel('Value', fontsize=12) ax[1].set_ylabel('Frequency', fontsize=12); # the distribution is the same, but scales change fig.tight_layout()

1.1.2.2 MinMaxScaler（按数值范围缩放）

假设我们要缩放的特征数值范围为 (a, b)。

公式：

其中，Min 是 X 的最小值，Max 是 X 的最大值。此方法也对异常值非常敏感，因为异常值同时影响到 Min 和 Max。

from sklearn.preprocessing import MinMaxScaler# in order to mimic the operation in real-world, we shall fit the MinMaxScaler # on the trainset and transform the testset # we take the top ten samples in the first column as test set # take the rest samples in the first column as train settest_set = X[0:10,0] # return array([8.3252, 8.3014, 7.2574, 5.6431, 3.8462, 4.0368, 3.6591, 3.12, 2.0804, 3.6912]) train_set = X[10:,0]model = MinMaxScaler(feature_range=(0,1)) # set the range to be (0,1)model.fit(train_set.reshape(-1,1)) # fit on the train set and transform the test set # top ten numbers for simplification result = model.transform(test_set.reshape(-1,1)).reshape(-1) # return array([0.53966842, 0.53802706, 0.46602805, 0.35469856, 0.23077613, # 0.24392077, 0.21787286, 0.18069406, 0.1089985 , 0.22008662]) # result is the same as (X[0:10,0] - X[10:,0].min())/(X[10:,0].max()-X[10:,0].min()) # visualize the distribution after the scaling # fit and transform the entire first featureimport seaborn as sns import matplotlib.pyplot as pltfig, ax = plt.subplots(2,1, figsize = (13,9)) sns.distplot(X[:,0], hist = True, kde=True, ax=ax[0]) ax[0].set_title('Histogram of the Original Distribution', fontsize=12) ax[0].set_xlabel('Value', fontsize=12) ax[0].set_ylabel('Frequency', fontsize=12); # this feature has long-tail distributionmodel = MinMaxScaler(feature_range=(0,1)) model.fit(X[:,0].reshape(-1,1)) result = model.transform(X[:,0].reshape(-1,1)).reshape(-1)# show the distribution of the entire feature sns.distplot(result, hist = True, kde=True, ax=ax[1]) ax[1].set_title('Histogram of the Transformed Distribution', fontsize=12) ax[1].set_xlabel('Value', fontsize=12) ax[1].set_ylabel('Frequency', fontsize=12); # the distribution is the same, but scales change fig.tight_layout() # now the scale change to [0,1]

1.1.2.3 RobustScaler（抗异常值缩放）

使用对异常值稳健的统计（分位数）来缩放特征。假设我们要将缩放的特征分位数范围为 (a, b)。

公式：

这种方法对异常点鲁棒性更强。

import numpy as np from sklearn.preprocessing import RobustScaler# in order to mimic the operation in real-world, we shall fit the RobustScaler # on the trainset and transform the testset # we take the top ten samples in the first column as test set # take the rest samples in the first column as train settest_set = X[0:10,0] # return array([8.3252, 8.3014, 7.2574, 5.6431, 3.8462, 4.0368, 3.6591, 3.12, 2.0804, 3.6912]) train_set = X[10:,0]model = RobustScaler(with_centering = True, with_scaling = True, quantile_range = (25.0, 75.0)) # with_centering = True => recenter the feature by set X' = X - X.median() # with_scaling = True => rescale the feature by the quantile set by user # set the quantile to the (25%, 75%)model.fit(train_set.reshape(-1,1)) # fit on the train set and transform the test set # top ten numbers for simplification result = model.transform(test_set.reshape(-1,1)).reshape(-1) # return array([ 2.19755974, 2.18664281, 1.7077657 , 0.96729508, 0.14306683, # 0.23049401, 0.05724508, -0.19003715, -0.66689601, 0.07196918]) # result is the same as (X[0:10,0] - np.quantile(X[10:,0], 0.5))/(np.quantile(X[10:,0],0.75)-np.quantile(X[10:,0], 0.25)) # visualize the distribution after the scaling # fit and transform the entire first featureimport seaborn as sns import matplotlib.pyplot as pltfig, ax = plt.subplots(2,1, figsize = (13,9)) sns.distplot(X[:,0], hist = True, kde=True, ax=ax[0]) ax[0].set_title('Histogram of the Original Distribution', fontsize=12) ax[0].set_xlabel('Value', fontsize=12) ax[0].set_ylabel('Frequency', fontsize=12); # this feature has long-tail distributionmodel = RobustScaler(with_centering = True, with_scaling = True, quantile_range = (25.0, 75.0)) model.fit(X[:,0].reshape(-1,1)) result = model.transform(X[:,0].reshape(-1,1)).reshape(-1)# show the distribution of the entire feature sns.distplot(result, hist = True, kde=True, ax=ax[1]) ax[1].set_title('Histogram of the Transformed Distribution', fontsize=12) ax[1].set_xlabel('Value', fontsize=12) ax[1].set_ylabel('Frequency', fontsize=12); # the distribution is the same, but scales change fig.tight_layout()

1.1.2.4 幂次变换（非线性变换）

以上介绍的所有缩放方法都保持原来的分布。但正态性是许多统计模型的一个重要假设。我们可以使用幂次变换将原始分布转换为正态分布。

Box-Cox 变换：

Box-Cox 变换只适用于正数，并假设如下分布：

考虑了所有的 λ 值，通过最大似然估计选择稳定方差和最小化偏度的最优值。

from sklearn.preprocessing import PowerTransformer# in order to mimic the operation in real-world, we shall fit the PowerTransformer # on the trainset and transform the testset # we take the top ten samples in the first column as test set # take the rest samples in the first column as train settest_set = X[0:10,0] # return array([8.3252, 8.3014, 7.2574, 5.6431, 3.8462, 4.0368, 3.6591, 3.12, 2.0804, 3.6912]) train_set = X[10:,0]model = PowerTransformer(method='box-cox', standardize=True) # apply box-cox transformationmodel.fit(train_set.reshape(-1,1)) # fit on the train set and transform the test set # top ten numbers for simplification result = model.transform(test_set.reshape(-1,1)).reshape(-1) # return array([ 1.91669292, 1.91009687, 1.60235867, 1.0363095 , 0.19831579, # 0.30244247, 0.09143411, -0.24694006, -1.08558469, 0.11011933]) # visualize the distribution after the scaling # fit and transform the entire first featureimport seaborn as sns import matplotlib.pyplot as pltfig, ax = plt.subplots(2,1, figsize = (13,9)) sns.distplot(X[:,0], hist = True, kde=True, ax=ax[0]) ax[0].set_title('Histogram of the Original Distribution', fontsize=12) ax[0].set_xlabel('Value', fontsize=12) ax[0].set_ylabel('Frequency', fontsize=12); # this feature has long-tail distributionmodel = PowerTransformer(method='box-cox', standardize=True) model.fit(X[:,0].reshape(-1,1)) result = model.transform(X[:,0].reshape(-1,1)).reshape(-1)# show the distribution of the entire feature sns.distplot(result, hist = True, kde=True, ax=ax[1]) ax[1].set_title('Histogram of the Transformed Distribution', fontsize=12) ax[1].set_xlabel('Value', fontsize=12) ax[1].set_ylabel('Frequency', fontsize=12); # the distribution now becomes normal fig.tight_layout()

Yeo-Johnson 变换：

Yeo Johnson 变换适用于正数和负数，并假设以下分布：

考虑了所有的 λ 值，通过最大似然估计选择稳定方差和最小化偏度的最优值。

from sklearn.preprocessing import PowerTransformer# in order to mimic the operation in real-world, we shall fit the PowerTransformer # on the trainset and transform the testset # we take the top ten samples in the first column as test set # take the rest samples in the first column as train settest_set = X[0:10,0] # return array([8.3252, 8.3014, 7.2574, 5.6431, 3.8462, 4.0368, 3.6591, 3.12, 2.0804, 3.6912]) train_set = X[10:,0]model = PowerTransformer(method='yeo-johnson', standardize=True) # apply box-cox transformationmodel.fit(train_set.reshape(-1,1)) # fit on the train set and transform the test set # top ten numbers for simplification result = model.transform(test_set.reshape(-1,1)).reshape(-1) # return array([ 1.90367888, 1.89747091, 1.604735 , 1.05166306, 0.20617221, # 0.31245176, 0.09685566, -0.25011726, -1.10512438, 0.11598074]) # visualize the distribution after the scaling # fit and transform the entire first featureimport seaborn as sns import matplotlib.pyplot as pltfig, ax = plt.subplots(2,1, figsize = (13,9)) sns.distplot(X[:,0], hist = True, kde=True, ax=ax[0]) ax[0].set_title('Histogram of the Original Distribution', fontsize=12) ax[0].set_xlabel('Value', fontsize=12) ax[0].set_ylabel('Frequency', fontsize=12); # this feature has long-tail distributionmodel = PowerTransformer(method='yeo-johnson', standardize=True) model.fit(X[:,0].reshape(-1,1)) result = model.transform(X[:,0].reshape(-1,1)).reshape(-1)# show the distribution of the entire feature sns.distplot(result, hist = True, kde=True, ax=ax[1]) ax[1].set_title('Histogram of the Transformed Distribution', fontsize=12) ax[1].set_xlabel('Value', fontsize=12) ax[1].set_ylabel('Frequency', fontsize=12); # the distribution now becomes normal fig.tight_layout()

1.1.3 正则化

以上所有缩放方法都是按列操作的。但正则化在每一行都有效，它试图“缩放”每个样本，使其具有单位范数。由于正则化在每一行都起作用，它会扭曲特征之间的关系，因此不常见。但是正则化方法在文本分类和聚类上下文中是非常有用的。

假设 X[i][j] 表示样本 i 中特征 j 的值。

L1 正则化公式：

L2 正则化公式：

L1 正则化：

from sklearn.preprocessing import Normalizer# Normalizer performs operation on each row independently # So train set and test set are processed independently###### for L1 Norm sample_columns = X[0:2,0:3] # select the first two samples, and the first three features # return array([[ 8.3252, 41., 6.98412698], # [ 8.3014 , 21., 6.23813708]])model = Normalizer(norm='l1') # use L2 Norm to normalize each samplesmodel.fit(sample_columns) result = model.transform(sample_columns) # test set are processed similarly # return array([[0.14784762, 0.72812094, 0.12403144], # [0.23358211, 0.59089121, 0.17552668]]) # result = sample_columns/np.sum(np.abs(sample_columns), axis=1).reshape(-1,1)

L2 正则化：

###### for L2 Norm sample_columns = X[0:2,0:3] # select the first three features # return array([[ 8.3252, 41., 6.98412698], # [ 8.3014 , 21., 6.23813708]])model = Normalizer(norm='l2') # use L2 Norm to normalize each samplesmodel.fit(sample_columns) result = model.transform(sample_columns) # return array([[0.19627663, 0.96662445, 0.16465922], # [0.35435076, 0.89639892, 0.26627902]]) # result = sample_columns/np.sqrt(np.sum(sample_columns**2, axis=1)).reshape(-1,1) # visualize the difference in the distribuiton after Normalization # compare it with the distribuiton after RobustScaling # fit and transform the entire first & second featureimport seaborn as sns import matplotlib.pyplot as plt# RobustScaler fig, ax = plt.subplots(2,1, figsize = (13,9))model = RobustScaler(with_centering = True, with_scaling = True, quantile_range = (25.0, 75.0)) model.fit(X[:,0:2]) result = model.transform(X[:,0:2])sns.scatterplot(result[:,0], result[:,1], ax=ax[0]) ax[0].set_title('Scatter Plot of RobustScaling result', fontsize=12) ax[0].set_xlabel('Feature 1', fontsize=12) ax[0].set_ylabel('Feature 2', fontsize=12);model = Normalizer(norm='l2')model.fit(X[:,0:2]) result = model.transform(X[:,0:2])sns.scatterplot(result[:,0], result[:,1], ax=ax[1]) ax[1].set_title('Scatter Plot of Normalization result', fontsize=12) ax[1].set_xlabel('Feature 1', fontsize=12) ax[1].set_ylabel('Feature 2', fontsize=12); fig.tight_layout() # Normalization distort the original distribution

1.1.4 缺失值的估算

在实际操作中，数据集中可能缺少值。然而，这种稀疏的数据集与大多数 scikit 学习模型不兼容，这些模型假设所有特征都是数值的，而没有丢失值。所以在应用 scikit 学习模型之前，我们需要估算缺失的值。

但是一些新的模型，比如在其他包中实现的 XGboost、LightGBM 和 Catboost，为数据集中丢失的值提供了支持。所以在应用这些模型时，我们不再需要填充数据集中丢失的值。

1.1.4.1 单变量特征插补

假设第 i 列中有缺失值，那么我们将用常数或第 i 列的统计数据（平均值、中值或模式）对其进行估算。

from sklearn.impute import SimpleImputertest_set = X[0:10,0].copy() # no missing values # return array([8.3252, 8.3014, 7.2574, 5.6431, 3.8462, 4.0368, 3.6591, 3.12, 2.0804, 3.6912])# manully create some missing values test_set[3] = np.nan test_set[6] = np.nan # now sample_columns becomes # array([8.3252, 8.3014, 7.2574, nan, 3.8462, 4.0368, nan, 3.12 ,2.0804, 3.6912])# create the test samples # in real-world, we should fit the imputer on train set and tranform the test set. train_set = X[10:,0].copy() train_set[3] = np.nan train_set[6] = np.nanimputer = SimpleImputer(missing_values=np.nan, strategy='mean') # use mean # we can set the strategy to 'mean', 'median', 'most_frequent', 'constant' imputer.fit(train_set.reshape(-1,1)) result = imputer.transform(test_set.reshape(-1,1)).reshape(-1) # return array([8.3252 , 8.3014 , 7.2574 , 3.87023658, 3.8462 , # 4.0368 , 3.87023658, 3.12 , 2.0804 , 3.6912 ]) # all missing values are imputed with 3.87023658 # 3.87023658 = np.nanmean(train_set) # which is the mean of the trainset ignoring missing values

1.1.4.2 多元特征插补

多元特征插补利用整个数据集的信息来估计和插补缺失值。在 scikit-learn 中，它以循环迭代的方式实现。

在每一步中，一个特征列被指定为输出 y，其他特征列被视为输入 X。一个回归器适用于已知 y 的（X，y）。然后，回归器被用来预测 y 的缺失值。这是以迭代的方式对每个特征进行的，然后对最大值插补回合重复进行。

使用线性模型（以 BayesianRidge 为例）：

from sklearn.experimental import enable_iterative_imputer # have to import this to enable # IterativeImputer from sklearn.impute import IterativeImputer from sklearn.linear_model import BayesianRidgetest_set = X[0:10,:].copy() # no missing values, select all features # the first columns is # array([8.3252, 8.3014, 7.2574, 5.6431, 3.8462, 4.0368, 3.6591, 3.12, 2.0804, 3.6912])# manully create some missing values test_set[3,0] = np.nan test_set[6,0] = np.nan test_set[3,1] = np.nan # now the first feature becomes # array([8.3252, 8.3014, 7.2574, nan, 3.8462, 4.0368, nan, 3.12 ,2.0804, 3.6912])# create the test samples # in real-world, we should fit the imputer on train set and tranform the test set. train_set = X[10:,:].copy() train_set[3,0] = np.nan train_set[6,0] = np.nan train_set[3,1] = np.nanimpute_estimator = BayesianRidge() imputer = IterativeImputer(max_iter = 10, random_state = 0, estimator = impute_estimator)imputer.fit(train_set) result = imputer.transform(test_set)[:,0] # only select the first column to revel how it works # return array([8.3252 , 8.3014 , 7.2574 , 4.6237195 , 3.8462 , # 4.0368 , 4.00258149, 3.12 , 2.0804 , 3.6912 ])

使用基于树的模型（以 ExtraTrees 为例）：

from sklearn.experimental import enable_iterative_imputer # have to import this to enable # IterativeImputer from sklearn.impute import IterativeImputer from sklearn.ensemble import ExtraTreesRegressortest_set = X[0:10,:].copy() # no missing values, select all features # the first columns is # array([8.3252, 8.3014, 7.2574, 5.6431, 3.8462, 4.0368, 3.6591, 3.12, 2.0804, 3.6912])# manully create some missing values test_set[3,0] = np.nan test_set[6,0] = np.nan test_set[3,1] = np.nan # now the first feature becomes # array([8.3252, 8.3014, 7.2574, nan, 3.8462, 4.0368, nan, 3.12 ,2.0804, 3.6912])# create the test samples # in real-world, we should fit the imputer on train set and tranform the test set. train_set = X[10:,:].copy() train_set[3,0] = np.nan train_set[6,0] = np.nan train_set[3,1] = np.nanimpute_estimator = ExtraTreesRegressor(n_estimators=10, random_state=0) # parameters can be turned in CV though sklearn pipeline imputer = IterativeImputer(max_iter = 10, random_state = 0, estimator = impute_estimator)imputer.fit(train_set) result = imputer.transform(test_set)[:,0] # only select the first column to revel how it works # return array([8.3252 , 8.3014 , 7.2574 , 4.63813, 3.8462 , 4.0368 , 3.24721, # 3.12 , 2.0804 , 3.6912 ])

使用 K 近邻（KNN）：

from sklearn.experimental import enable_iterative_imputer # have to import this to enable # IterativeImputer from sklearn.impute import IterativeImputer from sklearn.neighbors import KNeighborsRegressortest_set = X[0:10,:].copy() # no missing values, select all features # the first columns is # array([8.3252, 8.3014, 7.2574, 5.6431, 3.8462, 4.0368, 3.6591, 3.12, 2.0804, 3.6912])# manully create some missing values test_set[3,0] = np.nan test_set[6,0] = np.nan test_set[3,1] = np.nan # now the first feature becomes # array([8.3252, 8.3014, 7.2574, nan, 3.8462, 4.0368, nan, 3.12 ,2.0804, 3.6912])# create the test samples # in real-world, we should fit the imputer on train set and tranform the test set. train_set = X[10:,:].copy() train_set[3,0] = np.nan train_set[6,0] = np.nan train_set[3,1] = np.nanimpute_estimator = KNeighborsRegressor(n_neighbors=10, p = 1) # set p=1 to use manhanttan distance # use manhanttan distance to reduce effect from outliers# parameters can be turned in CV though sklearn pipeline imputer = IterativeImputer(max_iter = 10, random_state = 0, estimator = impute_estimator)imputer.fit(train_set) result = imputer.transform(test_set)[:,0] # only select the first column to revel how it works # return array([8.3252, 8.3014, 7.2574, 3.6978, 3.8462, 4.0368, 4.052 , 3.12 , # 2.0804, 3.6912])

1.1.4.3 标记估算值

有时，某些缺失值可能是有用的。因此，scikit learn 还提供了将缺少值的数据集转换为相应的二进制矩阵的功能，该矩阵指示数据集中缺少值的存在。

from sklearn.impute import MissingIndicator# illustrate this function on trainset only # since the precess is independent in train set and test set train_set = X[10:,:].copy() # select all features train_set[3,0] = np.nan # manully create some missing values train_set[6,0] = np.nan train_set[3,1] = np.nanindicator = MissingIndicator(missing_values=np.nan, features='all') # show the results on all the features result = indicator.fit_transform(train_set) # result have the same shape with train_set # contains only True & False, True corresponds with missing valueresult[:,0].sum() # should return 2, the first column has two missing values result[:,1].sum(); # should return 1, the second column has one missing value

1.1.5 特征变换

1.1.5.1 多项式变换

有时我们希望在模型中引入非线性特征，从而增加模型的复杂度。对于简单的线性模型，这将大大增加模型的复杂度。但是对于更复杂的模型，如基于树的 ML 模型，它们已经在非参数树结构中包含了非线性关系。因此，这种特性转换可能对基于树的 ML 模型没有太大帮助。

例如，如果我们将阶数设置为 3，形式如下：

from sklearn.preprocessing import PolynomialFeatures# illustrate this function on one synthesized sample train_set = np.array([2,3]).reshape(1,-1) # shape (1,2) # return array([[2, 3]])poly = PolynomialFeatures(degree = 3, interaction_only = False) # the highest degree is set to 3, and we want more than just intereaction termsresult = poly.fit_transform(train_set) # have shape (1, 10) # array([[ 1., 2., 3., 4., 6., 9., 8., 12., 18., 27.]])

1.1.5.2 自定义变换

from sklearn.preprocessing import FunctionTransformer# illustrate this function on one synthesized sample train_set = np.array([2,3]).reshape(1,-1) # shape (1,2) # return array([[2, 3]])transformer = FunctionTransformer(func = np.log1p, validate=True) # perform log transformation, X' = log(1 + x) # func can be any numpy function such as np.exp result = transformer.transform(train_set) # return array([[1.09861229, 1.38629436]]), the same as np.log1p(train_set)

好了，以上就是关于静态连续变量的数据预处理介绍。建议读者结合代码，在 Jupyter 中实操一遍。

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